loj 1406(状态压缩)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1406

思路：首先可以预处理出在每个顶点的状态的合法状态vis[u][state], 然后标记那些合法状态mark[state]。最后就是记忆化搜索了，对于当前状态state,我们有res = min(res, 1 + Solve(state ^ substate)), 其中substate为state的子状态，并且substate = (substate  - 1) & state.那么最终就是要求Solve((1 << n) - 1)了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 using namespace std;

 int n, m;
 bool vis[][ << ];
 bool mark[ << ];
 int dp[ << ];
 vector<int > g[];

 void dfs(int u, int state)
 {
     vis[u][state] = true;
     mark[state] = true;
     for (int i = ; i < (int)g[u].size(); i++) {
         int v = g[u][i];
         if (!vis[v][state | ( << v)]) {
             dfs(v, state | ( << v));
         }
     }
 }

 int Solve(int state)
 {
     if (state == ) return ;
     if (dp[state] != -) return dp[state];
     int res = ;
     for (int i = state; i > ; i = (i - ) & state) {
         if (mark[i]) {
             res = min(res,  + Solve(state ^ i));
         }
     }
     return dp[state] = res;
 }

 int main()
 {
     int _case, t = ;
     scanf("%d", &_case);
     while (_case--) {
         scanf("%d %d", &n, &m);
         for (int i = ; i < n; i++) g[i].clear();
         while (m--) {
             int u, v;
             scanf("%d %d", &u, &v);
             u--, v--;
             g[u].push_back(v);
         }
         memset(vis, false, sizeof(vis));
         memset(mark, false, sizeof(mark));
         for (int i = ; i < n; i++) dfs(i,  << i);
         memset(dp, -, sizeof(dp));
         printf("Case %d: %d\n", t++, Solve(( << n) -));
     }
     return ;
 }