POJ 2594 Treasure Exploration （最小路径覆盖+传递闭包（解决可重点））

题目地址：点击打开链接

题意：有n个地点，现在给出m条单向道路，问最少放多少人可以覆盖所有点，一个地点可以重复经过。

思路：这题和POJ1422（点击打开）很像，但有个关键的差别，1422每个点只能经过一次，而这题可以重复经过。

如果只能经过一次，那么直接匈牙利就行，但如果可以经过多次，就需要先利用floyd求解下传递闭包加一些

新边再来求解。

#include
#include
#include
#include
using namespace std;
const int maxn = 505;
vector g[maxn];
int n, m, match[maxn];
int link[maxn][maxn];
bool vis[maxn];

void floyd()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            if(link[i][k])
                for(int j = 1; j <= n; j++)
                    if(link[k][j])
                        link[i][j] = 1;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(link[i][j])
                g[i].push_back(j);
}

bool dfs(int x)
{
    for(int i = 0; i < g[x].size(); i++)
    {
        int v = g[x][i];
        if(!vis[v])
        {
            vis[v] = 1;
            if(match[v] == -1 || dfs(match[v]))
            {
                match[v] = x;
                return 1;
            }
        }
    }
    return 0;
}

int Hungary()
{
    int cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        memset(vis, 0, sizeof(vis));
        cnt += dfs(i);
    }
    return cnt;
}

int main(void)
{
    while(cin >> n >> m, n+m)
    {
        memset(link, 0, sizeof(link));
        memset(match, -1, sizeof(match));
        for(int i = 0; i < maxn; i++)
            g[i].clear();
        for(int i = 1; i <= m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            link[u][v] = 1;
        }
        floyd();
        printf("%d\n", n-Hungary());
    }
    return 0;
}

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10967		Accepted: 6105

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

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2