1003. Emergency (25)

原题连接：https://www.patest.cn/contests/pat-a-practise/1003

题目如下：

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

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

Sample Output

2 4

这道题一开始我并没有想到用Dijkra,只是想到了利用递归，但是有好几个case都过不去。代码如下：

#include<stdio.h>
#define MaxCity 501

int MostTeams,ShortestL,Teams,Length,Amount;
int G[MaxCity]={0};
int R[MaxCity][MaxCity];
int Already[MaxCity][MaxCity]={0};
int N,M,C1,C2;

void DFS(int Last,int In)
{
    int i;
    Teams+=G[Last];
    Length+=R[Last][In];
    Already[Last][In]=Already[In][Last]=1;

    if (In==C2)
    {
        Teams+=G[In];
        if(Length<ShortestL)
        {
            Amount=1;
            ShortestL=Length;
            MostTeams=Teams;
        }
        else if (Length==ShortestL)
        {
            Amount++;
            if (MostTeams<Teams)MostTeams=Teams;
        }
        else if (Length>ShortestL)
        {
            Length-=R[Last][In];
            return ;
        }
    }

    for (i=0;i<N;i++)
    {
        if (i!=C1 && R[In][i]!=-1 && (Already[In][i]==0 || Already[i][In]==0 ))DFS(In,i);
    }

   // return ;
}

int main()
{
    int i,j,v,w,len;
    MostTeams=Teams=Length=Amount=0;
    ShortestL=0x7FFFFF;
    for (i=0;i<MaxCity;i++)
    {
        for (j=0;j<MaxCity;j++)R[i][j]=R[j][i]=-1;
    }
    scanf("%d %d %d %d",&N,&M,&C1,&C2);
    for (i=0;i<N;i++)
    {
        scanf("%d",&G[i]);
    }
    for (i=0;i<M;i++)
    {
        scanf("%d %d %d",&v, &w, &len);
        R[v][w]=R[w][v]=len;
    }

    if (C1==C2){printf("%d %d",1,G[C1]);}
    else {
    for (i=0;i<N;i++)
    {
        for (v=0;v<N;v++)
        {
            for (w=0;w<N;w++)Already[v][w]=Already[w][v]=0;
        }
        Teams=0;Length=0;
        if (R[C1][i]!=-1)DFS(C1,i);
    }

    printf("%d %d",Amount,MostTeams);
    }

    return 0;
}

看了其他人的解法，发现核心还是最短路径，不管有没有使用递归，于是，用了其他人的样例，终于知道在递归的条件判断中，点是否在最短路径上这一条件及其重要！

几个备用的测试样例：

4 5 0 3                                    
1 4 1 2
0 1 1
0 2 2
0 3 4
1 2 1
2 3 2  

——————

6 9 0 2

1 2 1 5 3 4

0 1 1

0 2 2

0 3 1

1 2 1

2 3 1

2 4 1

2 5 1

3 4 1

4 5 1

AC的解法：

#include<stdio.h>
#define INFINITY 65535
#define MaxCity 503
int N,M,C1,C2;
int Team[MaxCity];
int G[MaxCity][MaxCity];
int collect[MaxCity]={0};
int dist[MaxCity];
int NumberR=0,NumberP=0;

void init();
int FindMin();
void Dijkstra();
void DFS(int,int );

int main()
{
    scanf("%d %d %d %d",&N,&M,&C1,&C2);
    int i;
    for (i=0;i<N;i++)
    {
        scanf("%d",&Team[i]);
    }

    init();
    Dijkstra();

    for (i=0;i<N;i++)collect[i]=0;
    collect[C1]=1;

    DFS(C1,Team[C1]);
    printf("%d %d",NumberR,NumberP);
    return 0;
}

void init()
{
    int i,j,v,w,len;
    for (i=0;i<N;i++)
    {
        for (j=0;j<N;j++)
        {
            if (i==j)G[i][j]=G[j][i]=0;
            else G[i][j]=G[j][i]=INFINITY;
        }
        dist[i]=INFINITY;
    }
    for (i=0;i<M;i++)
    {
        scanf("%d %d %d",&v,&w,&len);
        G[v][w]=G[w][v]=len;
    }
}

int FindMin()
{
    int MinV,v;
    int MinDist=INFINITY;
    for (v=0;v<N;v++)
    {
        if (!collect[v] && dist[v]<MinDist)
        {
            MinV=v;
            MinDist=dist[v];
        }
    }
    if (MinDist<INFINITY)
    {
        return MinV;
    }
    else return -1;
}

void Dijkstra()
{
    int v,i;

    collect[C1]=1;
    dist[C1]=0;

    for (v=0;v<N;v++)
    {
        if (!collect[v] && G[C1][v]<INFINITY)dist[v]=G[C1][v];
    }

    while(1)
    {
        v=FindMin();
        if (v==-1)break;
        collect[v]=1;
        for (i=0;i<N;i++)
        {
            if (!collect[i] && dist[i]>dist[v]+G[v][i])
            {
                dist[i]=dist[v]+G[v][i];
            }
        }
    }
}

void DFS(int src,int Numberp)
{
    if (src==C2)
    {
        if (Numberp>NumberP)NumberP=Numberp;
        NumberR++;
        return ;
    }

    int i;
    for (i=0;i<N;i++)
    {
        if (collect[i]==0 && (dist[i]==dist[src]+G[src][i]))
        {
            collect[i]=1;
            DFS(i,Numberp+Team[i]);
            collect[i]=0;
        }
    }
}