hdu 4650 Minimum Average Weight Path（最短路，5级）

全源最短路径平均边权问题

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本文探讨了一个图论问题，即寻找图中所有顶点对之间的路径，使得这些路径的平均边权最小。文章提供了一种解决方法，并通过代码示例详细解释了实现过程。

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Minimum Average Weight Path

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 94 Accepted Submission(s): 46

Problem Description

In mathematics, a graph is a representation of a set of objects where some pairs of the objects are connected by links. The interconnected objects are represented by mathematical abstractions called vertices, and the links that connect some pairs of vertices are called edges. A path in a graph is a sequence of vertices, and for any 2 adjacent u, v, there is a edge from u to v in graph. A path contains at least one edge. In the graph in Sample 2, {3, 3, 2, 2} can form a path from 3 to 2.

One of the common problem is to find the shortest path between two certain vertices, or all of them. They've been well studied as the single source shortest path problem (SSSP) and the all pairs shortest paths problem (APSP).

In this problem, we'll provide you a derivation analogous to APSP. You've been given a directed graph with positive or negative edge weights. We define the average weight of a path, as the sum of the edge weights divide the edges number of path. Now you need to find the minimum average weight between all pairs of vertices (APMAWP).

Input

Muiltcases. The first line contains two integer n, m, (1 ≤ n ≤ 10², 1 ≤ m ≤ 10⁴ ) the number of the vertices and the number of the edges.

The next m lines, each line contains three intergers u, v, w, representing a directed edge from u to v with weight w. (|w| ≤ 10³)

There is no multi-edge. It can contain self-loops.

Output

A n × n matrix representing the APMAWP. The j's element of the i's row represents the minimum average weight of all the paths from vertex i to vertex j. If no such path exists, you need to output "NO" instead (DO NOT output quote please). For each real number, you need to keep exactly 3 digits after digit point.

Sample Input

Sample Output

NO NO -8.000 NO
2.000 NO -3.000 -6.000
NO NO NO NO
NO NO 1.000 NO
NO NO NO NO NO
-990.000 -990.000 NO NO NO
-990.000 -990.000 735.000 NO -156.000
-990.000 -990.000 NO -18.000 NO
NO NO NO NO -156.000

Source

2013 Multi-University Training Contest 5

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zhuyuanchen52

思路：全源最短路，求平均边权最小，这个还是比较简单的，当时怎么没去看这题呢？

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=103;
const int oo=0x3f3f3f;
int g[mm][mm],fg[mm][mm],t[mm][mm];
int vis[mm][mm];//过几条边相见
double ans[mm][mm];
int n,m;
void acopy(int(*a)[mm],int(*b)[mm])
{
  FOR(i,1,n)FOR(j,1,n)a[i][j]=b[i][j];
}
void mul(int(*a)[mm],int(*b)[mm],int lol)
{ acopy(t,a);FOR(i,1,n)FOR(j,1,n)a[i][j]=oo;
  FOR(i,1,n)FOR(j,1,n)FOR(k,1,n)
  {
    if(vis[i][k]==-1||lol<vis[i][k])continue;
    if(vis[k][j]==-1||lol<vis[k][j])continue;
    a[i][j]=min(a[i][j],t[i][k]+b[k][j]);
  }
}
void debug(int(*a)[mm])
{ puts("+++++++++++++++++++++++");
  FOR(i,1,n)FOR(j,1,n)
  {
    printf("%d ",a[i][j]);
    if(j==n)puts("");
  }
  puts("+++++++++++++++++++++++++");
}
int main()
{ int a,b,c;
  while(~scanf("%d%d",&n,&m))
  {
    FOR(i,1,n)FOR(j,1,n)ans[i][j]=oo;
    clr(vis,-1);clr(g,0x3f);
    FOR(i,1,m)
    {
      scanf("%d%d%d",&a,&b,&c);
      g[a][b]=c;vis[a][b]=1;
    }
    FOR(k,1,n)FOR(i,1,n)if(vis[i][k]!=-1)
    FOR(j,1,n)if(vis[k][j]!=-1)
    {
      vis[i][j]=vis[i][j]==-1?vis[i][k]+vis[k][j]:min(vis[i][j],vis[i][k]+vis[k][j]);
    }///求出最少边相连
    acopy(fg,g);
    FOR(k,1,n){FOR(i,1,n)FOR(j,1,n)
    if(vis[i][j]!=-1&&k>=vis[i][j])///边数需要大于等相遇最小边数
     ans[i][j]=min(ans[i][j],(double)fg[i][j]/k);
     mul(fg,g,k);//debug(fg);
    }///K连边
    FOR(k,1,n)FOR(i,1,n)if(vis[i][k]!=-1)FOR(j,1,n)
    if(vis[k][j]!=-1&&vis[k][k]!=-1)
    {
      ans[i][j]=min(ans[i][j],ans[k][k]);
    }
    FOR(i,1,n)FOR(j,1,n)
    {
      if(j>1)putchar(' ');
      if(vis[i][j]==-1)printf("NO");
      else //cout<<ans[i][j];
        printf("%.3lf",ans[i][j]);
      if(j==n)printf("\n");
    }
  }
  return 0;
}