标签:struct add void ret names eof while size for
亲戚只有五个,可以把它们看成2,3,4,5,6号点,分别跑最短路,记录一下距离,然后DFS一下
这题非常玄学,我开了一个\(12*12\)的数组,没有离散化,竟然过了,开到\(5050*5050\)就RE,玄学
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int n, m, num, cnt, ans = INF;
int head[N], dis[N], home[N], d[12][12];
bool vis[N], mark[N];
struct node {
int v, nx, w;
} e[N];
struct edge {
int id, dis;
bool operator <(const edge &oth) const {
return this -> dis > oth.dis;
}
};
inline void add(int u, int v, int w) {
e[++num] = (node) {v, head[u], w}, head[u] = num;
}
priority_queue<edge>q;
void dijkstra(int S) {
memset(dis, INF, sizeof dis);
memset(vis, 0, sizeof vis);
dis[S] = 0;
q.push((edge) {S, 0});
while (!q.empty()) {
edge d = q.top();
q.pop();
int u = d.id;
if (vis[u]) continue;
vis[u] = 1;
for (int i = head[u]; ~i; i = e[i].nx) {
int v = e[i].v;
if (dis[v] > dis[u] + e[i].w)
q.push((edge) {v, dis[v] = dis[u] + e[i].w});
}
}
}
void dfs(int u, int sum, int cnt) {
if (cnt == 5) {
ans = min(ans, sum);
return;
}
if (sum > ans) return;
for (int i = 2; i <= 6; ++i) {
if (!mark[i]) {
mark[i] = 1;
dfs(i, sum + d[u][i], cnt + 1);
mark[i] = 0;
}
}
}
int main() {
memset(head, -1, sizeof head);
cin >> n >> m;
for (int i = 1; i <= 5; ++i) cin >> home[i];
for (int i = 1, x, y, z; i <= m; ++i)
cin >> x >> y >> z, add(x, y, z), add(y, x, z);
dijkstra(1);
for (int i = 1; i <= 5; ++i)
d[1][i + 1] = d[i + 1][1] = dis[home[i]];
for (int i = 1; i <= 5; ++i) {
dijkstra(home[i]);
for (int j = i + 1; j <= 5; ++j)
d[i + 1][j + 1] = d[j + 1][i + 1] = dis[home[j]];
}
mark[1] = 1;
dfs(1, 0, 0);
cout << ans << endl;
}
标签:struct add void ret names eof while size for
原文地址:https://www.cnblogs.com/lykkk/p/11438007.html