Travelling（HDU3001+状压dp+三进制+最短路）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3001

题目：

题意：n个城市，m条边，每条边都有一个权值，问你经过所有的城市且每条边通过次数不超过两次的最短距离。

思路：状压dp+三进制，dp[i][j]表示在状态i下以j为目标城市的最短距离，转移方程为nw = i + fi[k];dp[nw][k] = min(dp[nw][k], dp[i][j] + mp[j][k])。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn =  + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, m, u, v, w;
 int mp[][], dp[maxn][], fi[], state[maxn][];

 void init() {
     fi[] = ;
     for(int i = ; i <= ; i++) {
         fi[i] = fi[i-] * ;
     }
     for(int i = ; i < ; i++) {
         int t = i;
         for(int j = ; j <= ; j++) {
             state[i][j] = t % ;
             t /= ;
             if(t == ) break;
         }
     }
 }

 int main() {
     //FIN;
     init();
     while(~scanf("%d%d", &n, &m)) {
         memset(mp, inf, sizeof(mp));
         memset(dp, inf, sizeof(dp));
         for(int i = ; i <= n; i++) {
             mp[i][i] = , dp[fi[i]][i] = ;
         }
         for(int i = ; i <= m; i++) {
             scanf("%d%d%d", &u, &v, &w);
             mp[u][v] = mp[v][u] = min(mp[u][v], w);
         }
         int ans = inf;
         for(int i = ; i <  * fi[n]; i++) {
             int flag = ;
             for(int j = ; j <= n; j++) {
                 if(state[i][j] == ) flag = ;
                 if(dp[i][j] >= inf) continue;
                 for(int k = ; k <= n; k++) {
                     if(j == k) continue;
                     if(mp[j][k] >= inf || state[i][k] >= ) continue;
                     int nw = i + fi[k];
                     dp[nw][k] = min(dp[nw][k], dp[i][j] + mp[j][k]);
                 }
             }
             if(flag) {
                 for(int j = ; j <= n; j++) {
                     ans = min(ans, dp[i][j]);
                 }
             }
         }
         if(ans >= inf) printf("-1\n");
         else printf("%d\n", ans);
     }
     return ;
 }