POJ - 3660 Cow Contest(flod)

题意：有N头牛，M个关系，每个关系A B表示编号为A的牛比编号为B的牛强，问若想将N头牛按能力排名，有多少头牛的名次是确定的。

分析：

1、a[u][v]=1表示牛u比牛v强，flod扫一遍，可以将所有牛的大小关系都存入a。

2、对于每一头牛，cntfront表示比它强的牛的个数，cntrear表示比它弱的牛的个数，若两者相加等于N-1，那该牛的名次自然可以确定。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){if(fabs(a - b) < eps) return 0;return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN][MAXN];
int ans;
int N, M;
int solve(){for(int i = 1; i <= N; ++i){int cntfront = 0, cntrear = 0;for(int j = 1; j <= N; ++j){if(a[j][i]) ++cntfront;if(a[i][j]) ++cntrear;}if(cntfront + cntrear == N - 1) ++ans;}return ans;
}
int main(){scanf("%d%d", &N, &M);for(int i = 0; i < M; ++i){int u, v;scanf("%d%d", &u, &v);a[u][v] = 1;}for(int k = 1; k <= N; ++k){for(int i = 1; i <= N; ++i){for(int j = 1; j <= N; ++j){if(a[i][k] && a[k][j]) a[i][j] = 1;}}}printf("%d\n", solve());return 0;
}

转载于:https://www.cnblogs.com/tyty-Somnuspoppy/p/6480016.html

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