ZOJ 1655 Transport Goods

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The HERO country is attacked by other country. The intruder is attacking the capital so other cities must send supports to the capital. There are some roads between the cities and the goods must be transported along the roads.

According the length of a road and the weight of the goods, there will be some cost during transporting. The cost rate of each road is the ratio of the cost to the weight of the goods transporting on the road. It is guaranteed that the cost rate is less than 1.

On the other hand, every city must wait till all the goods arrive, and then transport the arriving goods together with its own goods to the next city. One city can only transport the goods to one city.

Your task is find the maximum weight of goods which can arrive at the capital.

Input

There are several cases.

For each case, there are two integers N (2 <= N <= 100) and M in the first line, where N is the number of cities including the capital (the capital is marked by N, and the other cities are marked from 1 to N-1), and M is the number of roads.

Then N-1 lines follow. The i-th (1 <= i <= N - 1) line contains a positive integer (<= 5000) which represents the weight of goods which the i-th city will transport to the capital.

The following M lines represent M roads. There are three numbers A, B, and C in each line which represent that there is a road between city A and city B, and the cost rate of this road is C.

Process to the end of the file.

Output

For each case, output in one line the maximum weight which can be transported to the capital, accurate up to 2 demical places.

Sample Input

5 6
10
10
10
10
1 3 0
1 4 0
2 3 0
2 4 0
3 5 0
4 5 0

Sample Output

40.00

额，最开始不怎么懂，想不好者类问题=。=

后来看了别人的，敲了个dijkstra，。。。

dis存的是最大的运输率，所以求最大的路径了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int eps=1e-6;
const int maxn=110;
double dis[maxn];
double cost[maxn];
int visit[maxn];
double mp[maxn][maxn];
int n,m;
void dijkstra()
{
    int pos;
    memset(visit,0,sizeof(visit));
    for(int i=1;i<=n;i++)
        dis[i]=mp[n][i];
    dis[n]=1;
    visit[n]=1;//起点设为最大
    for(int i=1;i<=n;i++)
    {
        pos=-1;
        for(int j=1;j<=n;j++)
        {
            if(!visit[j]&&(pos==-1||dis[pos]<dis[j]-eps))//求最大的pos
                pos=j;
        }
        visit[pos]=1;
        for(int j=1;j<=n;j++)
        {
            if(!visit[j]&&dis[j]<dis[pos]*mp[pos][j]-eps)//更新最大路径
                dis[j]=dis[pos]*mp[pos][j];
        }
    }
}
int main()
{
    int u,v;
    double w;
    while(~scanf("%d%d",&n,&m))
    {
         for(int i=1;i<n;i++)
            scanf("%lf",&cost[i]);
         memset(mp,0,sizeof(mp));
         for(int i=1;i<=m;i++)
         {
             scanf("%d%d%lf",&u,&v,&w);
             if(mp[u][v]<1.0-w)
                mp[u][v]=mp[v][u]=1.0-w;//拿1.0减去它求最大路径
         }
         dijkstra();
         double sum=0.0;
         for(int i=1;i<n;i++)
             sum+=cost[i]*dis[i];
         printf("%.2f\n",sum);
    }
    return 0;
}

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ZOJ 1655 Transport Goods

标签：style   os   strong   io   for   问题   ar   line   

原文地址：http://blog.csdn.net/u013582254/article/details/38358685