zoj 1655 Transport Goods

Transport Goods

Time Limit: 2 Seconds      Memory Limit: 65536 KB



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


求最长路径

#include <iostream>

#include <stdio.h>

#include <queue>

#include <stack>

#include <set>

#include <vector>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

#define N 102

int value[N];

double map[N][N];

int n;

bool b[N];

void dijskla()

{

    memset(b,0,sizeof(b));

    b[n]=1;

    int t=n;

    int i,k;

    double Max;

    while(--t)

    {

        Max=-1.0;

        for(i=1;i<n;i++)

        if(!b[i]&&Max<map[n][i])

         {

             Max=map[n][i];

             k=i;

         }

         if(Max==-1.0) break;//有些点不可达

         b[k]=1;

        for(i=1;i<n;i++)

        if(!b[i]&&map[n][i]<map[n][k]*map[k][i])

         {

            map[n][i]=map[n][k]*map[k][i];

         }

    }

}

int main()

{

    int m;

    int i,j;

    double r;

    while(scanf("%d %d",&n,&m)==2)

    {

        for(i=1;i<=n;i++)

         for(j=1;j<=n;j++)

          map[i][j]=0;

        for(i=1;i<n;i++)

         scanf("%d",&value[i]);

        while(m--)

        {

            scanf("%d %d %lf",&i,&j,&r);

            if(map[i][j]<1.0-r) map[i][j]=map[j][i]=1.0-r;//有重边呀、、、郁闷

        }

        dijskla();

        double ans=0;

        for(i=1;i<n;i++)

         ans+=map[n][i]*value[i];

        printf("%.2lf\n",ans);

    }

    return 0;

}