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。

 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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