2017 ACM/ICPC Asia Regional Beijing Online

A Visiting Peking University

直接模拟

B Reverse Suffix Array

unsolved

C Matrix

预处理每列前缀和以及每个每个行区间的每列的最小值，枚举行的所有区间，求最长连续字段和。

D Agent Communication

unsolved

E Territorial Dispute

n=1,2时不行

n=3时必须三点共线

n>3一定可以，任意找四个点判断即可。

好像也可以跑凸包，不在凸包上的点单独分开，如果全部在凸包上，找两个不相邻的点即可。

  1 #include <bits/stdc++.h>
  2 #define lson (id*2)
  3 #define rson (id*2+1)
  4 #define mid ((l+r)/2)
  5 using namespace std;
  6 
  7 typedef long long LL;
  8 const LL MAXN=1e3+5;
  9 
 10 LL n;
 11 struct PoLL{
 12     LL x,y,id;
 13 } p[MAXN];
 14 bool cmp(PoLL a,PoLL b){
 15     if(a.x==b.x) return a.y<b.y;
 16     return a.x<b.x;
 17 }
 18 LL vis[MAXN];
 19 
 20 LL cross(PoLL a,PoLL b){
 21     return a.x*b.y-a.y*b.x;
 22 }
 23 
 24 int main(){
 25     LL t;
 26     scanf("%lld",&t);
 27     while(t--){
 28         memset(vis,0,sizeof(vis));
 29         scanf("%lld",&n);
 30         for(LL i=1;i<=n;i++){
 31             scanf("%lld%lld",&p[i].x,&p[i].y);
 32             p[i].id=i;
 33         }
 34         sort(p+1,p+1+n,cmp);
 35         if(n<=2){
 36             printf("NO\n");
 37             continue;
 38         }
 39         else if(n==3){
 40             PoLL a=(PoLL){p[2].x-p[1].x,p[2].y-p[1].y};
 41             PoLL b=(PoLL){p[3].x-p[2].x,p[3].y-p[2].y};
 42             if(cross(a,b)==0) vis[p[2].id]=1;
 43             else{
 44                 printf("NO\n");
 45                 continue;
 46             }
 47         }
 48         else{
 49             LL flag=0;
 50             for(int tt=1;tt<=4;tt++){
 51                 int i,j,k;
 52                 if(tt==1) j=1,i=2,k=3;
 53                 if(tt==2) j=1,i=2,k=4;
 54                 if(tt==3) j=1,i=3,k=4;
 55                 if(tt==4) j=2,i=3,k=4;
 56                 PoLL ak=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y};
 57                 PoLL bk=(PoLL){p[k].x-p[i].x,p[k].y-p[i].y};
 58                 if(cross(ak,bk)==0){
 59                     vis[p[i].id]=1;
 60                     flag=1;
 61                     break;
 62                 }
 63             }
 64 
 65             if(flag==0)
 66             for(LL i=1;i<=4;i++)
 67             for(LL j=1;j<=4;j++)
 68             for(LL k=1;k<=4;k++)
 69             for(LL q=1;q<=4;q++){
 70                 if(flag) break;
 71                 if(i==j||i==k||i==q) continue;
 72                 if(j==k||j==q||k==q) continue;
 73                 PoLL aa=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y};
 74                 PoLL bb=(PoLL){p[i].x-p[k].x,p[i].y-p[k].y};
 75                 PoLL cc=(PoLL){p[i].x-p[q].x,p[i].y-p[q].y};
 76                 LL s1=abs(cross(aa,bb))+abs(cross(aa,cc))+abs(cross(bb,cc));
 77 
 78                 PoLL dd=(PoLL){p[j].x-p[k].x,p[j].y-p[k].y};
 79                 PoLL ee=(PoLL){p[q].x-p[k].x,p[q].y-p[k].y};
 80                 LL s2=abs(cross(dd,ee));
 81                 if(s1==s2){
 82                     flag=1;
 83                     vis[p[i].id]=1;
 84                     break;
 85                 }
 86             }
 87 
 88             if(flag==0)
 89             for(LL i=1;i<=4;i++)
 90             for(LL j=1;j<=4;j++)
 91             for(LL k=1;k<=4;k++)
 92             for(LL q=1;q<=4;q++){
 93                 if(flag) break;
 94                 if(i==j||i==k||i==q) continue;
 95                 if(j==k||j==q||k==q) continue;
 96                 PoLL aa=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y};
 97                 PoLL bb=(PoLL){p[i].x-p[k].x,p[i].y-p[k].y};
 98                 PoLL cc=(PoLL){p[i].x-p[q].x,p[i].y-p[q].y};
 99                 if(cross(aa,bb)*cross(aa,cc)<0){
100                     vis[p[i].id]=vis[p[j].id]=1;
101                     flag=1;
102                     break;
103                 }
104             }
105             if(!flag){
106                 printf("NO\n");
107                 continue;
108             }
109         }
110         printf("YES\n");
111         for(LL i=1;i<=n;i++){
112             if(vis[i]) printf("A");
113             else printf("B");
114         }
115         printf("\n");
116     }
117 
118     return 0;
119 }
Psong

F Cake

unsolved

G Bounce

经过的总数：res=(n-1)*(m-1)/gcd(n-1,m-1)+1

交点与拐点：(res-1)/gcd(n-1,m-1)

拐点个数：(n-1)/gcd(n-1,m-1)+(m-1)/gcd(n-1,m-1)-1

 1 #include <bits/stdc++.h>
 2 #define lson (id*2)
 3 #define rson (id*2+1)
 4 #define mid ((l+r)/2)
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 
 9 LL gcd(LL x,LL y){
10     return y==0?x:gcd(y,x%y);
11 }
12 
13 int main(){
14     LL n,m,k;
15     while(cin>>n>>m){
16         LL G=gcd(n-1,m-1);
17         LL res=(n-1)*(m-1)/G+1;
18         res=res-(res-1)/G+(n-1)/G+(m-1)/G-1;
19         cout<<res<<endl;
20     }
21 
22     return 0;
23 }
Psong

H Polynomial Product

unsolved

I Minimal

线段树维护最大最小正负四个值即可

  1 #include <bits/stdc++.h>
  2 #define lson (id*2)
  3 #define rson (id*2+1)
  4 #define mid ((l+r)/2)
  5 using namespace std;
  6 
  7 typedef long long LL;
  8 const int MAXN=2e5+5;
  9 const int inf=0x3f3f3f3f;
 10 
 11 int n,q;
 12 int a[MAXN];
 13 
 14 struct node{
 15     int val;
 16     int mizhen,mafu;
 17     int mi,ma;
 18 } tre[MAXN*4],ans;
 19 
 20 void push_up(int id,int ls,int rs){
 21     tre[id].mizhen=min(tre[ls].mizhen,tre[rs].mizhen);
 22     tre[id].mafu=max(tre[ls].mafu,tre[rs].mafu);
 23     tre[id].mi=min(tre[ls].mi,tre[rs].mi);
 24     tre[id].ma=max(tre[ls].ma,tre[rs].ma);
 25 }
 26 void build(int id,int l,int r){
 27     if(l==r){
 28         tre[id].mi=tre[id].ma=tre[id].val=a[l];
 29         if(a[l]>0){
 30             tre[id].mizhen=a[l];
 31             tre[id].mafu=-inf;
 32         }
 33         if(a[l]<0){
 34             tre[id].mizhen=inf;
 35             tre[id].mafu=a[l];
 36         }
 37         if(a[l]==0){
 38             tre[id].mizhen=a[l];
 39             tre[id].mafu=a[l];
 40         }
 41         return;
 42     }
 43     build(lson,l,mid);
 44     build(rson,mid+1,r);
 45     push_up(id,lson,rson);
 46 }
 47 
 48 void update(int id,int l,int r,int pos,int tt){
 49     if(l==r){
 50         tre[id].mi=tre[id].ma=tre[id].val=tt;
 51         if(tt>=0){
 52             tre[id].mizhen=tt;
 53             tre[id].mafu=-inf;
 54         }
 55         if(tt<=0){
 56             tre[id].mizhen=inf;
 57             tre[id].mafu=tt;
 58         }
 59         return;
 60     }
 61     if(pos<=mid)
 62         update(lson,l,mid,pos,tt);
 63     else
 64         update(rson,mid+1,r,pos,tt);
 65     push_up(id,lson,rson);
 66 }
 67 
 68 void query(int id,int l,int r,int ql,int qr){
 69     if(ql<=l&&r<=qr){
 70         ans.mizhen=min(ans.mizhen,tre[id].mizhen);
 71         ans.mafu=max(ans.mafu,tre[id].mafu);
 72         ans.ma=max(ans.ma,tre[id].ma);
 73         ans.mi=min(ans.mi,tre[id].mi);
 74         return;
 75     }
 76     if(ql<=mid)
 77         query(lson,l,mid,ql,qr);
 78     if(mid+1<=qr)
 79         query(rson,mid+1,r,ql,qr);
 80 }
 81 
 82 void solve(){
 83     LL res=(LL)ans.mi*ans.ma;
 84     res=min(res,(LL)ans.mi*ans.mi);
 85     res=min(res,(LL)ans.ma*ans.ma);
 86     if(ans.mafu!=-inf){
 87         res=min(res,(LL)ans.mafu*ans.mafu);
 88         res=min(res,(LL)ans.mafu*ans.mi);
 89         res=min(res,(LL)ans.mafu*ans.ma);
 90     }
 91     if(ans.mizhen!=inf){
 92         res=min(res,(LL)ans.mizhen*ans.mizhen);
 93         res=min(res,(LL)ans.mizhen*ans.mi);
 94         res=min(res,(LL)ans.mizhen*ans.ma);
 95     }
 96     if(ans.mafu!=-inf&&ans.mizhen!=inf) res=min(res,(LL)ans.mafu*ans.mizhen);
 97     printf("%lld\n",res);
 98 }
 99 
100 int main(){
101     int t;
102     scanf("%d",&t);
103     while(t--){
104         scanf("%d",&n);
105         n=(1<<n);
106         for(int i=1;i<=n;i++)
107             scanf("%d",&a[i]);
108         build(1,1,n);
109         scanf("%d",&q);
110         while(q--){
111             int op,x,y;
112             scanf("%d%d%d",&op,&x,&y);
113             if(op==1){
114                 x++; y++;
115                 ans.mi=ans.mizhen=inf;
116                 ans.ma=ans.mafu=-inf;
117                 query(1,1,n,x,y);
118                 solve();
119             }
120             else{
121                 x++;
122                 update(1,1,n,x,y);
123             }
124         }
125     }
126 
127     return 0;
128 }
Psong

J Typist's Problem

unsolved

posted on 2017-09-27 18:02  Psong  阅读(250)  评论(0)  编辑  收藏  举报

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