「NOI2007」 社交网络 - 最短路

题目描述

输入格式

输出格式

样例输入

4 4
1 2 1
2 3 1
3 4 1
4 1 1

样例输出

1.000
1.000
1.000
1.000

分析

没什么好说的,floyd暴力就行了,再注意一下范围.

代码

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#define inf 0x7fffffff/2
using namespace std;
long long n,m,d[105][105],g[105][105];
long double ans[105];
int main() {
	scanf("%lld%lld",&n,&m);
	for (int i = 1;i <= n;i++)
		for (int j = 1;j <= n;j++)
			d[i][j]=(i==j?0:inf);
	for (int i = 1;i <= m;i++) {
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		d[a][b]=d[b][a]=c;
	}
	for (int i = 1;i <= n;i++)
		for (int j = 1;j <= n;j++)
			if (d[i][j]!=inf&&d[i][j]!=0) g[i][j]=1;
	for (int k = 1;k <= n;k++)
		for (int i = 1;i <= n;i++)
			for (int j = 1;j <= n;j++) {
				if (d[i][j]==d[i][k]+d[k][j]) g[i][j]+=g[i][k]*g[k][j];
				if (d[i][j]>d[i][k]+d[k][j]) {
					d[i][j]=d[i][k]+d[k][j];
					g[i][j]=g[i][k]*g[k][j];
				}
			}
	for (int k = 1;k <= n;k++)
		for (int i = 1;i <= n;i++) {
			if (i==k) continue;
			for (int j = 1;j <= n;j++) {
				if (j==k) continue;
				if (d[i][k]+d[k][j]==d[i][j]) {
					ans[k]+=((long double)(g[i][k])*g[k][j])/g[i][j];
				}
			}
		}
	for (int i = 1;i <= n;i++)
		printf("%.3Lf\n",ans[i]);
    return 0;
}