1003 Emergency (25分)（一个字短！）

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本文介绍了一个基于Dijkstra算法的紧急救援路线规划问题。在给定的城市地图上，算法需找到从当前位置到目标城市的所有最短路径，并计算这些路径上可集结的最大救援队伍数量。通过使用priority_queue实现Dijkstra算法，文章提供了具体的代码实现。

1003 Emergency (25分)

题目

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

题意

求无向图中两点的最短路径长度，并记录沿路最大的救援队总和

分析

Dijkstra算法，用priority_queue实现

代码

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int n, m, c1, c2, a, b, c, distTo[510], team[510], maxteam[510], roads[510], marked[510];
vector<pair<int,int> > edge[510];
struct node {
 	int v, d;
 	node(int vv, int dd) :v(vv), d(dd) {}
 	bool operator < (node either) const {
  	return d > either.d;
 	}
};
int main() {
 	fill(distTo, distTo + 510, 99999999);
 	fill(marked, marked + 510, 0);
 	scanf("%d %d %d %d", &n, &m, &c1, &c2);
 	for (int i = 0; i < n; ++i)scanf("%d", &team[i]);
 	for (int i = 0; i < m; ++i) {
  		scanf("%d %d %d", &a, &b, &c);
  		edge[a].push_back(pair<int, int>(b, c));
  		edge[b].push_back(pair<int, int>(a, c));
 	}
 	distTo[c1] = 0;
 	roads[c1] = 1;
 	maxteam[c1] = team[c1];
 	priority_queue<node> q;
 	q.push(node(c1, 0));
 	while (!q.empty()) {
  		int f = q.top().v;
  		if (f == c2)break;
  		if (marked[f])continue;
  		marked[f] = 1;
  		q.pop();
  		for (int i = 0; i < edge[f].size(); ++i) {
   			if (marked[edge[f][i].first])continue;
   			int v = edge[f][i].first, d = edge[f][i].second;
   			if (distTo[v] > distTo[f] + d) {
    				distTo[v] = distTo[f] + d;
    				maxteam[v] = maxteam[f] + team[v];
    				roads[v] = roads[f];
    				q.push(node(v, distTo[v]));
   			}
   			else if (distTo[v] == distTo[f] + d) {
    				roads[v] += roads[f];
    				maxteam[v] = max(maxteam[v], maxteam[f] + d);
   			}
  		}
 	}
 	printf("%d %d", roads[c2], maxteam[c2]);
 	return 0;
}