Codeforces #303 div2 (only)
2015-05-20 11:44:30
总结：水场... 赛中 AK 了，但是发现 E 题没开、优先队列打残了，惨遭 FST...。难得的 AK 机会就这么溜走辣。
来总结一下吧～
A题：水题。
检查每一行，如果之存在0、-1 、2 则答案+1
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; int n; int g[110][110]; int main(){ scanf("%d",&n); int cnt = 0; vector<int> y; for(int i = 1; i <= n; ++i){ int f = 1; for(int j = 1; j <= n; ++j){ scanf("%d",&g[i][j]); if(j != i && g[i][j] != 0 && g[i][j] != 2){ f = 0; } } if(f){ cnt += f; y.PB(i); } } printf("%d\n",cnt); for(int i = 0; i < y.size(); ++i) printf("%d ",y[i]); puts(""); return 0; }
B题：水题。
找到一个距离两个串（s1和s2）海明距离相等的串 (str)，先统计两串的海明距离 cnt，如果 cnt 为奇数显然 impossible
否则，对于每个位置 i，若 s1[i] == s2[i]，则 str[i] = s1[i] = s2[i]；再考虑所有的位置 i （s1[i] != s2[i]）前 cnt/2 个取 s1，其余取 s2
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; char s1[100010]; char s2[100010]; int vis[100010]; char s[100010]; int main(){ scanf("%s%s",s1,s2); int len = strlen(s1); int cnt = 0; for(int i = 0; i < len; ++i){ if(s1[i] != s2[i]) ++cnt; } if(cnt & 1) printf("impossible\n"); else{ cnt /= 2; for(int i = 0; i < len; ++i){ if(s1[i] != s2[i]){ if(cnt){ s[i] = s1[i]; --cnt; } else{ s[i] = s2[i]; } } else s[i] = s1[i]; } s[len] = '\0'; printf("%s\n",s); } return 0; }
C题：贪心题。
从左道右以此考虑，尽量左倒，不能左倒考虑右倒，都不能就不倒
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; int n; struct node{ int x,h; }p[100010]; int main(){ scanf("%d",&n); for(int i = 1; i <= n; ++i){ scanf("%d%d",&p[i].x,&p[i].h); } int ans = 0; int pos = -INF; for(int i = 1; i <= n; ++i){ if(p[i].x - p[i].h > pos){ pos = p[i].x; ans++; } else if(i == n || p[i].x + p[i].h < p[i + 1].x){ pos = p[i].x + p[i].h; ans++; } pos = max(pos,p[i].x); } printf("%d\n",ans); return 0; }
D题：贪心题。
按照 t 排序，然后贪心考虑即可。符合条件就把该人加入队列，并把答案个数+1，否则不加入队列。
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; int n; int t[100010]; int main(){ scanf("%d",&n); for(int i = 1; i <= n; ++i){ scanf("%d",&t[i]); } sort(t + 1,t + n + 1); int ans = 0; ll sum = 0; for(int i = 1; i <= n; ++i){ if(sum <= (ll)t[i]){++ans; sum += (ll)t[i]; } } printf("%d\n",ans); return 0; }
E题：最短路，路径记录。
由于数据范围比较大，直接从给出的起点开始跑一边堆优化的 Dijstra（优先队列实现），过程中需要维护每个点的最优前驱边。
最后只要输出所有前驱边集即可。注意 long long 和优先队列的写法！
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<long long,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; const int MAXN = 600010; int n,m,st; int first[MAXN],ecnt; int pre[MAXN]; int vis[MAXN]; ll dis[MAXN],ans; vector<int> g; struct edge{ int next,u,v,id; ll w; }e[MAXN]; void add_edge(int u,int v,ll w,int id){ e[++ecnt].next = first[u]; e[ecnt].u = u; e[ecnt].v = v; e[ecnt].w = w; e[ecnt].id = id; first[u] = ecnt; } struct cmp{ //将优先队列改为小根堆 bool operator()(pii a,pii b){ return a.first>b.first; } }; void Dijstra(){ priority_queue<pii,vector<pii>,cmp> Q; fill(dis + 1,dis + n + 1,1LL << 60); dis[st] = 0; Q.push(MP(dis[st],st)); ans = 0; while(!Q.empty()){ pii x = Q.top(); Q.pop(); if(dis[x.second] < x.first) continue; for(int i = first[x.second]; ~i; i = e[i].next){ int v = e[i].v; if(dis[v] > dis[x.second] + e[i].w){ dis[v] = dis[x.second] + e[i].w; pre[v] = i; Q.push(MP(dis[v],v)); } else if(dis[v] == dis[x.second] + e[i].w){ if(e[i].w < e[pre[v]].w){ pre[v] = i; } } } } } int main(){ memset(first,-1,sizeof(first)); ecnt = 0; scanf("%d%d",&n,&m); int a,b,c; for(int i = 1; i <= m; ++i){ scanf("%d%d%d",&a,&b,&c); add_edge(a,b,c,i); add_edge(b,a,c,i); } scanf("%d",&st); Dijstra(); for(int i = 1; i <= n; ++i) vis[pre[i]] = 1; for(int i = 1; i <= ecnt; ++i) if(vis[i]){ ans += e[i].w; g.PB(e[i].id); } printf("%I64d\n",ans); for(int i = 0; i < g.size(); ++i) printf("%d ",g[i]); puts(""); return 0; }
