Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 16619 | Accepted: 6325 |
Description
Input
Output
Sample Input
5 4 PHPP PPHH PPPP PHPP PHHP
Sample Output
6
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 7 int n,m; 8 char tom[103][12]; 9 int a[103]; 10 int state[62],len; 11 int Knum[62]; 12 int dp1[61][61],dp2[61][61]; 13 14 void make_init() 15 { 16 int i,k,ans,x; 17 k=1<<10; 18 len=0; 19 for(i=0;i<k;i++) 20 { 21 if( (i&(i<<1)) || (i&(i>>1)) || (i&(i<<2)) || (i&(i>>2)) ) continue; 22 state[len]=i; 23 x=i; 24 ans=0; 25 while(x)//每一种状态能放几个兵。 26 { 27 ++ans; 28 x=(x-1)&x; 29 } 30 Knum[len++]=ans; 31 } 32 } 33 void make_then() 34 { 35 int i,j,k,s,hxl,x; 36 k=1<<m; 37 memset(dp1,0,sizeof(dp1)); 38 memset(dp2,0,sizeof(dp2)); 39 for(i=1;i<=n;i++) 40 { 41 for(j=0;j<len&&state[j]<k;j++) 42 { 43 if( (a[i]&state[j])==0 ) //限制了,山地。 44 for(s=0;s<len&&state[s]<k;s++)//前一xing状态 45 { 46 if( (state[j]&state[s])>0 ) continue; // 当前状态和前一矛盾 47 hxl=0; 48 for(x=0;x<len&&state[x]<k;x++)//前2行的状态 49 { 50 if( (state[s]&state[x])>0 || (state[j]&state[x])>0 ) continue; 51 if(hxl<dp1[s][x]) 52 hxl=dp1[s][x]; 53 } 54 dp2[j][s]=hxl+Knum[j]; 55 } 56 } 57 for(j=0;j<len&&state[j]<k;j++) 58 { 59 for(s=0;s<len&&state[s]<k;s++) 60 { 61 dp1[j][s]=dp2[j][s]; 62 dp2[j][s]=0; 63 } 64 } 65 } 66 } 67 int main() 68 { 69 int i,j,k,hxl; 70 make_init();//预处理. 71 while(scanf("%d%d",&n,&m)>0) 72 { 73 for(i=1;i<=n;i++) 74 scanf("%s",tom[i]+1); 75 for(i=1;i<=n;i++) 76 { 77 a[i]=0; 78 for(j=1;j<=m;j++) 79 { 80 a[i]=a[i]<<1; 81 if(tom[i][j]=='H') 82 a[i]+=1; 83 } 84 }//! 85 make_then(); 86 hxl=0; 87 k=1<<m; 88 for(i=0;i<len&&state[i]<k;i++) 89 for(j=0;j<len&&state[j]<k;j++) 90 if(dp1[i][j]>hxl) hxl=dp1[i][j]; 91 printf("%d\n",hxl); 92 } 93 return 0; 94 }