思路
最直观的解法就是，建完图之后，每读入一个pick-up spots，就以它为起点，进行一次dijkstra，获得从它到所有点（包括救护车派遣中心）的最短距离、前驱。之后，从每个救护车派遣中心出发，进行dfs，得到从该救护车派遣中心到当前pick-up spots的完整路径，比较路径即可，获得最优路径。
使用优先队列优化的dijkstra，第四个数据点用时200ms。
(睡觉
#include <bits/stdc++.h>
using namespace std;
struct node {
int v, dis;
node () {
}
node (int _v, int _d) : v(_v), dis(_d) {
}
bool operator < (const node &b) const {
return dis > b.dis;
}
};
const int maxn = 1020;
const int INF = 1e9;
int d[maxn], G[maxn][maxn];
int acnt[maxn];
vector<int> temp, ans;
vector<node> adj[maxn];
vector<int> pre[maxn];
int optdis, optcnt, optstreet;
int ns, na, m, k, total = 0, st;
bool vis[maxn];
void dijkstra() {
fill(d, d + maxn, INF);
d[st] = 0;
memset(vis, false, sizeof(vis));
priority_queue<node> q;
q.push(node(st, 0));
while (q.size()) {
int u = q.top().v;
q.pop();
if (vis[u]) continue;
vis[u] = true;
for (auto cur : adj[u]) {
int v = cur.v, dis = cur.dis;
if (!vis[v]) {
if (d[v] > d[u] + dis) {
d[v] = d[u] + dis;
pre[v].clear();
pre[v].push_back(u);
q.push(node(v, d[v]));
} else if (d[v] == d[u] + dis) {
pre[v].push_back(u);
}
}
}
}
}
int get_dis () {
int r = 0;
for (int i = 1; i < temp.size(); i++) {
int u = temp[i-1], v = temp[i];
r += G[u][v];
}
return r;
}
void dfs(int u) {
if (u == st) {
temp.push_back(u);
int cur_dis = get_dis();
if ((cur_dis < optdis) || (cur_dis == optdis && acnt[temp[0]] > optcnt) || (cur_dis == optdis && acnt[temp[0]] == optcnt && temp.size() < optstreet)) {
optdis = cur_dis;
optcnt = acnt[temp[0]];
optstreet = temp.size();
ans = temp;
}
temp.pop_back();
return;
}
temp.push_back(u);
for (auto v : pre[u]) {
dfs(v);
}
temp.pop_back();
}
void show() {
for (int i = 0; i < ans.size(); i++) {
if (i) printf(" ");
int u = ans[i];
if (u > ns) printf("A-%d", u - ns);
else printf("%d", u);
}
printf("\n");
}
int main()
{
scanf("%d%d", &ns, &na);
for (int i = ns + 1; i <= ns + na; i++) {
scanf("%d", &acnt[i]);
total += acnt[i];
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
string s1, s2;
int dis;
cin >> s1 >> s2 >> dis;
int u, v;
if (s1[0] == 'A') u = ns + stoi(s1.substr(2));
else u = stoi(s1);
if (s2[0] == 'A') v = ns + stoi(s2.substr(2));
else v = stoi(s2);
adj[u].push_back(node(v, dis));
adj[v].push_back(node(u, dis));
G[u][v] = G[v][u] = dis;
}
scanf("%d" ,&k);
while (k--) {
scanf("%d", &st);
if (total < 1) printf("All Busy\n");
else {
for (int i = 0; i < maxn; i++) pre[i].clear();
dijkstra();
optdis = INF, optcnt = 0, optstreet = INF;
temp.clear(), ans.clear();
for (int cur = ns + 1; cur <= ns + na; cur++) {
if (acnt[cur] == 0) continue;
dfs(cur);
}
show();
printf("%d\n", optdis);
acnt[ans[0]]--;
}
total--;
}
return 0;
}