hdu4085 Peach Blossom Spring 斯坦纳树，状态dp

（1）集合中元素表示（1<<i）, i从0开始

（2）注意dp[i][ss] = min(dp[i][ss], dp[i][rr | s[i]] + dp[i][(ss ^ rr) | s[i]]);，后面的要 |s[i],保证状态的正确

（3）INF初始化CLR（dp， 0x3f）

（4）注意斯坦纳树状态理解，分层松弛的理解

参考：http://endlesscount.blog.163.com/blog/static/821197872012525113427573/

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)

typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;
const int MAXN = 1000010;

struct node{
    int to, next, l;
};
int tol;
node adj[2010];
int head[55];
void init_adj()
{
    tol = 0;
    CLR(head, -1);
}
void add_edge(int x, int y, int ll)
{
    adj[tol].to = y;
    adj[tol].l = ll;
    adj[tol].next = head[x];
    head[x] = tol++;
}
int dp[55][1 << 11];
queue<int>q;
bool inq[55][1 << 11];
int d[1 << 11];
int s[55];
int n, m, k;
int cnt, ALL;

bool check(int ss)
{
    int num = 0;
    for (int i = 0; i < cnt; i++)
        if (ss & (1 << i))
            num += i < k ? 1 : -1;
    if (!num) return 1;
    return 0;
}

void spfa()
{
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        int ui = u / 10000;
        int uss = u % 10000;
        inq[ui][uss] = 0;
        for (int r = head[ui]; r != -1; r = adj[r].next)
        {
            int vi = adj[r].to;
            int vss = uss | s[vi];
            if (dp[vi][vss] > dp[ui][uss] + adj[r].l)
            {
                dp[vi][vss] = dp[ui][uss] + adj[r].l;
                if (vss == uss && !inq[vi][vss]) /// 同层且不再队列中
                {
                    inq[vi][vss] = 1;
                    q.push(vi * 10000 + vss);
                }
            }
        }
    }
}

void solve()
{
    while (!q.empty()) q.pop();
    CLR(inq, 0);
    FE(ss, 0, ALL)
    {
        FE(i, 1, n)
        {
            for (int rr = ss & (ss - 1); rr; rr = (rr - 1) & ss)///ss
                    dp[i][ss] = min(dp[i][ss], dp[i][rr | s[i]] + dp[i][(ss ^ rr) | s[i]]);
            if (dp[i][ss] < INF)
                q.push(i * 10000 + ss), inq[i][ss] = 1;
        }
        spfa();
    }

    CLR(d, 0x3f);
    FE(ss, 0, ALL) FE(j, 1, n)
        d[ss] = min(d[ss], dp[j][ss]);

    REP(ss, ALL + 1)
        for(int rr = ss & (ss - 1); rr; rr = (rr - 1) & ss)
            if (check(rr) && check(ss ^ rr))
                d[ss] = min(d[ss], d[rr] + d[ss ^ rr]);

    if (d[ALL] == INF) printf("No solution\n");
    else printf("%d\n", d[ALL]);
}

void init()
{
    int x, y, l;
    init_adj();
    RIII(n, m, k);
    while (m--)
    {
        RIII(x, y, l);
        add_edge(x, y, l);
        add_edge(y, x, l);
    }

    cnt = 2 * k;
    ALL = (1 << cnt) - 1;

    CLR(s, 0);
    CLR(dp, 0x3f);

    FE(i, 1, k)///集合
    {
        s[i] = 1 << (i - 1);
        dp[i][s[i]] = 0;
        s[n - k + i] = 1 << (k + i - 1);
        dp[n - k + i][s[n - k + i]] = 0;
    }
}

int main ()
{
    int T;
    RI(T);
    while (T--)
    {
        init();
        solve();
    }
    return 0;
}