【2020 HDU 多校训练第三场 1007 Tokitsukaze and Rescue】暴力删边&最短路

Tokitsukaze and Rescue

题意

给出一个有 n 个顶点的完全图，现在要删除 k 条边，使得删完边之后，从顶点 1 到达顶点 n 的最短路最长，输出这个距离。

(3≤n≤50,1≤k≤min(n−2,5))

题解

比赛的时候把 k 看成50了。。。。

代码

#include <bits/stdc++.h>
#define fuck system("pause")
#define emplace_back push_back
#define pb push_back
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 55;

int g[N][N], dis[N], vis[N], back[N], ans, n, k;
void dfs(int cnt)
{
    int tmp[N];
    for (int i = 1; i <= n; i++) {
        tmp[i] = vis[i] = 0, dis[i] = inf;
    }
    dis[1] = 0;
    for (int i = 1; i <= n; i++) {
        int u, minn = inf;
        for (int j = 1; j <= n; j++) {
            if (!vis[j] && dis[j] < minn) {
                minn = dis[j];
                u = j;
            }
        }
        vis[u] = 1;
        for (int j = 1; j <= n; j++) {
            if (!vis[j] && dis[u] + g[u][j] < dis[j]) {
                tmp[j] = u;
                dis[j] = dis[u] + g[u][j];
            }
        }
    }
    if (!cnt) {
        if (dis[n] != inf)
            ans = max(ans, dis[n]);
        return;
    }
    int now = n;
    while (tmp[now]) {//tmp数组一定要在函数里开，否则进行dfs回溯的时候数组发生变化
        int temp = g[tmp[now]][now];
        g[tmp[now]][now] = g[now][tmp[now]] = inf;
        dfs(cnt - 1);
        g[tmp[now]][now] = g[now][tmp[now]] = temp;
        now = tmp[now];
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        ans = 0;
        memset(g, inf, sizeof(g));
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= (n - 1) * n / 2; i++) {
            int x, y, v;
            scanf("%d%d%d", &x, &y, &v);
            g[x][y] = g[y][x] = v;
        }
        dfs(k);
        printf("%d
", ans);
    }
    // fuck;
    return 0;
}
/*
*/