标签:poj2139
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3131 | Accepted: 1455 |
Description
Input
Output
Sample Input
4 2 3 1 2 3 2 3 4
Sample Output
100
Hint
Source
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
#define maxn 302
#define inf 0x3f3f3f3f
int G[maxn][maxn], N, M;
int sta[maxn];
void Floyd(){
int i, j, k;
for(k = 1; k <= N; ++k)
for(i = 1; i <= N; ++i)
for(j = 1; j <= N; ++j)
if(G[i][k] + G[k][j] < G[i][j])
G[i][j] = G[i][k] + G[k][j];
}
int main() {
// freopen("stdin.txt", "r", stdin);
int i, j, num, ans = inf;
memset(G, 0x3f, sizeof(G));
scanf("%d%d", &N, &M);
for(i = 1; i <= N; ++i)
G[i][0] = G[i][i] = 0;
while(M--) {
scanf("%d", &num);
for(i = 0; i < num; ++i) {
scanf("%d", sta + i);
for(j = 0; j < i; ++j)
G[sta[i]][sta[j]] = G[sta[j]][sta[i]] = 1;
}
}
Floyd();
for(i = 1; i <= N; ++i) {
for(j = 1; j <= N; ++j)
G[i][0] += G[i][j];
if(ans > G[i][0]) ans = G[i][0];
}
printf("%d\n", ans * 100 / (N - 1));
return 0;
}POJ2139 Six Degrees of Cowvin Bacon 【Floyd】
标签:poj2139
原文地址:http://blog.csdn.net/chang_mu/article/details/41677247