1003 Emergency

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该博客介绍了如何运用迪杰斯特拉算法计算从起点到终点的最短路径，并在此基础上通过深度优先搜索策略，找出能够集结最多救援队伍的方案。在给定的输入数据中，包括城市数量、道路数、起点和终点，以及各城市的救援队伍数量和道路长度。通过算法实现，得出不同最短路径的数量以及最大可能集结的救援队伍总数。

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 Specification:

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 Specification:

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

利用迪杰斯特拉算法计算出距离终点的最短路，再利用dfs找出答案。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,c1,c2;
const int N = 510;
int peo[510];int g[510][510];bool st[510];int dist[510];
int res=0;int maxa=0;
bool sign[N];

int dijkstra()
{
	memset(dist,0x3f,sizeof dist);
	dist[c1]=0;
	for(int i=0;i<n;i++)
	{
		int t=-1;
		for(int j=0;j<n;j++)
		 if(!st[j]&&(t==-1||dist[t]>dist[j]))
		 			t=j;
		for(int j=0;j<n;j++)
			dist[j]=min(dist[j],dist[t]+g[t][j]);
		
		st[t]=true;
	}
}
int dfs(int x,int step,int p)
{
	if(step>dist[c2])return 0;
	if(x==c2&&step==dist[c2])
	{
	res++;
	maxa=max(maxa,p);
	return 0;
	}
	for(int i=0;i<n;i++)
	{
		if(i==x)continue;
		if(g[x][i]&&!sign[i])
		{
			sign[i]=true;
			dfs(i,step+g[x][i],p+peo[i]);
			sign[i]=false;
		}
	}
}
int main()
{
	cin>>n>>m>>c1>>c2;
	for(int i=0;i<n;i++)
	{
		cin>>peo[i];
	}
	memset(g,0x3f,sizeof g);
	for(int i=0;i<m;i++)
	{
		int a,b,c;cin>>a>>b>>c;
		g[a][b]=g[b][a]=min(c,g[a][b]);
	}
	dijkstra();sign[c1]=true;
	dfs(c1,0,peo[c1]);cout<<res<<" "<<maxa;
	return 0;
}