题目链接
思路
亲戚很少，可以每个点都算一遍单源最短路
然后dfs
错误原因
算错复杂度
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxm = 2e5 + 7;
const int maxn = 5e4 + 7;
struct edge {
int v, q, nxt;
} e[maxm];
struct node {
int x, y;
node(int a, int b) {
x = a, y = b;
}
bool operator < (const node &b) const {
return y > b.y;
}
};
int head[maxm], tot;
void add_edge(int u, int v, int q) {
e[++tot].v = v;
e[tot].q = q;
e[tot].nxt = head[u];
head[u] = tot;
}
int n, m;
int qinqi[7];
int a[10][10];
int dis[6][maxn];
int read() {
int x = 0, f = 1; char s = getchar();
for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
return x * f;
}
void dijstra(int s) {
priority_queue<node> q;
q.push(node(qinqi[s], 0));
dis[s][qinqi[s]] = 0;
while (!q.empty()) {
node u = q.top();
q.pop();
if (dis[s][u.x] != u.y) continue;
for (int i = head[u.x]; i; i = e[i].nxt) {
int v = e[i].v;
if (dis[s][v] > dis[s][u.x] + e[i].q) {
dis[s][v] = dis[s][u.x] + e[i].q;
q.push(node(v, dis[s][v]));
}
}
}
}
int ans=inf;
int vis[10];
void dfs(int last,int js,int tot)
{
if(js==5) {
ans=min(ans,tot);
return;
}
if(tot > ans) {
return;
}
for(int i=1;i<=5;++i)
{
if(!vis[i]) {
vis[i]=1;
dfs(i,js+1,tot+a[last][i]);
vis[i]=0;
}
}
}
int main() {
freopen("a.in", "r", stdin);
memset(dis, inf, sizeof(dis));
n = read(), m = read();
qinqi[0] = 1;
for (int i = 1; i <= 5; ++i)
qinqi[i] = read();
for (int i = 1; i <= m; ++i) {
int x = read(), y = read(), z = read();
add_edge(x, y, z);
add_edge(y, x, z);
}
for (int i = 0; i <= 5; ++i) {
dijstra(i);
}
for (int i = 0; i <= 5; ++i) {
for (int j = 0; j <= 5; ++j) {
a[i][j] = dis[i][qinqi[j]];
}
}
dfs(0,0,0);
cout<<ans;
return 0;
}