PAT 1003 Emergency 最短路

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 diferent 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

题目意思：有n个城市m条路相连，每个城市都有一些救援队，给定起点城市和C1和终点城市C2，求从起点到终点最短路径的条数以及最短路径上能够调动来的救援队的数目。

解题思路：和一般求最短路的题目不同，这里所求的是最短路的数目和最短路上点的权值和（将城市看做点，那么每个城市中救援队的数目就可以看成点的权值）。其实既然是最短路，无非就是那几个算法，这里我们来考虑最短路数目不唯一的原因，从起点到终点的边的权值和是一样的，但经过的点是不一样的，这可以用Dijkstra来对每一个点逐步贪心，判断是否需要纳入到最短路结点的集合中。这里用dis[i]表示从起点C1到i点最短路的路径长度，用num[i]表示从起点到i点最短路的个数，用w[i]表示从起点到i点救援队的数目之和。当判定dis[u] + e[u][v] < dis[v]的时候，也就是说新纳入的u点可以减短到v的路径长度，是符合要求的，不仅仅要更新dis[v]，还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] ==dis[v]，也就是说新纳入的u点，经过u到达v的路径长度和不经过u到v的路径长度一致，这就产生了两条最短路了，这时候要更新num[v] += num[u]，⽽且判断⼀下是否权重w[v]更⼩，如果更⼩了就更新w[v] = weight[v] + w[u];

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
const int inf= 99999999;
using namespace std;
int n,m,c1,c2;
int e[510][510];
int w[510];//救援队数目之和
int dis[510];//从起点到i点最短路径的长度
int num[510];//最短路径的条数
int weight[510];//第i个城市救援队数目，点权
bool vis[510];
int main()
{
    int a,b,c,i,j,mins,u,v;
    scanf("%d%d%d%d",&n,&m,&c1,&c2);
    for(i=0; i<n; i++)
    {
        scanf("%d",&weight[i]);
    }
    fill(e[0], e[0] + 510 * 510, inf);
    fill(dis, dis + 510, inf);
    for(i=0; i<m; i++)//建图
    {
        scanf("%d%d%d",&a,&b,&c);
        e[a][b]=e[b][a]=c;
    }
    dis[c1]=0;
    w[c1]=weight[c1];
    num[c1]=1;//至少会有一条最短路
    for(i=0; i<n; i++)//遍历所有的点
    {
        u=-1;//因为是正权通路，这里设置为-1
        mins=inf;
        for(j=0; j<n; j++)//找到距离已纳入点集合中的最近点
        {
            if(vis[j]==false&&dis[j]<mins)
            {
                u=j;
                mins=dis[j];
            }
            //printf("%d\n",mins);
        }

        if(u==-1)
        {
            break;
        }
        vis[u]=true;//标记该点以纳入S集
        for(v=0; v<n; v++)//更新刚纳入的u点与其他尚未纳入点之间的距离
        {
            if(vis[v]==false&&e[u][v]!=inf)
            {
                if(dis[u]+e[u][v]<dis[v])//Dijkstra
                {
                    dis[v]=dis[u]+e[u][v];//更新边权
                    num[v]=num[u];//最短路条数不会变
                    w[v]=w[u]+weight[v];//最短路沿路的救援队数
                }
                else if(dis[u]+e[u][v]==dis[v])//出现边权相等
                {
                    num[v]=num[v]+num[u];//产生了两条分路
                    if(w[u]+weight[v]>w[v])//取点权最大的表示为到v的救援队数目
                    {
                        w[v]=w[u]+weight[v];
                    }
                }
            }
        }
    }
    printf("%d %d",num[c2],w[c2]);
    return 0;
}