PAT 1003 Emergency

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本文探讨了作为城市应急救援团队领导者时，在面对紧急情况时如何迅速集结资源并进行有效救援。通过应用Dijkstra算法解决最短路径问题，确保救援队伍能够尽快到达指定地点，同时最大化沿途集结的救援人员数量。

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

Sample Output

2 4

解题思路：

简单的最短路算法题目，应用Dijkstra算法即可解答。

#include<stdio.h>

int map[510][510];//邻接矩阵

int max;//到目的地的最大救援人员数

#define INF 0x7fffffff//表示两点之间无路
int Dijkstra(int n,int c1,int c2,int rescue[])
{
   int v[510],index,dist[510],i,min,j,sum[510],temp,num[510];
   for(i=0;i<n;i++)//初始化距离，访问标记，各点到原点的最短路径数，最大救援人员数
   {
        v[i]=0;
       dist[i]=INF;
       sum[i]=0;
       num[i]=0;
   }

dist[c1]=0;num[c1]=rescue[c1];sum[c1]=1;max=0;

   for(i=0;i<n;i++)
   {
       min=INF;
       for(j=0;j<n;j++)
           if(!v[j]&&dist[j]<min)
           {
               min=dist[j];
               index=j;
           }
       if(v[index]==1)break;
        v[index]=1;
       for(j=0;j<n;j++)
       {
           if(map[index][j]==INF)
               continue;
           temp=dist[index]+map[index][j];

           if(temp<dist[j])
           {
               dist[j]=temp;
               num[j]=num[index]+rescue[j];
               sum[j]=sum[index];
           }
           else if(temp==dist[j])
           {
               sum[j]+=sum[index];
               if(num[j]<num[index]+rescue[j])
                   num[j]=num[index]+rescue[j];
           }
       }
   }
   max=num[c2];
   return sum[c2];
}
int main()
{
   int n,m,c1,c2,i,j,k,rescue[510]={0},cnt=0;
   for(i=0;i<510;i++)
       for(j=0;j<510;j++)
           map[i][j]=INF;
   scanf("%d%d%d%d",&n,&m,&c1,&c2);
   for(i=0;i<n;i++){
       scanf("%d",&rescue[i]);
   }
   for(i=0;i<m;i++)
   {
       scanf("%d%d",&j,&k);
       scanf("%d",&map[j][k]);
       map[k][j]=map[j][k];
   }
   cnt=Dijkstra(n,c1,c2,max,rescue);
   printf("%d %d\n",cnt,max);
   return 0;
}