POJ3660 Cow Contest (传递闭包)

最新推荐文章于 2022-03-20 11:58:43 发布

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#传递闭包

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本文介绍了一个竞赛场景下的算法问题，即通过一系列比赛结果确定牛的技能排名。利用邻接矩阵和Floyd算法来解决该问题，确保算法的有效性和准确性。

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9308		Accepted: 5240

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

Sample Output

Source

USACO 2008 January Silver

题意：告诉你总共有n头牛和m组战斗力强弱的关系，并且当两头牛的战斗力关系确定以后他们的胜负一定确定，问给定这些关系以后让他们打架，有多少头牛的排名是我们可以算出来的。

弄一个邻接矩阵,一组数据相当于一条边,然后我们用floyd求一下传递闭包,最后扫一遍有几头牛能打败他的和他能打败的是全部即可。

#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define foreach(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
#define cl(a) memset((a),0,sizeof(a))
#define sc(x) scanf("%d",&x)
#define dbg(x) cout << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1);
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
int gcd(int a,int b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★

const int maxn = 105;
int mp[maxn][maxn];
int n,m,ans;
void init(){
	cl(mp);
	fur(i,0,n)mp[i][i] = 1;
}

void froyd(){
	fur(k,1,n)
		fur(i,1,n)
			fur(j,1,n)
				if(mp[i][k]&&mp[k][j])
					mp[i][j] = 1;
}

void check(){
	fur(i,1,n){
		int tot = 0;
		fur(j,1,n) 
			if(mp[i][j]|mp[j][i])
				tot++;
		if(tot == n)
			ans++;
	}
}

int main(){
	//freopen("E:\\data.in","r",stdin);
	//freopen("E:\\data.out","w",stdout);
	while(scanf("%d %d",&n,&m) != EOF){
		int a,b;
		init();
		fur(i,1,m){
			gn(a);gn(b);
			mp[a][b] = 1;
		}
		froyd();
		ans = 0;
		check();
		printf("%d\n",ans);
	}	
	return 0;
}