lightoj 1086 状态压缩DP

题意：给定一个无相图，求从任意点为起点遍历所有边最后回到起点的最短距离

思路：

这题首先要知道，是要构造欧拉回路，然后我们找出所有奇度数的点，添加边使之变为偶度数

然后添加边就用状态压缩DP来做

AC代码如下：

#include 
#include 
#include 
#include 
#include 
using namespace std;

#define MAX 0x3f3f3f3f

vector maps[17][17];
int dis[17][17];
int N, M;
int cnt[16];
int dp[17][1<<17];
vector vis;
int K;

int solve( int pos, int statu ){
    if( pos >= K ){
        return 0;
    }
    if( statu & ( 1 << pos ) ){
        return solve( pos + 1, statu );
    }
    if( dp[pos][statu] != -1 ){
        return dp[pos][statu];
    }
    int ans = MAX;
    for( int i = pos + 1; i < K; i++ ){
        if( statu & ( 1 << i ) )    continue;
        ans = min( ans, solve( pos + 1, statu | ( 1 << pos ) | ( 1 << i ) ) + dis[vis[i]][vis[pos]] );
    }
    return dp[pos][statu] = ans;
}

int main(){
    int T, Case = 1, sum;

    scanf( "%d", &T );
    while( T-- ){
        scanf( "%d%d", &N, &M );
        for( int i = 1; i <= N; i++ ){
            for( int j = 1; j <= N; j++ ){
                maps[i][j].clear();
                maps[i][j].push_back( MAX );
            }
        }
        sum = 0;
        memset( cnt, 0, sizeof( cnt ) );
        for( int i = 0; i < M; i++ ){
            int temp1, temp2, temp3;
            scanf( "%d%d%d", &temp1, &temp2, &temp3 );
            maps[temp1][temp2].push_back( temp3 );
            maps[temp1][temp2][0] = min( maps[temp1][temp2][0], temp3 );
            maps[temp2][temp1].push_back( temp3 );
            maps[temp2][temp1][0] = min( maps[temp2][temp1][0], temp3 );
            cnt[temp1]++;
            cnt[temp2]++;
            sum += temp3;
        }
        for( int i = 1; i <= N; i++ ){
            for( int j = 1; j <= N; j++ ){
                if( i == j ){
                    dis[i][j] = 0;
                }else{
                    dis[i][j] = maps[i][j][0];
                }
            }
        }
      //  cout << dis[1][2] << " " << dis[1][3] << " " << dis[2][3] << endl;
        for( int k = 1; k <= N; k++ ){
            for( int i = 1; i <= N; i++ ){
                for( int j = 1; j <= N; j++ ){
                    dis[i][j] = min( dis[i][j], dis[i][k] + dis[k][j] );
                }
            }
        }
     //   cout << dis[1][2] << " " << dis[1][3] << " " << dis[2][3] << endl;
        K = 0;
        vis.clear();
        for( int i = 1; i <= N; i++ ){
            if( cnt[i] % 2 == 1 ){
                vis.push_back( i );
              //  cout << i << endl;
                K++;
            }
        }
        int ans = 0;
        memset( dp, -1, sizeof( dp ) );
        ans = sum + solve( 0, 0 );
        printf( "Case %d: %d\n", Case++, ans );
    }
    return 0;
}
/*
1
3 4
1 3 3
1 3 12
1 2 10
2 3 13
*/