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poj 1975 Median Weight Bead

2015-07-02 22:28 423 查看
需要floyd实现闭环计算,然后统计每个元素大于其他元素的次数,如次数大于(N / 2) + 1, 那么说明该元素一定小于中间值。

#include<iostream>
#include<string.h>
#include<math.h>
#include<fstream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<vector>
#define MAXSIZE 100
using namespace std;
const int INF = 1<<29;
int t = 0, N = 0, M = 0;
int dp[110][110];
int ans = 0;
void floyd_dp()
{
for (int k = 0; k < N; k++)
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
dp[i][j] = min (dp[i][j], dp[i][k] + dp[k][j]);
}
}
}
}
int main()
{
//freopen("data_1975.txt","r",stdin);
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &N, &M);
ans = 0;
memset(dp, 0, sizeof(dp));
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
dp[i][j] = INF;
}
}
for (int i = 0; i < M; i++)
{
int a = 0, b = 0;
scanf("%d%d", &a, &b);
a--;
b--;
dp[a][b] = 1;
}
floyd_dp();
int judge = (N / 2) + 1;
int a = 0, b = 0;
for (int i = 0; i < N; i++)
{
a = 0;
b = 0;
for (int j = 0; j < N; j++)
{
if (dp[i][j] < INF)
{
a++;
}else if (dp[j][i] < INF)
{
b++;
}

}
if (a >= judge || b >= judge)
{
ans++;
}
}
printf("%d\n", ans);
}

return 0;
}
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