bzoj4006 [JLOI2015]管道连接

Posted 2020-09-15 逢山开路 遇水架桥

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4006

【题解】

即求斯坦纳森林……

然后我们发现可以先把所有点全当做重要点做一遍steiner tree。

然后我们可以求出联通状态为S的时候的最小花费

然后子集更新即可。（再一遍dp）

注意只需要做一遍就够了！！！

而且斯坦纳树板子要用对（大概快4倍左右）

# include <queue>
# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, STATUS = 1027, N = 1000 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, m, p;
int head[M], nxt[M], to[M], tot, w[M];

inline void add(int u, int v, int _w) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; w[tot] = _w;
}
inline void adde(int u, int v, int _w) {
    add(u, v, _w);
    add(v, u, _w);
}

int f[N][STATUS], st[STATUS];
bool vis[N];
queue <int> q;

int cp[N], cx[N];
int s[STATUS], dp[STATUS];

int cnt[23], all[23];
inline bool ok(int status) {
    memset(cnt, 0, sizeof cnt);
    for (int i=1; i<=p; ++i)
        if(status & (1<<(i-1))) cnt[cp[i]] ++;
    for (int i=1; i<=p; ++i)
        if(cnt[i] && cnt[i] != all[i]) return 0;
    return 1;
}

inline void spfa(int status) {
    for (int i=1; i<=n; ++i) vis[i] = 1, q.push(i);
    while(!q.empty()) {
        int top = q.front(); q.pop(); vis[top] = 0;
        for (int i=head[top]; i; i=nxt[i]) {
            if(f[to[i]][status] > f[top][status] + w[i]) {
                f[to[i]][status] = f[top][status] + w[i];
                if(!vis[to[i]]) {
                    vis[to[i]] = 1;
                    q.push(to[i]);
                }
            }
        }
    }
}

int main() {
    scanf("%d%d%d", &n, &m, &p);
    for (int i=1, u, v, _w; i<=m; ++i) {
        scanf("%d%d%d", &u, &v, &_w);
        adde(u, v, _w);
    }
    for (int i=1; i<=p; ++i) scanf("%d%d", &cp[i], &cx[i]), all[cp[i]] ++;
    int status_size = (1<<p)-1;
    for (int i=1; i<=n; ++i) for (int j=1; j<=status_size; ++j) f[i][j] = 1e9;
    for (int i=1; i<=p; ++i) f[cx[i]][1<<(i-1)] = 0, st[cx[i]] = (1<<(i-1));
    for (int status = 0; status <= status_size; ++status) {
        for (int i=1; i<=n; ++i) 
            for (int sub = status-1 & status; sub; sub = sub-1 & status)
                f[i][status] = min(f[i][status], f[i][sub] + f[i][status^sub]);
        spfa(status);
    }
    for (int status = 0; status <= status_size; ++status) {
        s[status] = 1e9;
        for (int i=1; i<=n; ++i) s[status] = min(s[status], f[i][status]);
//        printf("%d\n", s[status]);
    }
    for (int status = 0; status <= status_size; ++status) {
        dp[status] = s[status];
        if(ok(status))
            for (int sub = status-1 & status; sub; sub = sub-1 & status) 
                if(ok(sub)) dp[status] = min(dp[status], dp[sub] + dp[status^sub]);
    }
    printf("%d\n", dp[status_size]);
    return 0;
}

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