POJ 1185 炮兵阵地(状态压缩DP)

题目链接

写了久。DEBUG很久。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <cstdlib>
  5 #include <map>
  6 #include <queue>
  7 #include <vector>
  8 #include <algorithm>
  9 using namespace std;
 10 int dp1[101][101],dp2[101][101];
 11 int que[101];
 12 int mat[101];
 13 int sum[101];
 14 char str[101][101];
 15 int m;
 16 int getsum(int x)
 17 {
 18     int i,ans = 0;
 19     for(i = 0; i < m; i ++)
 20     {
 21         if(x&(1<<i))
 22             ans ++;
 23     }
 24     return ans;
 25 }
 26 int main()
 27 {
 28     int n,i,j,k,u,num,ans;
 29 
 30     while(scanf("%d%d",&n,&m)!=EOF)
 31     {
 32 
 33         memset(dp1,-1,sizeof(dp1));
 34         memset(dp2,-1,sizeof(dp2));
 35         memset(sum,0,sizeof(sum));
 36         memset(mat,0,sizeof(mat));
 37         num = 1;
 38         for(i = 1; i <= n; i ++)
 39             scanf("%s",str[i]);
 40         for(i = 1; i <= n; i ++)
 41         {
 42             for(j = 0; j < m; j ++)
 43             {
 44                 if(str[i][j] == 'H')
 45                     mat[i] = mat[i] + (1<<j);
 46             }
 47         }
 48         for(i = 0; i < 1<<m; i ++)
 49         {
 50             if(i&(i<<1))
 51                 continue;
 52             if(i&(i>>1))
 53                 continue;
 54             if(i&(i<<2))
 55                 continue;
 56             if(i&(i>>2))
 57                 continue;
 58             sum[num] = getsum(i);
 59             que[num++] = i;
 60         }
 61         for(i = 1; i < num; i ++)
 62         {
 63             if(que[i]&mat[1])
 64                 continue;
 65             dp1[i][1] = sum[i];
 66         }
 67         for(i = 2; i <= n; i ++)
 68         {
 69             for(j = 1; j < num; j ++)
 70             {
 71                 if(mat[i]&que[j]) continue;
 72                 for(k = 1; k < num; k ++)
 73                 {
 74                     if(que[j]&que[k]) continue;
 75                     for(u = 1; u < num; u ++)
 76                     {
 77                         if(que[j]&que[u]) continue;
 78                         if(dp1[k][u] == -1) continue;
 79                         if(dp2[j][k] < dp1[k][u] + sum[j])
 80                             dp2[j][k] = dp1[k][u] + sum[j];
 81                     }
 82                 }
 83             }
 84             for(j = 1; j < num; j ++)
 85             {
 86                 for(k = 1; k < num; k ++)
 87                 {
 88                     dp1[j][k] = dp2[j][k];
 89                     dp2[j][k] = -1;
 90                 }
 91             }
 92         }
 93         ans = 0;
 94         for(j = 1; j < num; j ++)
 95         {
 96             for(k = 1; k < num; k ++)
 97             {
 98                 ans = max(ans,dp1[j][k]);
 99             }
100         }
101         printf("%d\n",ans);
102     }
103     return 0;
104 }

 

posted @ 2013-04-01 11:04  Naix_x  阅读(164)  评论(0)  编辑  收藏  举报