题目链接：1488. 最短距离 - AcWing题库

输入样例：

7 7
1 2 5
1 4 3
2 3 2
2 5 1
3 6 7
5 6 8
6 7 6
3
7
5
4
7
1
2
3
4
5
6
7

输出样例：

3
1
3
0
0
6
0

题意：给一个有n个点和m条带权边的无向图，图中共有k个商店，问这些点去最近的商店的距离

思路：堆优化的迪杰斯特拉，用一个点连接所有商店，并且边权设为0，这样相当于每个商店都是起点，就可以算出从商店到各个点的最短距离。

代码1（邻接表）：

#include<bits/stdc++.h>
#define pii pair<int, int>
using namespace std;
const int N = 1e5 + 5;
int n, m, k, d[N];
struct node{int to, w;
};
vector<node> v[N];void dij(){memset(d, 127, sizeof(d));d[n + 1] = 0;priority_queue<pii, vector<pii>, greater<pii> > q;q.push({0, n + 1});while(!q.empty()){pii p = q.top();q.pop();int sec = p.second;for(int i = 0; i < v[sec].size(); i++){node e = v[sec][i];if(d[e.to] > d[sec] + e.w){d[e.to] = d[sec] + e.w;q.push({d[e.to], e.to});}}
//        printf("***%d %d\n", d[sec], sec);}
}int main(){cin >> n >> m;for(int i = 1; i <= m; i++){int x, y, w;  cin >> x >> y >> w;v[x].push_back({y, w});v[y].push_back({x, w});}cin >> k;for(int i = 1; i <= k; i++){int x;  cin >> x;v[n + 1].push_back({x, 0});}dij();int q;  cin >> q;while(q--){int x;  cin >> x;printf("%d\n", d[x]);}
}

代码2（链式前向星）：

好像会快一点

#include<bits/stdc++.h>
#define pii pair<int, int>
using namespace std;
const int N = 3e5 + 5;
int n, m, k, d[N], vis[N];
int h[N], ne[N], w[N], e[N], idx;
void add(int a, int b, int c){e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}void dij(){memset(d, 127, sizeof(d));d[n + 1] = 0;priority_queue<pii, vector<pii>, greater<pii> > q;q.push({0, n + 1});while(!q.empty()){pii p = q.top();q.pop();int sec = p.second;if(vis[sec])  continue;vis[sec] = 1;for(int i = h[sec]; i != -1; i = ne[i]){int j = e[i];if(d[j] > d[sec] + w[i]){d[j] = d[sec] + w[i];q.push({d[j], j});}}}
}int main(){scanf("%d%d", &n, &m);memset(h, -1, sizeof(h));for(int i = 1; i <= m; i++){int x, y, w;  scanf("%d%d%d", &x, &y, &w);add(x, y, w);add(y, x, w);}cin >> k;for(int i = 1; i <= k; i++){int x;  scanf("%d", &x);add(n + 1, x, 0);}dij();int q;  scanf("%d", &q);while(q--){int x;  scanf("%d", &x);printf("%d\n", d[x]);}
}