POJ 1975 Median Weight Bead floyd求传递闭包 || bfs

题目：

http://poj.org/problem?id=1975

题意：

有n个珠子，n是奇数，每个珠子都有一个重量，有m个一对珠子之间的大小关系，问通过当前已知信息，有多少珠子一定不是重量为中位数的珠子

思路

对于一个珠子，只需要求出重量大于它的珠子数量和重量小于它的珠子数量，两者中有任何一个大于等于(n+1)/2，那么这个珠子肯定不是重量为中位数的珠子。用floyd求传递闭包后可以统计这个数量，或者直接两次bfs。做了这些题发现，能用floyd求传递闭包求解的，都能用bfs求出来
floyd:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100 + 10, INF = 0x3f3f3f3f;

bool mp[N][N];

void floyd(int n)
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            if(mp[i][k])
                for(int j = 1; j <= n; j++)
                    mp[i][j] = mp[i][j] || (mp[i][k] && mp[k][j]);
}

int main()
{
    int t, n, m;
    scanf("%d", &t);
    while(t--)
    {
        memset(mp, 0, sizeof mp);
        scanf("%d%d", &n, &m);
        int v, u;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &v, &u);
            mp[v][u] = true;
        }
        floyd(n);
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            int t1 = 0, t2 = 0;
            for(int j = 1; j <= n; j++)
            {
                if(mp[i][j]) t1++;
                if(mp[j][i]) t2++;
            }
            if(t1 >= (n+1)/2 || t2 >= (n+1)/2) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}

bfs:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 100 + 10, INF = 0x3f3f3f3f;

struct edge
{
    int to, next;
}g[N*N*2];

int cnt, head[N];
int a[N*N], b[N*N];
bool vis[N];

void add_edge(int v, int u)
{
    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
int bfs(int s)
{
    queue<int> que;
    memset(vis, 0, sizeof vis);
    que.push(s), vis[s] = true;
    int tot = 0;
    while(! que.empty())
    {
        int v = que.front(); que.pop();
        for(int i = head[v]; i != -1; i = g[i].next)
        {
            int u = g[i].to;
            if(! vis[u]) que.push(u), vis[u] = true, tot++;
        }
    }
    return tot;
}
int main()
{
    int t, n, m;
    scanf("%d", &t);
    while(t--)
    {
        cnt = 0;
        memset(head, -1, sizeof head);
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &a[i], &b[i]);
            add_edge(a[i], b[i]);
        }
        int t1[N], t2[N];
        for(int i = 1; i <= n; i++) t1[i] = bfs(i);
        cnt = 0;
        memset(head, -1, sizeof head);
        for(int i = 1; i <= m; i++) add_edge(b[i], a[i]);
        for(int i = 1; i <= n; i++) t2[i] = bfs(i);
        int ans = 0;
        for(int i = 1; i <= n; i++)
            if(t1[i] >= (n+1)/2 || t2[i] >= (n+1)/2) ans++;
        printf("%d\n", ans);
    }
    return 0;
}