BZOJ3590【状压DP】

SNOI2013竟然出了一道很有难度的状压DP.实在是出人意料.

而且网上似乎并没有题解.我就来写一篇好了.

HYF菊苣似乎写过这道题的题解.

这道题大意就是:给出一个无向图.求一个权值最小的包含所有点的双联通子图.

定义一些状态:

f[i]:集合状态为i.且使在i中的点双联通的最小权值.

h[i][j][0]:一个端点是j.另一个端点在点集i中的边的最小权值.

h[i][j][1]:一个端点是j.另一个端点在点集i中的边的次小权值.

g[i][j][k].集合状态为i.且使在i中的点构成一条链.两端点分别是j和k的最小权值.

容易发现.一个双联通图.必然是由一个双联通的子图和一条链构成的.那么就可以枚举这条链进行转移.

要注意的是:一个点是一个权值为0的双联通图.同时也是一条权值为0的链.

在转移的时候.如果这条链是一个点.转移方程是:

	f[i] = min(f[i], f[t] + g[s][u][u] + h[t][u][0] + h[t][u][1]);
如果这条链的两端点不同.转移方程是:

	f[i] = min(f[i], f[t] + g[s][u][v] + h[t][u][0] + h[t][v][0]);
先预处理出h和m数组就好了.

关于h数组的处理.直接暴力枚举就可以了.

关于g数组的处理.也是一个状压DP.枚举当前点集i能转移出的状态更新就好了.

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <string>
#define make make_pair
#define fi first
#define se second
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
 
const int maxn = 20;
const int maxm = 10100;
const int maxs = 26;
const int inf = 1 << 28;
const int P = 1000000007;
const double error = 1e-9;
 
inline int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch <= 47 || ch >= 58)
        f = (ch == 45 ? -1 : 1), ch = getchar();
    while (ch >= 48 && ch <= 57)
        x = x * 10 + ch - 48, ch = getchar();
    return x * f;
}
 
struct edge
{
    int u, v, w, next;
} e[maxm];
 
int n, m, cnt, head[maxn], bin[maxn],
    f[maxm], g[maxm][maxn][maxn], h[maxm][maxn][2];
 
void insert(int u, int v, int w)
{
    e[cnt] = (edge) {u, v, w, head[u]}, head[u] = cnt++;
    e[cnt] = (edge) {v, u, w, head[v]}, head[v] = cnt++;
}   
 
int fac(int x)
{
    int ans = 1;
    for (x = x & (x - 1); x; x = x & (x - 1))
        ans++;
    return ans;
}
 
void init()
{
    for (int i = 0; i < (1 << n); i++)
        for (int j = 0; j <= n; j++)
            for (int k = 0; k <= n; k++)
                g[i][j][k] = inf;
 
    for (int i = 1; i <= n; i++) {
        bin[i] = 1 << (i - 1);
        g[bin[i]][i][i] = 0;
    }
 
    for (int i = 0; i < cnt; i++) {
        int u = e[i].u, v = e[i].v, s = bin[u] + bin[v];
        g[s][u][v] = min(g[s][u][v], e[i].w);
    }
 
    for (int i = 1; i < (1 << n); i++)
        for (int u = 1; u <= n; u++)
            for (int v = 1; v <= n; v++)
                if ((bin[u] | i) == i && (bin[v] | i) == i)
        for (int k = head[v]; k != -1; k = e[k]. next) {
            int fv = e[k].v;
            if ((bin[fv] | i) != i) {
                int s = i + bin[fv];
                g[s][u][fv] = min(g[s][u][fv], g[i][u][v] + e[k].w);
            }
        }

    for (int i = 0; i < (1 << n); i++)
        for (int j = 0; j <= n; j++)
            for (int k = 0;k <= 1; k++)
                h[i][j][k] = inf;
 
    for (int i = 1; i < (1 << n); i++)
        for (int u = 1; u <= n; u++) 
            if ((bin[u] | i) != i)
                for (int j = head[u]; j != -1; j = e[j].next) {
                    int v = e[j].v;
                    if ((bin[v] | i) == i) {
                        if (e[j].w <= h[i][u][0]) {
                            h[i][u][1] = h[i][u][0];
                            h[i][u][0] = e[j].w;
                        }
                        else if (e[j].w < h[i][u][1])
                            h[i][u][1] = e[j].w;
                    }       
                }
}
 
int work()
{
    for (int i = 1; i < (1 << n); i++)
        f[i] = inf;
     
    for (int i = 1; i <= n; i++)
        f[bin[i]] = 0;
 
    for (int i = 1; i < (1 << n); i++)
        if (fac(i) >= 2)
            for (int s = i & (i - 1); s; s = (s - 1) & i) {
                int t = i - s;
                for (int u = 1; u <= n; u++)
                    for (int v = 1; v <= n; v++)
        if ((bin[u] | s) == s && (bin[v] | s) ==s) {
            if (u == v)
                f[i] = min(f[i], f[t] + g[s][u][u] + h[t][u][0] + h[t][u][1]);
            else
                f[i] = min(f[i], f[t] + g[s][u][v] + h[t][u][0] + h[t][v][0]);
        }
            }
}
 
int main()
{
    int t = read();
 
    while (t--) {
        n = read(), m = read();
         
        memset(head, - 1, sizeof head);
     
        cnt = 0;
 
        for (int i = 1; i <= m; i++) {
            int u = read(), v = read(), w = read();
            insert(u, v, w);
        }
 
        init(), work();
 
        if (f[(1 << n) - 1] < inf)
            printf("%d\n", f[(1 << n) - 1]);
        else
            printf("impossible\n");
    }
 
    return 0;
}

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