POJ3660 - Cow Contest - 关系传递闭包（最短路变形）+思维

1.题目描述：

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11280		Accepted: 6262

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

2.题意概述：

有n只奶牛，有n个连续的实力，如果u的实力大于v的实力，就能打赢它，
然后给定m种关系，求最后能确定其排名的奶牛个数。

3.解题思路：

离散的知识——传递闭包：有向图的传递闭包就是指有向图所有传递关系中的最小关系

如果一头牛被x头牛打败，并且可以打败y头牛，如果x+y=n-1，则我们容易知道这头牛的排名就被确定了，所以我们只要将任一头牛，可以打败其他的牛的个数x，和能打败该牛的牛的个数y求出来，在遍历所有牛判断一下是否满足x+y=n-1，就知道这个牛的排名是否能确定了（而传递闭包，正好将所有能得出关系都求出来了），再将满足这个条件的牛数目加起来就是所求解。 x可以看成是入度， y是出度。

在floyd-warshall求每对顶点间的最短路径算法中，可以通过O(v^3)的方法求出图的传递闭包。可以位每条边赋以权值1,然后运行Floyd-Wareshall。如果从 i 到 j 存在一条路径，则mp(i,j)<N,否则mp(i,j)=MAX。

一种改进的算法是：由于我们需要的只是判断是否从i到j存在一条通路，所以在Floyd-Wareshall中的动态规划比较中，我们可以把min和+操作改为逻辑or( || )和逻辑（&&）。也就是将 mp[i][j] = min(mp[i][j], mp[i][k]+mp[k][j]); 改成 if(mp[i][j] == 1 || (mp[i][k] == 1 && mp[k][j] == 1)) mp[i][j] = 1;
设 mp(i,j) ＝ 1表示从 i 到 j 存在一条通路 p，且 p 的所有中间节点都在0,1,2,...,k中, 否则mp(i,j)=0。我们把边(i,j)加入到E*中当且仅当mp(i,j)=1。

4.AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <ctime>
#define INF 0x3f3f3f3f
#define maxn 100100
#define N 111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
bool mp[N][N];
void floyd(int n)
{for (int k = 1; k <= n; k++)for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (mp[i][j] || (mp[i][k] && mp[k][j]))mp[i][j] = 1;
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();
#endifint n, m;while (~scanf("%d%d", &n, &m)){memset(mp, 0, sizeof(mp));while (m--){int u, v;scanf("%d%d", &u, &v);mp[u][v] = 1;}floyd(n);int cnt = 0;for (int i = 1; i <= n; i++){int win = 0, lose = 0;for (int j = 1; j <= n; j++)if (i != j){if (mp[i][j])win++;if (mp[j][i])lose++;}if (win + lose == n - 1)cnt++;}printf("%d\n", cnt);}
#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);
#endifreturn 0;
}

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