Cow Contest POJ - 3660 :http://poj.org/problem?id=3660
题意:
n头牛,有m对牛进行了比赛,现在告诉你每队牛比赛的结果,A胜B,问有几头牛的排名可以确定。
思路:
题目给出了m对的相对关系,求有多少个排名是确定的。
使用floyed求一下传递闭包。如果这个点和其余的关系都是确定的,那么这个点的排名就是确定的。
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000") //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull; typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行
#define REP(i , j , k) for(int i = j ; i < k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0); template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/
const int maxn = ;
int mp[maxn][maxn];
int main(){
int n,m;
scanf("%d%d", &n, &m);
for(int i=; i<=m; i++){
int u,v;
scanf("%d%d", &u, &v);
mp[u][v] = ;
}
for(int k=; k<=n; k++){
for(int i=; i<=n; i++){
for(int j=; j<=n; j++){
mp[i][j] = (mp[i][j] || (mp[i][k] && mp[k][j]));
}
}
}
int ans = ,cnt;
for(int i=; i<=n; i++){
cnt = ;
for(int j=; j<=n; j++){
if(mp[i][j] || mp[j][i])cnt++;
}
if(cnt>=n-)ans++;
}
printf("%d\n", ans);
return ;
}
POJ - 3660