loj#10078. 新年好(最短路)

Posted 2022-09-12 lykkk

题目：

loj#10078. 新年好

解析：

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

代码：

#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;