loj 1406(状态压缩)

来源：互联网发布：dj香烟淘宝链接编辑：程序博客网时间：2024/06/03 15:34

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。

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 using namespace std; 7  8 int n, m; 9 bool vis[16][1 << 16];10 bool mark[1 << 16];11 int dp[1 << 16];12 vector<int > g[16];13 14 void dfs(int u, int state)15 {16     vis[u][state] = true;17     mark[state] = true;18     for (int i = 0; i < (int)g[u].size(); i++) {19         int v = g[u][i];20         if (!vis[v][state | (1 << v)]) {21             dfs(v, state | (1 << v));22         }23     }24 }25 26 int Solve(int state)27 {28     if (state == 0) return 0;29     if (dp[state] != -1) return dp[state];30     int res = 16;31     for (int i = state; i > 0; i = (i - 1) & state) {32         if (mark[i]) {33             res = min(res, 1 + Solve(state ^ i));34         }35     }36     return dp[state] = res;37 }38 39 40 int main()41 {42     int _case, t = 1;43     scanf("%d", &_case);44     while (_case--) {45         scanf("%d %d", &n, &m);46         for (int i = 0; i < n; i++) g[i].clear();47         while (m--) {48             int u, v;49             scanf("%d %d", &u, &v);50             u--, v--;51             g[u].push_back(v);52         }53         memset(vis, false, sizeof(vis));54         memset(mark, false, sizeof(mark));55         for (int i = 0; i < n; i++) dfs(i, 1 << i);56         memset(dp, -1, sizeof(dp));57         printf("Case %d: %d\n", t++, Solve((1 << n) -1));58     }59     return 0;60 }

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