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题意：给一个图，有n个城市，m条路，每个城市都有一定数目的紧急救援队伍，求从c1到c2的最短路有多少条，且从c1向c2赶过去的路上召集经过城市的救援队，求能够召集的救援队的最大数目为多少。
解法：用BFS求最短路，复杂度O（n*n），记录当前救援队的数目。将到达c2的最短路长度和救援队数目保存，最后排序求出答案即可，细节见代码。
代码：
#include<bits/stdc++.h>
using namespace std;
int n,m,c1,c2;
int a,b,d;
int team[500];
int road[500][500];
int shortest[500];
struct Node{
int num;
int d;
int team;
Node(int num = 0,int d = 0,int team = 0):num(num),d(d),team(team) {
}
bool operator < (const Node &r1) const {
return d > r1.d;
}
};
struct node {
int d;
int team;
node(int d = 0,int team = 0):d(d),team(team) {
}
bool operator < (const node &r) const {
if(d != r.d) return d > r.d;
return team > r.team;
}
};
vector<node> vn;
void BFS() {
vn.clear();
priority_queue<Node> Q;
Q.push(Node(c1,0,team[c1]));
shortest[c1] = 0;
while(!Q.empty()) {
Node now = Q.top();
Q.pop();
if(shortest[now.num] < now.d) continue;
if(now.num == c2) vn.push_back(node(now.d,now.team));
for(int i = 0;i < n;++i) {
if(i != now.num && now.d + road[now.num][i] <= shortest[i]) {
shortest[i] = now.d + road[now.num][i];
Q.push(Node(i,shortest[i],now.team + team[i]));
}
}
}
}
int main()
{
memset(shortest,0x3f,sizeof shortest);
memset(road,0x3f,sizeof road);
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,&d);
road[a][b] = road[b][a] = d;
}
BFS();
sort(vn.begin(),vn.end());
int sd = vn[0].d;
int num = 0;
for(int i = 0;i < vn.size();++i) {
if(sd == vn[i].d) num++;
else break;
}
printf("%d %d",num,vn[0].team);
return 0;
}