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;}