山东省第三届ACM省赛
| ID | Title | Hint |
|---|---|---|
| A | Impasse (+) | |
| B | Chess | |
| C | An interesting game | 最小费用最大流 |
| D | n a^o7 ! | |
| E | Fruit Ninja I | dp |
| F | Pixel density | 无 |
| G | Mine Number | dfs |
| H | The Best Seat in ACM Contest | 无 |
| I | Pick apples | 贪心，背包 |
| J | Fruit Ninja II | 积分 |
B.
d.国际象棋？就是求能被0~16个棋子吃掉的棋子分别有几个。
s.当时有点思路，感觉对。但是时间好像不大够，也没敲。
用二分查找，查询周围16个位置有棋子没，能不能把它吃掉。
棋子个数n <= 10^5。
10^5*16*log(10^5)<=4*10^7
C.
d.
s.这个题感觉也可以，不知道为啥wa了。
感觉直接暴力贪心就行啊，题意理解错了吗？
c.上个***丝测试代码。。。数据比较大的时候可以找到不同的，但是还是不懂为什么贪心不行。。
#include <iostream> #include <cstdio> #include <algorithm> #include<string.h> #include <vector> #include <queue> #include<time.h> using namespace std; const int maxn=2200; const int oo=0x3f3f3f3f; struct Edge { int u, v, cap, flow, cost; Edge(int u, int v, int cap, int flow, int cost):u(u), v(v), cap(cap), flow(flow), cost(cost) {} }; struct MCMF { int n, m, s, t; vector<Edge> edge; vector<int> G[maxn]; int inq[maxn], d[maxn], p[maxn], a[maxn]; void init(int n) { this->n=n; for(int i=0; i<n; i++) G[i].clear(); edge.clear(); } void AddEdge(int u, int v, int cap, int cost) { edge.push_back(Edge(u, v, cap, 0, cost)); edge.push_back(Edge(v, u, 0, 0, -cost)); m=edge.size(); G[u].push_back(m-2); G[v].push_back(m-1); } bool spfa(int s, int t, int& flow, int& cost) { memset(d, 0x3f, sizeof d); memset(inq, 0, sizeof inq); d[s]=0, inq[s]=1, p[s]=0, a[s]=oo; queue<int> q; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); inq[u]=0; for(int i=0; i<G[u].size(); i++) { Edge& e=edge[G[u][i]]; if(e.cap>e.flow && d[e.v]>d[u]+e.cost) { d[e.v]=d[u]+e.cost; p[e.v]=G[u][i]; a[e.v]=min(a[u], e.cap-e.flow); if(!inq[e.v]) { q.push(e.v); inq[e.v]=1; } } } } if(d[t]==oo)return false; flow+=a[t]; cost+=d[t]*a[t]; int u=t; while(u!=s) { edge[p[u]].flow+=a[t]; edge[p[u]^1].flow-=a[t]; u=edge[p[u]].u; } return true; } int MinCost(int s, int t) { int flow=0, cost=0; while(spfa(s, t, flow, cost)); return cost; } } net; int a[maxn], b[maxn]; struct node{ int h; int id; int h2; }a2[1005],a3[1005]; bool cmp(node a,node b){ if(a.h!=b.h)return a.h<b.h; return a.h2<b.h2; } bool cmp2(node a,node b){ if(a.h!=b.h)return a.h<b.h; return a.h2<b.h2; } int test1[]={14,20,20,3,10,28,1,19,3,26,29,17,27,11,5,24,3,20,2,8,8,0,11,0,19,29, 10,23,17,3,14,28,14,8,6,6,8,5,15,1,23,28,17,9,15,20,13,6,3,11,8,6,23,3,1,21,13,20,17,0,16, 14,4,1,18,27,25,21,29,13,19,11,10,22,2,27,0,19,1,8,7,18,7,5,11,22,23,24,22,3,27,3, 17,21,26,8,17,17,17,19,12,28,14,17,28,19,25,12,10,1,30,6,1,1,29,16,0,6,22,27,25, 25,3,0,0,29,8,29,19,24,9,27,15,1,11,28,28,7,0,24,8,17,16,13,14,29,2,12,23,22,12, 25,13,19,21,11,9,10,9,16,16,13,20,22,10,30,29,27,12,23,16,11,22,24,9,15,4,26,4,15, 19,16,16,19,27,27,30,11,27,27,4,22,19,24,29,10,28,27,2,30,29,19,3,8,21,4,2,4,2, 23,13,6,20,21,5,10,2,8,3,2,3,17,10,7,25,16,13,12,10,23,30,27,22,24,6,10,0,0,30, 7,0,1,1,18,29,5,19,10,8,30,30,0,15,3,17,2,12,22,1,8,20,14,2,15,10,17,4,26,23,19, 15,25,30,17,23,13,18,24,15,1,15,14,5,13,18,10,28,11,11,30,15,0,2,28,4,27,29,21,30, 1,29,8,6,1,21,13,8,29,24,13,22,1,3,20,8,7,25,26,21,11,3,3,10,9,24,27,5,5,7,9,16, 17,18,0,5,9,12,10,29,9,20,24,11,4,3,30,12,30,26,6,17,14,20,15,26,17,27,28,15,21, 15,27,12,24,25,24,17,3,25,9,0,3,15,12,28,6,0,3,17,27,7,26,10,9,13,24,4,21,21,19, 20,29,15,15,0,24,22,17,30,2,6,21,12,21,16,24,4,10,25,6,18,26,19,5,7,17,18,7,1, 12,13,15,17,21,14,13,17,9,14,4,6,1,13,12,28,1,0,11,29,22,3,25,3,11,21,12,22,5,18,7, 5,0,12,26,11,17,8,14,0,18,6,7,8,30,23,19,0,26,6,15,23,24,30,29,11,20,2,21,14, 4,20,24,1,15,28,30,14,19,8,22,2,23,3,14,20,30,1,1,1,12,17,12,0,2,1,29,18,17,17,5, 17,21,8,8,23,15,11,2,18,30,12,26,9,13,15,11,30,12,23,21,5,18,12,4,20,8,15,19,19, 16,26,24,8,6,23,11,11,9,4,7,22,27,1,24,1,29,19,12,4,18,25,1,5,4,3,25,25,6,0,3,6, 23,23,15,4,6,21,10,20,25,19,11,4,26,4,28,8,8,22,26,12,6,20,28,24,7,4,11,25,28,23, 0,25,1,3,0,2,10,11,20,4,21,8,15,11,10,6,1,30,2,6,2,6,0,9,19,13,27,30,21,13,18, 17,10,21,27,23,2,5,13,15,10,4,4,18,10,9,5,7,10,24,19,12,2,12,26,6,25,8,18,3,13,19, 16,19,15,28,11,16,8,29,15,13,23,24,2,2,19,0,7,23,14,2,28,22,2,0,26,16,18,11,19, 0,10,19,5,26,8,8,8,13,5,30,20,29,21,30,3,7,29,27,19,28,13,2,15,18,28,13,5,24,13, 15,27,9,29,0,7,16,28,12,1,9,4,8,6,12,11,12,14,22,28,20,14,8,17,25,20,13,16,6,8, 3,7,7,28,17,23,17,13,16,0,19,23,5,17,6,20,21,9,14,30,2,26,7,5,30,24,2,10,23,22,8, 19,4,9,27,29,11,4,13,23,8,18,6,6,1,18,5,17,6,18,13,6,30,6,28,9,22,9,11,3,3,4,25, 29,3,10,19,26,14,11,29,9,23,4,0,23,29,14,2,21,29,18,10,10,3,2,4,16,20,23,2,5,0, 11,10,1,19,14,10,4,15,22,25,8,10,12,25,15,30,5,2,30,10,5,30,7,3,23,23,19,18,3,10, 13,2,30,16,30,2,6,22,30,3,4,28,6,2,22,3,3,5,20,14,16,12,14,28,12 }; int test2[]={ 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3, 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4, 4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6, 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7, 7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8, 8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10, 10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,11,11, 11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13, 13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14, 14,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15, 15,15,15,15,15,15,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,16,16,16,16,16, 16,16,16,16,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,17,17,17,17,17, 17,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,18,18,18,18,18, 18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19, 19,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20, 20,20,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,21,21,21,21, 21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22, 22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23, 23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24, 24,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25 ,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26, 27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28, 28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29, 29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30, 30,30,30,30,30,30,30,30,30,30,30,30,30 }; int temp; int f(int N,int M,int K,int b[]){ int T; //int N,M,K; //int a[1005],b[1005]; int i; int ca=0; int sum; //int a2[1005],a3[1005]; int p1,p2,p3,p4,q1,q2; int t1,t2; bool use[1005]; //scanf("%d",&T); //while(T--){ //scanf("%d%d%d",&N,&M,&K); //for(i=0;i<N;++i){ // scanf("%d",&a[i]); //} sum=0; for(i=0;i<N-1;++i){ sum=sum+abs(a[i]-a[i+1]); temp=sum; if(a[i]>a[i+1]){ a2[i].h=a[i]; a2[i].h2=a[i+1]; a3[i].h=a[i+1]; a3[i].h2=a[i]; a2[i].id=i; a3[i].id=i; } else{ a2[i].h=a[i+1]; a2[i].h2=a[i]; a3[i].h=a[i]; a3[i].h2=a[i+1]; a2[i].id=i; a3[i].id=i; } /* cout<<"i: "<<i<<" a2[i].h "<<a2[i].h<<endl; cout<<"i: "<<i<<" a2[i].h2 "<<a2[i].h2<<endl; */ } /* i=1; cout<<"##"<<endl; cout<<"i: "<<i<<" a2[i].h "<<a2[i].h<<endl; cout<<"i: "<<i<<" a2[i].h2 "<<a2[i].h2<<endl; */ /* i=1; cout<<"##"<<endl; cout<<"i: "<<i<<" a2[i].h "<<a2[i].h<<endl; cout<<"i: "<<i<<" a2[i].h2 "<<a2[i].h2<<endl; */ sort(a2,a2+(N-1),cmp); sort(a3,a3+(N-1),cmp2); /* i=1; cout<<"##"<<endl; cout<<"i: "<<i<<" a2[i].h "<<a2[i].h<<endl; cout<<"i: "<<i<<" a2[i].h2 "<<a2[i].h2<<endl; */ /* i=1; cout<<"##"<<endl; cout<<"i: "<<i<<" a2[i].h "<<a2[i].h<<endl; cout<<"i: "<<i<<" a2[i].h2 "<<a2[i].h2<<endl; */ //for(i=0;i<M;++i){ // scanf("%d",&b[i]); //} sort(b,b+M); p1=0;p2=N-2; p3=0;p4=N-2; q1=0;q2=M-1; memset(use,false,sizeof(use)); for(i=0;i<K;++i){ if(p1<0||p2<0||p3<0||p4<0||q1<0||q2<0){ cout<<"越界"<<endl; } if(p1>N-2||p2>N-2||p3>N-2||p4>N-2||q1>M-1||q2>M-1){ cout<<"越界2"<<endl; } while(use[a2[p1].id]==true){ ++p1; } //cout<<"p1 "<<p1<<endl; t1=b[q2]-a2[p1].h; //cout<<"t1 "<<t1<<endl; if(t1<0){ t1=0; } while(use[a3[p4].id]==true){ --p4; } //cout<<"p4 :"<<p4<<endl; t2=a3[p4].h-b[q1]; //cout<<"t2 "<<t2<<endl; if(t2<0){ t2=0; } if(t1>t2){ sum=sum+2*t1; use[a2[p1].id]=true; // //cout<<a2[p1].id<<"### t1: "<<t1<<endl; ++p1; --q2; } else{ sum=sum+2*t2; use[a3[p4].id]=true; // //cout<<a3[p4].id<<"else t2: "<<t2<<endl; --p4; ++q1; } //cout<<"i:"<<i<<",sum:"<<sum<<endl; } //printf("Case %d: %d\n",++ca,sum); return sum; // } } int main() { int ccc[maxn]; int tot; int T, kase=0; //scanf("%d", &T); T=100; T=100; srand(time(NULL)); while(T--) { int n, m, k, ans=0; n=3; m=2; k=2; n=900; m=800; k=800; //scanf("%d%d%d", &n, &m, &k); net.init(n+100); memset(b, 0, sizeof b); for(int i=0; i<n; i++){ a[i]=rand()%31; //a[i]=test1[i]; //scanf("%d", a+i); } tot=0; for(int i=0; i<m; i++) { int x; /* x=rand()%31; //scanf("%d", &x); b[x]++; ccc[tot++]=x; */ // /* int fuck=rand()%31; b[fuck]++; ccc[tot++]=fuck; // */ } for(int i=1; i<n; i++) { ans+=abs(a[i]-a[i-1]); net.AddEdge(0, i, 1, 0); for(int j=0; j<=30; j++) if(b[j]) { int dis=abs(a[i]-j)+abs(a[i-1]-j)-abs(a[i]-a[i-1]); net.AddEdge(i, n+j, 1, -dis); } } int S=n+50, T=S+1; for(int i=0; i<=30; i++) if(b[i]) net.AddEdge(i+n, T, b[i], 0); net.AddEdge(S, 0, k, 0); int ans2=net.MinCost(S, T); int ans1=ans-ans2; //printf("Case %d: %d\n", ++kase, ); //cout<<"贪心："<<endl; int tanxin=f(n,m,k,ccc); if(ans1!=tanxin){ cout<<"ans:"<<ans<<endl; cout<<"ans2:"<<ans2<<endl; cout<<"temp:"<<temp<<endl; cout<<tanxin-temp<<endl; cout<<"yes"<<endl; cout<<"a:"<<endl; for(int i=0;i<n;++i){ cout<<a[i]<<','; } cout<<endl; cout<<"ccc:"<<endl; for(int i=0;i<m;++i){ cout<<ccc[i]<<','; } cout<<endl; cout<<endl; cout<<ans1<<"网络流"<<endl; cout<<tanxin<<"贪心"<<endl; //return 0; } } return 0; } /* yes a: 14,20,20,3,10,28,1,19,3,26,29,17,27,11,5,24,3,20,2,8,8,0,11,0,19,29,10,23,17,3,1 4,28,14,8,6,6,8,5,15,1,23,28,17,9,15,20,13,6,3,11,8,6,23,3,1,21,13,20,17,0,16,14 ,4,1,18,27,25,21,29,13,19,11,10,22,2,27,0,19,1,8,7,18,7,5,11,22,23,24,22,3,27,3, 17,21,26,8,17,17,17,19,12,28,14,17,28,19,25,12,10,1,30,6,1,1,29,16,0,6,22,27,25, 25,3,0,0,29,8,29,19,24,9,27,15,1,11,28,28,7,0,24,8,17,16,13,14,29,2,12,23,22,12, 25,13,19,21,11,9,10,9,16,16,13,20,22,10,30,29,27,12,23,16,11,22,24,9,15,4,26,4,1 5,19,16,16,19,27,27,30,11,27,27,4,22,19,24,29,10,28,27,2,30,29,19,3,8,21,4,2,4,2 ,23,13,6,20,21,5,10,2,8,3,2,3,17,10,7,25,16,13,12,10,23,30,27,22,24,6,10,0,0,30, 7,0,1,1,18,29,5,19,10,8,30,30,0,15,3,17,2,12,22,1,8,20,14,2,15,10,17,4,26,23,19, 15,25,30,17,23,13,18,24,15,1,15,14,5,13,18,10,28,11,11,30,15,0,2,28,4,27,29,21,3 0,1,29,8,6,1,21,13,8,29,24,13,22,1,3,20,8,7,25,26,21,11,3,3,10,9,24,27,5,5,7,9,1 6,17,18,0,5,9,12,10,29,9,20,24,11,4,3,30,12,30,26,6,17,14,20,15,26,17,27,28,15,2 1,15,27,12,24,25,24,17,3,25,9,0,3,15,12,28,6,0,3,17,27,7,26,10,9,13,24,4,21,21,1 9,20,29,15,15,0,24,22,17,30,2,6,21,12,21,16,24,4,10,25,6,18,26,19,5,7,17,18,7,1, 12,13,15,17,21,14,13,17,9,14,4,6,1,13,12,28,1,0,11,29,22,3,25,3,11,21,12,22,5,18 ,7,5,0,12,26,11,17,8,14,0,18,6,7,8,30,23,19,0,26,6,15,23,24,30,29,11,20,2,21,14, 4,20,24,1,15,28,30,14,19,8,22,2,23,3,14,20,30,1,1,1,12,17,12,0,2,1,29,18,17,17,5 ,17,21,8,8,23,15,11,2,18,30,12,26,9,13,15,11,30,12,23,21,5,18,12,4,20,8,15,19,19 ,16,26,24,8,6,23,11,11,9,4,7,22,27,1,24,1,29,19,12,4,18,25,1,5,4,3,25,25,6,0,3,6 ,23,23,15,4,6,21,10,20,25,19,11,4,26,4,28,8,8,22,26,12,6,20,28,24,7,4,11,25,28,2 3,0,25,1,3,0,2,10,11,20,4,21,8,15,11,10,6,1,30,2,6,2,6,0,9,19,13,27,30,21,13,18, 17,10,21,27,23,2,5,13,15,10,4,4,18,10,9,5,7,10,24,19,12,2,12,26,6,25,8,18,3,13,1 9,16,19,15,28,11,16,8,29,15,13,23,24,2,2,19,0,7,23,14,2,28,22,2,0,26,16,18,11,19 ,0,10,19,5,26,8,8,8,13,5,30,20,29,21,30,3,7,29,27,19,28,13,2,15,18,28,13,5,24,13 ,15,27,9,29,0,7,16,28,12,1,9,4,8,6,12,11,12,14,22,28,20,14,8,17,25,20,13,16,6,8, 3,7,7,28,17,23,17,13,16,0,19,23,5,17,6,20,21,9,14,30,2,26,7,5,30,24,2,10,23,22,8 ,19,4,9,27,29,11,4,13,23,8,18,6,6,1,18,5,17,6,18,13,6,30,6,28,9,22,9,11,3,3,4,25 ,29,3,10,19,26,14,11,29,9,23,4,0,23,29,14,2,21,29,18,10,10,3,2,4,16,20,23,2,5,0, 11,10,1,19,14,10,4,15,22,25,8,10,12,25,15,30,5,2,30,10,5,30,7,3,23,23,19,18,3,10 ,13,2,30,16,30,2,6,22,30,3,4,28,6,2,22,3,3,5,20,14,16,12,14,28,12, ccc: 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3, 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4, 4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6, 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7, 7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8, 8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10 ,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,11,1 1,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13 ,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,1 4,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15, 15,15,15,15,15,15,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,16,16,16,16,16 ,16,16,16,16,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,17,17,17,17,1 7,17,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,18,18,18,18,18, 18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19 ,19,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,2 0,20,20,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,21,21,21,21, 21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22 ,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,2 3,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24, 24,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25 ,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,2 7,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28, 28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29 ,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,3 0,30,30,30,30,30,30,30,30,30,30,30,30,30, 25090网络流 25086贪心 Process returned 0 (0x0) execution time : 1.981 s Press any key to continue. */ /* #include <iostream> #include <cmath> #include <vector> #include <cstdlib> #include <cstdio> #include <cstring> #include <queue> #include <stack> #include <list> #include <algorithm> #include <map> #include <set> #define LL long long #define Pr pair<int,int> #define fread() freopen("data1.in","r",stdin) #define fwrite() freopen("out.out","w",stdout) using namespace std; const int INF = 0x3f3f3f3f; const int msz = 10000; const int mod = 1e9+7; const double eps = 1e-8; struct Edge { int v,f,w,next; }; Edge eg[66666]; int head[2333],h[2333],hcnt[2333]; int dis[2333],pre[2333]; bool vis[2333]; int tp,ans,minf; void Add(int u,int v,int w,int f) { eg[tp].v = v; eg[tp].w = w; eg[tp].f = f; eg[tp].next = head[u]; head[u] = tp++; } bool bfs(int st,int en) { memset(dis,-1,sizeof(dis)); memset(vis,0,sizeof(vis)); memset(pre,-1,sizeof(pre)); queue <int> q; q.push(st); dis[st] = 0; minf = INF; int u,v,w,f; while(!q.empty()) { u = q.front(); q.pop(); vis[u] = 0; for(int i = head[u]; i != -1; i = eg[i].next) { v = eg[i].v; w = eg[i].w; f = eg[i].f; if(f && (dis[v] == -1 || dis[v] < dis[u]+w)) { dis[v] = dis[u]+w; minf = min(f,minf); pre[v] = i; if(!vis[v]) { q.push(v); vis[v] = 1; } } } } if(pre[en] == -1) return false; ans += dis[en]; for(int i = pre[en]; i != -1; i = pre[eg[i^1].v]) { eg[i].f -= minf; eg[i^1].f += minf; } return true; } int main() { int t,n,m,k,x; scanf("%d",&t); for(int z = 1; z <= t; ++z) { scanf("%d%d%d",&n,&m,&k); memset(head,-1,sizeof(head)); memset(hcnt,0,sizeof(hcnt)); tp = 0; ans = 0; for(int i = 0; i < n; ++i) { scanf("%d",&h[i]); if(i) ans += abs(h[i]-h[i-1]); } while(m--) { scanf("%d",&x); hcnt[x]++; } //n为起点 0~n-2表示初始第i个山爬到i+1山路程 n+1~n+30表示高1~30的山数 n+31表示起点前的一个点(限流) n+32表示终点 Add(n+33,n+31,0,k); Add(n+31,n+33,0,0); for(int i = 0; i <= 30; ++i) { if(!hcnt[i]) continue; Add(n+31,n+i,0,hcnt[i]); Add(n+i,n+31,0,0); for(int j = 0; j < n-1; ++j) { Add(n+i,j,abs(h[j]-i)+abs(h[j+1]-i)-abs(h[j]-h[j+1]),hcnt[i]); Add(j,n+i,-abs(h[j]-i)-abs(h[j+1]-i)+abs(h[j]-h[j+1]),0); } } for(int j = 0; j < n-1; ++j) { Add(j,n+32,0,1); Add(n+32,j,0,0); } while(bfs(n+33,n+32)); printf("Case %d: %d\n",z,ans); } return 0; } */
s.正宗解法，最小费用流
参考：山东省第三届省赛 C 题 – An interesting game(费用流)
ps：参考的这个代码比较快，600ms。 我的好像得1000多。。。
大意：有n个山坡，排成一排，为了增加难度，现在要从m个山坡中挑选k个，插入到原有的n个山坡中，两个原有的山坡中只能插入一个山坡，原有山坡的两侧不能插入。使插入k个山坡后的难度最大，总的难度是相邻两个山坡差值绝对值的和。
思路：训练赛的时候根本没往图论上想，当时想贪心不太可能，还以为是dp。
如果把原问题看做是原有山坡的排列中，两个相邻山坡中间的间隙和m个山坡做匹配的话，就可以看做是，最大流量为k时，最大费用是多少，费用流可以搞。对于最大费用，可以先把费用取负值，求出最小费用后再取负值，就是最大费用。匹配的费用是插入之后的难度的增加量，最大费用加上原山坡的难度，就是答案。
因为n和m的范围是[2, 1000]，直接建图会TLE，发现插入的m个山坡的范围只有[0, 30]，这么小的数据范围一定能优化答案，把插入山坡的高度看做结点，建图。
设置源点S，超级源点SS，汇点T。
S对所有间隙连边，容量1，费用0.
所有间隙对所有高度连边，容量1，费用是难度增加量的相反数.
所有高度对T连边，容量为此高度的个数，费用0.
SS对S连边，容量k，费用0.
答案就是原图的难度-最小费用。
c.最小费用最大流
/* SPFA版费用流 最小费用最大流，求最大费用最大流只需要取相反数，结果取相反数即可。 点的总数为N，点的编号0~N-1 */ #include<iostream> #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; const int MAXN=1500; const int MAXM=80000; const int INF=0x3f3f3f3f; struct Edge{ int to,next,cap,flow,cost; }edge[MAXM]; int head[MAXN],tol; int pre[MAXN],dis[MAXN]; bool vis[MAXN]; int N;//节点总个数，节点编号从0~N-1 void init(int n){ N=n; tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int cap,int cost){ edge[tol].to=v; edge[tol].cap=cap; edge[tol].cost=cost; edge[tol].flow=0; edge[tol].next=head[u]; head[u]=tol++; edge[tol].to=u; edge[tol].cap=0; edge[tol].cost=-cost; edge[tol].flow=0; edge[tol].next=head[v]; head[v]=tol++; } bool spfa(int s,int t){ queue<int>q; for(int i=0;i<N;i++){ dis[i]=INF; vis[i]=false; pre[i]=-1; } dis[s]=0; vis[s]=true; q.push(s); while(!q.empty()){ int u=q.front(); q.pop(); vis[u]=false; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].to; if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost){ dis[v]=dis[u]+edge[i].cost; pre[v]=i; if(!vis[v]){ vis[v]=true; q.push(v); } } } } if(pre[t]==-1)return false; else return true; } //返回的是最大流，cost存的是最小费用 int minCostMaxflow(int s,int t,int &cost){ int flow=0; cost=0; while(spfa(s,t)){ int Min=INF; for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){ if(Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow; } for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){ edge[i].flow+=Min; edge[i^1].flow-=Min; cost+=edge[i].cost*Min; } flow+=Min; } return flow; } int main(){ int T; int N2,M,K; int X[1005],Y[1005]; int YNum[40]; int i,j; int sp,sp2;//源点，源点2 int sc;//汇点 int sum; int tmp; int mi_cost; int ma_flow; int ca=0; scanf("%d",&T); while(T--){ scanf("%d%d%d",&N2,&M,&K); init(N2+50); for(i=0;i<N2;++i){ scanf("%d",&X[i]); } memset(YNum,0,sizeof(YNum)); for(i=0;i<M;++i){ scanf("%d",&Y[i]); ++YNum[Y[i]]; } sum=0; for(i=0;i<N2-1;++i){//初始值 sum=sum+abs(X[i]-X[i+1]); } sp=N2+40; for(i=0;i<N2-1;++i){//加边 addedge(sp,i,1,0);//源点到空隙 for(j=0;j<=30;++j){ if(YNum[j]){ tmp=abs(X[i]-j)+abs(X[i+1]-j)-abs(X[i]-X[i+1]); addedge(i,N2+j,1,-tmp);//空隙到M个山丘 } } } sp2=N2+41; sc=N2+42; addedge(sp2,sp,K,0); for(i=0;i<=30;++i){ if(YNum[i]){ addedge(N2+i,sc,YNum[i],0); } } ma_flow=minCostMaxflow(sp2,sc,mi_cost); printf("Case %d: %d\n",++ca,sum-mi_cost); } return 0; }
D.
d.好像是按要求输出串
s.没写这个题
c.张
#include <iostream> #include <cstdio> #include <string> #include <cstring> #define MAX 500 using namespace std; int main () { char temp1[MAX],temp2[MAX],str[MAX]; strcpy(temp1," n5!wpuea^o7!usimdnaevoli"); strcpy(temp2," usimdnaevolin5!wpuea^o7!"); int Len = strlen(temp1); int T; scanf("%d",&T); getchar(); int coun=1; while(T--) { gets(str); int len=strlen(str); for(int i=0,j=0; i<len; i++) { for(j=0; j<Len; j++) if(str[i]==temp1[j]) break; if(j<Len) str[i]=temp2[j]; } printf("Case %d: ",coun++); for(int i=len-1; i>=0; i--) printf("%c",str[i]); printf("\n"); } return 0; }
E.
d.水果忍者，水果从上向下掉落，并且只能水平切。
给你n个水果的出现时间，出现的水平位置，和他是否是好水果，切到好水果+1，切到坏水果-1，连续切到三个以上好水果分数加倍。
两刀之间的间隔大于等于m，求能求得的最大分数。
s.先求出每一秒出手能得到的最大的得分，很简单。
再用DP求最大分数。
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MAXN 10010 int d[MAXN][210];//d[i][j]标志i秒j位置水果 int dp[MAXN];//dp[i]表示前i秒获得的最大分数 int main(){ int T; int n,m; int t,s,p; int i,j; int ma_t;//最大时间 int ma_p[MAXN];//每秒最大位置 int sum,num,ma;// int ma_sum[MAXN];//每秒可获得的最大分数 int ca=0; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&m); memset(d,-1,sizeof(d)); ma_t=0; memset(ma_p,0,sizeof(ma_p)); for(i=0;i<n;++i){ scanf("%d%d%d",&t,&s,&p); d[t][p]=s; if(t>ma_t){ ma_t=t; } if(p>ma_p[t]){ ma_p[t]=p; } } memset(ma_sum,0,sizeof(ma_sum)); for(i=1;i<=ma_t;++i){//先求出每秒可获得的最大分数 sum=0; num=0; ma=0; for(j=1;j<=ma_p[i];++j){ if(d[i][j]==0){ ++num; } else if(d[i][j]==1){ if(num>=3){ sum=sum+num*2; } else{ sum=sum+num; } if(sum>ma){ ma=sum; } num=0; --sum; if(sum<0){ sum=0; } } } if(num>0){ if(num>=3){ sum=sum+num*2; } else{ sum=sum+num; } if(sum>ma){ ma=sum; } } ma_sum[i]=ma; } memset(dp,0,sizeof(dp)); for(i=1;i<=m;++i){ dp[i]=max(dp[i-1],ma_sum[i]);//不出手，和出手 } for(i=m+1;i<=ma_t;++i){ dp[i]=max(dp[i-1],dp[i-m-1]+ma_sum[i]); } printf("Case %d: %d\n",++ca,dp[ma_t]); } return 0; }
F.
d.串中提取数字
s.比较坑的是里面空格只要一个就行了
c.当时臃肿的代码。虽然长，但是思路还可以
#include<iostream> #include<stdio.h> #include<string.h> #include<math.h> using namespace std; char str[12345]; char str1[12345]; char str2[12345]; char num1[12345]; char num2[12345]; char num3[12345]; int main(){ int T; int i; int len; int len1,len2; int p1,p2; int p3,p4; int len_num1,len_num2,len_num3; int j; double a,a1,a2,b,b1,b2,c,c1,c2; int p5; double Dp; double ans; int ca=0; int p6,p7; /* num1 num2 num3 The new iPad 0009.7 inches 2048*1536 PAD p3 p1 p2 p4 0009.7 2048*1536 p5 p6,07是考虑了后面的小数点，实际好像后面只能是整数 */ scanf("%d",&T); getchar(); while(T--){ scanf("%[^\n]",str); getchar(); len=strlen(str); len1=0; len2=0; len_num1=0; len_num2=0; len_num3=0; for(i=0;i<len;++i){ if(str[i]=='i'){ if(str[i+1]=='n'&&str[i+2]=='c'&&str[i+3]=='h'&& str[i+4]=='e'&&str[i+5]=='s'&&str[i+6]==' '){ p1=i; j=i-1; while(str[j]==' '){ --j; } while(str[j]!=' '){ num1[len_num1++]=str[j--]; } while(str[j]==' '){ --j; } p3=j; } } if(str[i]=='*'){ if(i>=1&&'0'<=str[i-1]&&str[i-1]<='9'){ if('0'<=str[i+1]&&str[i+1]<='9'){ p2=i; j=i-1; while(str[j]!=' '){ num2[len_num2++]=str[j--]; } j=i+1; while(str[j]!=' '){ num3[len_num3++]=str[j++]; } while(str[j]==' '){ ++j; } p4=j; } } } } for(i=0;i<=p3;++i){ if(str[i]!=' '){ break; } } str1[len1++]=str[i++]; for(;i<=p3;++i){ if(str[i]==' '&&str1[len1-1]==' '){ continue; } str1[len1++]=str[i]; } str1[len1]='\0'; for(i=p4;i<len;++i){ if(str[i]==' '&&str2[len2-1]==' '){ continue; } if('a'<=str[i]&&str[i]<='z'){ str2[len2++]=str[i]; } else if('A'<=str[i]&&str[i]<='Z'){ str2[len2++]=str[i]+32; } else{ str2[len2++]=str[i]; } } str2[len2]='\0'; for(i=len2-1;i>=0;--i){ if(str2[i]!=' '){ len2=i+1; break; } } str2[i+1]='\0'; p5=-1; for(i=0;i<len_num1;++i){ if(num1[i]=='.'){ p5=i; break; } } a=0; a1=0; a2=0; if(p5==-1){ for(i=0;i<len_num1;++i){ a1=a1+(num1[i]-'0')*pow(10,i); } } else{ for(i=0;i<p5;++i){ a2=a2/10+(num1[i]-'0')/10.0; } for(i=p5+1;i<len_num1;++i){ a1=a1+(num1[i]-'0')*pow(10,(i-(p5+1))); } } a=a1+a2; if(a==0){ printf("Case %d: The %s of %s's PPI is %.2f.\n",++ca,str2,str1,0.0); continue; } p6=-1; for(i=0;i<len_num2;++i){ if(num2[i]=='.'){ p6=i; break; } } if(p6==-1){ b=0; for(i=0;i<len_num2;++i){ b=b+(num2[i]-'0')*pow(10,i); } } else{ b=0; b1=0; b2=0; for(i=0;i<p6;++i){ b2=b2/10+(num2[i]-'0')/10.0; } for(i=p6+1;i<len_num2;++i){ b1=b1+(num2[i]-'0')*pow(10,(i-(p6+1))); } b=b1+b2; } p7=-1; for(i=0;i<len_num3;++i){ if(num3[i]=='.'){ p7=i; break; } } if(p7==-1){ c=0; for(i=0;i<len_num3;++i){ c=c*10+(num3[i]-'0'); } } else{ c=0; c1=0; c2=0; for(i=0;i<p7;++i){ c1=c1*10+(num3[i]-'0'); } for(i=p7+1;i<len_num3;++i){ c2=c2/10+(num3[i]-'0')/10.0; } c=c1+c2; } Dp=sqrt(b*b+c*c); ans=Dp/a; printf("Case %d: The %s of %s's PPI is %.2f.\n",++ca,str2,str1,ans); } return 0; }
G.
d.类似于扫雷的游戏，这里不是周围8个位置，而是上下左右加它自己5个位置。给出周围多少个雷的数组，求出雷的分布。
s.dfs。先确定第一行，下面的每个位置只看上面的位置即可。
#include<iostream> #include<stdio.h> using namespace std; int d[25][25]; char g[25][25]; int n,m; int dir[][2]={{-1,0},{1,0},{0,-1},{0,1},{0,0}}; bool stop; void dfs(int r,int c); void f1(int r,int c){ int i; int a,b; g[r][c]='*'; for(i=0;i<5;++i){ a=r+dir[i][0]; b=c+dir[i][1]; if( 0<=a&&a<n && 0<=b&&b<m ){ --d[a][b]; } } if(c==m-1){ a=r+1; b=0; } else{ a=r; b=c+1; } dfs(a,b); g[r][c]='\0'; for(i=0;i<5;++i){ a=r+dir[i][0]; b=c+dir[i][1]; if( 0<=a&&a<n && 0<=b&&b<m ){ ++d[a][b]; } } } void f2(int r,int c){ int a,b; g[r][c]='.'; if(c==m-1){ a=r+1; b=0; } else{ a=r; b=c+1; } dfs(a,b); g[r][c]='\0'; } void dfs(int r,int c){ int i,j; if(r==n){ for(i=0;i<m;++i){ if(d[n-1][i]!=0){ return; } } stop=true; for(i=0;i<n;++i){ for(j=0;j<m;++j){ printf("%c",g[i][j]); } printf("\n"); } return; } int a,b; if(r==0){ bool flag=true;//可以放雷 for(i=0;i<5;++i){ a=r+dir[i][0]; b=c+dir[i][1]; if( 0<=a&&a<n && 0<=b&&b<m ){ if(d[a][b]==0){ flag=false; break; } } } if(flag){ f1(r,c); if(stop){ return; } } f2(r,c); } else{ if(d[r-1][c]==1){ f1(r,c); if(stop){ return; } } else if(d[r-1][c]==0){ f2(r,c); } } } int main(){ int T; int i,j; int ca=0; char str[25]; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&m); for(i=0;i<n;++i){ scanf("%s",str); for(j=0;j<m;++j){ d[i][j]=str[j]-'0'; } } printf("Case %d:\n",++ca); stop=false; dfs(0,0); } return 0; }
H.
d.在一个N*M（N,M<=20）的矩阵中，某个位置的值取决于上下左右和它自己，求出所有位置的值。
s.双重循环，直接求一遍就行了。
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}}; int main(){ int T; int N,M; int s[25][25]; int v[25][25]; int i,j; int a,b; int ma,ma_i,ma_j; int ca=0; int k; scanf("%d",&T); while(T--){ scanf("%d%d",&N,&M); for(i=0;i<N;++i){ for(j=0;j<M;++j){ scanf("%d",&s[i][j]); } } memset(v,0,sizeof(v)); for(i=0;i<N;++i){ for(j=0;j<M;++j){ for(k=0;k<4;++k){ a=i+dir[k][0]; b=j+dir[k][1]; if(a<0||b<0||a>=N||b>=M){ v[i][j]-=1; continue; } if(s[a][b]>s[i][j]){ v[i][j]=v[i][j]+(s[a][b]-s[i][j]); } else{ v[i][j]=v[i][j]-(s[i][j]-s[a][b]); } } } } ma=v[0][0]; ma_i=0; ma_j=0; for(i=0;i<N;++i){ for(j=0;j<M;++j){ if(v[i][j]>=ma){ ma=v[i][j]; ma_i=i; ma_j=j; } } } printf("Case %d: %d %d %d\n",++ca,ma,ma_i+1,ma_j+1); } return 0; }
I.
d.比较直接的一个完全背包，但是大小有这么大V <= 100,000,000
s.很显然，直接来会爆内存了。
但是这个题物品大小不大1 <= S<= 100。然后很萎缩的5000以上贪心，剩余的用完全背包。。。
为什么过了？
#include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; struct node{ long long S; long long P; double q; }a[3]; bool cmp(node a,node b){ return a.q>b.q; } int main(){ int T; long long V; long long sum; long long dp[6000]; long long tmp; int i,j; int ca=0; scanf("%d",&T); while(T--){ scanf("%lld%lld",&a[0].S,&a[0].P); a[0].q=(double)(a[0].P)/a[0].S; //cout<<a[0].q<<endl; scanf("%lld%lld",&a[1].S,&a[1].P); a[1].q=(double)(a[1].P)/a[1].S; //cout<<a[1].q<<endl; scanf("%lld%lld",&a[2].S,&a[2].P); a[2].q=(double)(a[2].P)/a[2].S; //cout<<a[2].q<<endl; scanf("%lld",&V); sort(a,a+3,cmp); sum=V%a[0].S; for(i=1;;++i){ if(a[0].S*i>=5000){ break; } } sum=sum+a[0].S*i; memset(dp,0,sizeof(dp)); for(i=0;i<3;++i){ for(j=a[i].S;j<=sum;++j){ tmp=dp[j-a[i].S]+a[i].P; if(tmp>dp[j]){ dp[j]=tmp; } } } for(i=sum;;--i){ if(dp[i]>0){ break; } } printf("Case %d: %lld\n",++ca,((V-sum)/a[0].S)*a[0].P+dp[i]); } return 0; }
J.
d.好像是积分求椭球体的体积。
c.王
#include <iostream> #include <stdio.h> #include <cmath> using namespace std; int main() { int T; scanf("%d",&T); int i; for(i=1; i<=T; i++) { int a,b,h; scanf("%d%d%d",&a,&b,&h); double V; V=(4.0/3)*M_PI*a*b*b; if(h>=b) { printf("Case %d: %.3lf\n",i,V); } else { double v=M_PI*a*b*(b-h)-M_PI*(1.0*a/b)*(1.0/3)*(b*b*b-h*h*h); if(V-v-v>0) printf("Case %d: %.3lf\n",i,V-v); else printf("Case %d: %.3lf\n",i,v); } } return 0; }
