HDU 4085 斯坦纳树+DP

https://cn.vjudge.net/problem/HDU-4085

给你n,m,k ，分别表示有n个点，m条边，每条边有一个权值，表示修复这条边需要的代价

从前k个点中任取一个使其和后k个点中的某一个点，通过边连接，并且必须是一一对应，问最小的代价是多少。

先用斯坦纳树模板求出f[i][1<<k]    然后用dp[i]表示所有点为根的情况下连通状态为i的最小花费

这样我们就可以从1dp到1<<k得到答案 注意dp之前要先判总状态是否合法 再判子集是否合法 最后再进行dp更新

#include<bits/stdc++.h>
#define N 6003
#define inf 1000000000
using namespace std;
int n, m, k, tot;
int point[N], next1[N], v[N], len[N];
int f[53][(1 << 11)], mi[20], can[N], dp[(1 << 11)];
queue<int> p;
void add(int x, int y, int z) {
        tot++;
        next1[tot] = point[x];
        point[x] = tot;
        v[tot] = y;
        len[tot] = z;
        tot++;
        next1[tot] = point[y];
        point[y] = tot;
        v[tot] = x;
        len[tot] = z;
}
void spfa(int sta) {
        while (!p.empty()) {
                int now = p.front();
                p.pop();
                for (int i = point[now]; i; i = next1[i]) {
                        if (f[v[i]][sta] > f[now][sta] + len[i]) {
                                f[v[i]][sta] = f[now][sta] + len[i];
                                if (!can[v[i]]) {
                                        can[v[i]] = 1;
                                        p.push(v[i]);
                                }
                        }
                }
                can[now] = 0;
        }
}
bool check(int sta) { //判断当前的状态是否满足一一对应关系
        int ans = 0;
        for (int i = 0; i < k; i++) {
                if (sta & (1 << i))
                        ans++;
                if (sta & (1 << (i + k)))
                        ans--;
        }
        return (ans == 0);
}
int main() {
        int t;
        scanf("%d", &t);
        mi[0] = 1;
        for (int i = 1; i <= 11; i++)
                mi[i] = mi[i - 1] * 2;
        for (int T = 1; T <= t; T++) {
                scanf("%d%d%d", &n, &m, &k);
                tot = 0;
                memset(point, 0, sizeof(point));
                memset(next1, 0, sizeof(next1));
                for (int i = 1; i <= m; i++) {
                        int x, y, z;
                        scanf("%d%d%d", &x, &y, &z);
                        add(x, y, z);
                }
                for (int i = 1; i <= n; i++)
                        for (int j = 0; j < mi[10]; j++)
                                f[i][j] = inf;
                for (int i = 1; i <= k; i++)
                        f[i][mi[i - 1]] = 0;
                int t = k;
                for (int i = n - k + 1; i <= n; i++)
                        f[i][mi[t]] = 0, t++;
                for (int sta = 1; sta < mi[t]; sta++) {
                        for (int i = 1; i <= n; i++) {
                                for (int s = sta & (sta - 1); s; s = sta & (s - 1)) {
                                        int t = f[i][sta - s] + f[i][s];
                                        f[i][sta] = min(f[i][sta], t);
                                }
                                if (f[i][sta] != inf)
                                        p.push(i), can[i] = 1;
                        }
                        spfa(sta);
                }
                for (int sta = 1; sta < mi[t]; sta++) {
                        dp[sta] = inf;
                        for (int i = 1; i <= n; i++)
                        dp[sta] = min(dp[sta], f[i][sta]);
                }
                for (int sta = 1; sta < mi[t]; sta++)
                        if (check(sta))
                                for (int s = sta & (sta - 1); s; s = sta & (s - 1))
                                        if (check(s))
                                                dp[sta] = min(dp[sta], dp[s] + dp[sta - s]);
                if (dp[mi[t] - 1] == inf)
                        printf("No solution
");
                else
                        printf("%d
", dp[mi[t] - 1]);
        }
}