bzoj4006

斯坦纳树

比之前要求高了一些

其实利用斯坦纳树的dp[i][s]以i为根，S为状态就行了，先跑一遍斯坦纳树，预处理出dp数组，记住每个S的最小值，然后再dp，这里dp必须要求同一种颜色的状态都必须在S里，然后跑枚举子集就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 1005, inf = 0x3f3f3f3f;
struct edge {
    int nxt, to, w;
} e[N * 6];
struct points {
    int x, c;
} a[N];
int n, m, p, cnt = 1;
int head[N], sum[11], vis[N], ans[1 << 11], dp[N][1 << 11], tmp[11];
bool check(int S)
{
    memset(tmp, 0, sizeof(tmp));
    for(int i = 0; i < p; ++i) if(S >> i & 1) ++tmp[a[i].c];
    for(int i = 1; i <= p; ++i) if(tmp[i] && sum[i] != tmp[i]) return false;
    return true;
}
void link(int u, int v, int w)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].w = w;
}
int main()
{
    scanf("%d%d%d", &n, &m, &p);
    for(int i = 1; i <= m; ++i)
    {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        link(u, v, w);
        link(v, u, w);
    } 
    memset(dp, 0x3f3f, sizeof(dp));
    for(int i = 0; i < p; ++i) 
    {
        scanf("%d%d", &a[i].c, &a[i].x);
        dp[a[i].x][1 << i] = 0;
        ++sum[a[i].c];
    }
    for(int S = 1; S < (1 << p); ++S) 
    {
        queue<int> q; 
        for(int i = 1; i <= n; ++i)
        { 
            for(int S0 = S; S0; S0 = (S0 - 1) & S)
                dp[i][S] = min(dp[i][S], dp[i][S ^ S0] + dp[i][S0]);
            if(dp[i][S] < inf) 
            {
                vis[i] = 1;
                q.push(i);
            }
        }
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            vis[u] = 0;
            for(int i = head[u]; i; i = e[i].nxt) if(dp[u][S] + e[i].w < dp[e[i].to][S])
            {
                dp[e[i].to][S] = dp[u][S] + e[i].w;
                if(!vis[e[i].to])
                {
                    vis[e[i].to] = 1;
                    q.push(e[i].to);
                }
            }
        }
        ans[S] = inf;
        for(int i = 1; i <= n; ++i) ans[S] = min(ans[S], dp[i][S]);
    }
    for(int i = 1; i < (1 << p); ++i)
        if(check(i))
            for(int S = i; S; S = (S - 1) & i) 
                if(check(S))
                    ans[i] = min(ans[i], ans[i ^ S] + ans[S]);
    printf("%d\n", ans[(1 << p) - 1]);
    return 0;
}