ZOJ Monthly, July 2011 【B 】Cake Robbery
2011-07-25 14:30
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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3511
由于能力有限,想不出好的办法,只能模拟,但想不到暴力竟然可以过~~
很简单,用vector做,v[i]代表第几个多边形,而v[i].size()代表点数也就是边数,记得每次输入对角线都要查询然后修改。
由于能力有限,想不出好的办法,只能模拟,但想不到暴力竟然可以过~~
很简单,用vector做,v[i]代表第几个多边形,而v[i].size()代表点数也就是边数,记得每次输入对角线都要查询然后修改。
#include<iostream> #include<vector> #include<map> #include<stack> #include<algorithm> #include<queue> #include<list> #include<set> #include<string.h> #include<stdlib.h> #include<math.h> #include<stdio.h> #include<ctype.h> #include<iomanip> using namespace std; #define LL long long #define pi acos(-1) #define N 10010 #define INF 9999999999 #define eps 1e-8 //**************************************** //zoj3511 //Copyright@leolin. All rights reserved. //**************************************** vector<int> v ; int c ; int main() {//freopen("a.txt","r",stdin); int n,m; int i,j,k; while(scanf("%d%d",&n,&m)!=EOF) { int x,y; for(i=0;i<=N;i++)v[i].clear(); for(i=1;i<=n;i++)v[0].push_back(i); k=1;//the number of convex polygon while(m--) { scanf("%d%d",&x,&y); if(x>y)swap(x,y); for(i=0;i<k;i++)//第几个多边形 { int cnt=0; int tag1=-1,tag2=-1; int cn=0; for(j=0;j<v[i].size();j++) { if(x==v[i][j]) tag1=j,cnt++; if(y==v[i][j]) tag2=j,cnt++; c[cn++]=v[i][j]; } if(cnt==2) { for(j=tag1;j<=tag2;j++) v[k].push_back(v[i][j]); v[i].clear(); for(j=0;j<=tag1;j++) v[i].push_back(c[j]); for(j=tag2;j<cn;j++) v[i].push_back(c[j]); k++; break; } } } int ans=0; for(i=0;i<k;i++) if(ans<v[i].size()) ans=v[i].size(); printf("%d\n",ans); } return 0; }
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