POJ 3660 Cow Contest (Floyd求传递闭包)
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传送门:http://poj.org/problem?id=3660
题目大意:n(<=100)头奶牛,每两头之间有一个能力大小关系,表示能否战胜,战胜关系可以传递,问可以确定多少头牛的排名。
解题思路:关系可以传递,则可以借助Floyd求出两两之间的关系,然后对于每一头判断能否与之确定关系的牛数是否等于n-1(不包含自己),是的话,答案数加1。
Code:
/* W w w mm mm 222222222 7777777777777 *//* W w w w m m m m 222 22 7777 *//* w w w w m m m m 22 777 *//* w w w w m m m m 22 77 *//* w w w w m m m m 222 77 *//* w w w w m m m m 222 77 *//* w w w w m m m m 222 77 *//* w w w w m m m m 222 77 *//* w w w w m m m m 222 77 *//* ww ww m mm m 222222222222222 77 *///#pragma comment(linker, "/STACK:102400000,102400000")//C++//int size = 256 << 20; // 256MB//char *p = (char*)malloc(size) + size;//__asm__("movl %0, %%esp\n" :: "r"(p));//G++#include<set>#include<map>#include<queue>#include<stack>#include<ctime>#include<deque>#include<cmath>#include<vector>#include<string>#include<cctype>#include<cstdio>#include<cstdlib>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#define REP(i,s,t) for(int i=(s);i<=(t);i++)#define REP2(i,t,s) for(int i=(t);i>=s;i--)using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned long ul;const int MAXN=105;const int MAXM=4505;int n,m;bool d[MAXN][MAXN];int main(){ #ifdef ONLINE_JUDGE #else freopen("test.in","r",stdin); #endif while(~scanf("%d%d",&n,&m)) { memset(d,false,sizeof(d)); REP(i,1,m) { int u,v; scanf("%d%d",&u,&v); d[u][v]=true; } REP(k,1,n) { REP(i,1,n) { REP(j,1,n) { if(d[i][k]&&d[k][j]) { d[i][j]=true; } } } } int ans=0; REP(i,1,n) { int cnt=0; REP(j,1,n) { if(i==j) { continue; } if(d[i][j]) { cnt++; } if(d[j][i]) { cnt++; } } if(cnt==n-1) { ans++; } } printf("%d\n",ans); } return 0;}
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