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

;

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

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
, t2
;
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;
}