Poj 2404 状态压缩DP

本来是在light上看到的这题的。。。然后一直不过，没想到在poj上能过。。。。。。呵呵

题目是说每条边至少走一次然后回到原点，那么找出所有奇数度的节点两两相连得到欧拉路

AC代码如下：

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

#define MAX 0x3f3f3f3f

int minlength[16][16], weight[16][16];
int N, M, ans;
int deg[16];
int odddeg[16], tot;
int dp[1<<15];

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

//	cin >> T;
	while( cin >> N && N ){
		
		cin >> M;
		
		memset( deg, 0, sizeof( deg ) );
		memset( weight, 0x3f, sizeof( weight ) );
		for( int i = 1; i <= N; i++ )	weight[i][i] = 0;
		ans = 0;
		
		for( int i = 1; i <= M; i++ ){
			int temp1, temp2, temp3;
			cin >> temp1 >> temp2 >> temp3;
			ans += temp3;
			weight[temp2][temp1] = min( temp3, weight[temp2][temp1] );
			weight[temp1][temp2] = weight[temp2][temp1];
			deg[temp1]++;
			deg[temp2]++;
		}

		for( int i = 1; i <= N; i++ ){
			for( int j = 1; j <= N; j++ ){
				minlength[i][j] = weight[i][j];
			}
		}
		for( int i = 1; i <= N; i++ ){
			for( int j = 1; j <= N; j++ ){
				for( int k = 1; k <= N; k++ ){
					minlength[i][j] = min( minlength[i][j], minlength[i][k] + minlength[k][j] );
				}
			}
		}

		tot = 0;
		for( int i = 1; i <= N; i++ ){
			if( deg[i] % 2 ){
				odddeg[tot++] = i;
			}
		}

		memset( dp, 0x3f, sizeof( dp ) );
		dp[0] = 0;
		for( int stu = 0; stu < ( 1 << tot ); stu++ ){
			for( int i = 0; i < tot; i++ )if( !( stu & ( 1 << i ) ) ){
				for( int j = 0; j < tot; j++ )if( !( stu & ( 1 << j ) ) && i != j ){
					dp[stu|(1<