poj1185-炮兵阵地（状态压缩dp）

  1 #include <map>
  2 
  3 #include <cstdio>
  4 
  5 #include <iostream>
  6 
  7 #include <string.h>
  8 
  9 #include <algorithm>
 10 
 11 using namespace std;
 12 
 13 const int maxn=100;
 14 
 15 int n,m,cnt,ans;
 16 
 17 char ch;
 18 
 19 int a[maxn+1];
 20 
 21 int vald[maxn+1][1<<10+1];
 22 
 23 int valdcount[maxn+1];
 24 
 25 int dp[maxn+1][maxn+10][maxn+10];
 26 
 27 //map<int,int> ma[maxn+10];
 28 
 29 bool judge(int x){
 30     return (!(x&(x<<1))&&!(x&(x<<2)));
 31 }
 32 
 33 int BitCount(unsigned int n)
 34 {
 35     unsigned int c =0 ;
 36     for (c =0; n; ++c)
 37     {
 38         n &= (n -1) ;
 39     }
 40     return c ;
 41 }
 42 
 43 int main()
 44 {
 45         scanf("%d%d",&n,&m);
 46         for(int i=1;i<=n;i++){
 47             for(int j=1;j<=m;j++){
 48                 scanf(" %c",&ch);
 49                 if(ch==‘H‘){
 50                     a[i]|=(1<<(j-1));
 51                 }
 52             }
 53         }
 54         for(int i=1;i<=n;i++){
 55             cnt=0;
 56             for(int j=0;j<=((1<<m)-1);j++){
 57                 if((!(j&a[i]))&&judge(j)){
 58                     vald[i][cnt]=j;
 59                     //ma[i][j]=cnt;
 60                     cnt++;
 61                 }
 62             }
 63             valdcount[i]=cnt-1;
 64         }
 65         for(int i=0;i<=valdcount[1];i++){
 66             int x=vald[1][i];
 67             for(int j=0;j<=(1<<(m)-1);j++) dp[1][i][j]=BitCount(x);
 68         }
 69         for(int i=0;i<=valdcount[2];i++){
 70             int x=vald[2][i];
 71             int q=BitCount(x);
 72             for(int j=0;j<=valdcount[1];j++) {
 73                 int y=vald[1][j];
 74                 if(!(x&y)) dp[2][i][j]=max(dp[2][i][j],q+BitCount(y));
 75             }
 76         }
 77         for(int i=3;i<=n;i++){
 78             for(int j=0;j<=valdcount[i];j++){
 79                    int x=vald[i][j];
 80                    int q=BitCount(x);
 81                   for(int k=0;k<=valdcount[i-1];k++){
 82                       int y=vald[i-1][k];
 83                     for(int h=0;h<=valdcount[i-2];h++){
 84                             int z=vald[i-2][h];
 85                            if(!(x&y)&&!(y&z)&&!(x&z)){
 86                                  dp[i][j][k]=max(dp[i][j][k],dp[i-1][k][h]+q);
 87                            }
 88                     }
 89                   }
 90             }
 91         }
 92         for(int i=0;i<=valdcount[n];i++){
 93             int x=vald[n][i];
 94             for(int j=0;j<=valdcount[n-1];j++) {
 95                     ans=max(ans,dp[n][i][j]);
 96             }
 97         }
 98         printf("%d\n",ans);
 99     return 0;
100 }