PAT 1003 Emergency [dijkstra] [最短路径条数]

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本文介绍了一种使用Dijkstra算法解决城市间最短路径问题的方法，通过算法找到从当前城市到目标城市的最短路径及沿途能集结的最大救援队伍数量。文章详细展示了算法的实现过程，包括初始化、寻找最短路径、更新路径数量和救援队伍数量。

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 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:

Sample Output:

2 4

-------------------------------------这是题目和解题的分割线-------------------------------------

Dijkstra算法求解最短路径问题。

#include<cstdio>
#include<algorithm>
 
const int inf = 0x3fffffff;
const int maxN = 540;

using namespace std;

//dis记录所有结点到达指定点的最短距离 
int G[maxN][maxN],vis[maxN] = {},dis[maxN];
//num记录所有结点到达指定点的路径条数
//teams记录所有结点到达指定点的营救队数量
//value记录所有结点拥有的营救队数量 
int num[maxN] = {},teams[maxN] = {},value[maxN];
int n,m,start,en;

void dijkstra(int st)
{
	//想要赋值0和-1以外的数，看准fill函数~ 
	fill(dis,dis+maxN,inf);
	dis[st] = 0;  
	num[st] = 1;
	teams[st] = value[st];
	//以上三个，起点的值都要进行初始化 
	int i,j;
	for(i=0;i<n;i++)
	{
		//在未访问过的点中，寻找距离指定点最近的 
		int minI = -1,minN = inf;
		for(j=0;j<n;j++)
		{
			if(!vis[j]&&dis[j]<minN)
			{
				minI = j;
				minN = dis[j];  
			}
		}
		vis[minI] = 1; //标记为已访问
		//如果新加入的结点让最短距离发生了改变 
		for(j=0;j<n;j++)
		{
			//1.没访问过 2.存在边 3.最短距离可更新 
			if(!vis[j]&&G[minI][j]!=inf&&dis[minI]+G[minI][j]<dis[j])
			{
				dis[j] = dis[minI]+G[minI][j];
				num[j] = num[minI]; //该结点的num等于前驱结点的num 
				teams[j] = teams[minI]+value[j];
			}
			//如果距离相等 
			else if(dis[minI]+G[minI][j]==dis[j])
			{
				num[j] += num[minI]; //条数增加 
				if(teams[minI]+value[j]>teams[j]) //如果营救队数量更多 
					teams[j] = teams[minI]+value[j]; //才更新 
			}
		}
	}
}

int main()
{
	int i;
	scanf("%d%d%d%d",&n,&m,&start,&en);
	fill(G[0],G[0]+maxN*maxN,inf); //二维数组的fill写法 
	for(i=0;i<n;i++)
		scanf("%d",&value[i]);
	for(i=0;i<m;i++)
	{
		int a,b,x;
		scanf("%d%d%d",&a,&b,&x);
		G[a][b] = x;
		G[b][a] = x; //无向边！！！ 
	}
	dijkstra(start); 
	printf("%d %d\n",num[en],teams[en]);
	return 0;
}