1003. Emergency(Dijkstra算法)

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1003 Emergency (25 分)

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, C​1​​ and C​2​​ - 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 c​1​​, c​2​​ 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 C​1​​ to C​2​​.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C​1​​ and C​2​​, 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

此题Dijkstra算法中

救援人数 可以看成边的第二权值，

求出最短边权

求最短边权的数量

筛选出最短边权上最大的第二权值

dis数组规定点到指定点的最短距离

定义num数组存放最短的边数量，

定义w数组存放第二边权

核心代码：

当找到相对更小边的情况：

if(dis[flag]+edge[flag][i]

更新

dis数组，num数组，w数组

dis[i]=dis[flag]+edge[flag][i];
	num[i]=num[flag];
	w[i]=w[flag]+weight[i];

当找到相等的最小边

dis[flag]+edge[flag][i]==dis[i]

更新

dis数组，num数组，w数组

	num[i]+=num[flag];
	if(w[i]

AC边权为4 ABC边权和为4 A->C存在两条最短路径 因此从C->D也存在两条最短路径

   

代码实现：

//N (<= 500) maxsize开空间为520 
#include
#include
using namespace std;
int n,m,c1,c2;
const int inf=0x3f3f3f;
const int maxsize = 520;
int edge[maxsize][maxsize],dis[maxsize],weight[maxsize],num[maxsize],w[maxsize];
bool visit[maxsize];
int dijkstra(int c1){       //全处理的顶点
	
	dis[c1]=0;
	num[c1]=1;
	w[c1]=weight[c1];
	for(int i=0;i