POJ 1185 Artillery positions (state compression DP)

Description

Headquarters generals are planning to deploy their artillery units on the grid map of N*m. A n*m map consists of n rows of m columns, each of which may be mountainous (denoted by "H" ) or plain (with "P"

Input

The first line contains two positive integers separated by a space, representing N and M, and the next n rows, each containing a continuous m-character ('P' or 'H ' '  - .

Output

Only one row, containing an integer k, indicates the maximum number of artillery units that can be placed.

Sample Input

5 4 phpppphhppppphppphhp

Sample Output

6

Source

Noi State compression DP, similar to the previous poj3254

1 #pragmaComment (linker, "/stack:1024000000,1024000000")2#include <iostream>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <math.h>7#include <algorithm>8#include <queue>9#include <Set>Ten#include <bitset> One#include <map> A#include <vector> -#include <stdlib.h> - using namespacestd; the #defineMax (a) (a) > (b)? (a): (b) - #defineMin (a) (a) < (b)? (a): (b) - #definell Long Long - #defineEPS 1e-10 + #defineMOD 1000000007 - #defineN 106 + #defineM 16 A #defineINF 1e12 at intn,m; - CharMp[n][m]; - intCnt[n];//stored is the binary state of the given graph - intStk[n];//preprocessing all states that do not attack each other - intNum[n];//the number of 1 of STK stored - intDp[n][n][n];//Dp[i][j][k] indicates that the state at I is J, the maximum value when the state of I-1 is K in intTop//top indicates how many STK - BOOLOkintx) {//determine whether the state is legal, that is, there is no distance between adjacent 1 is less than 3 to     if(x& (x<<1))return false; +     if(x& (x<<2))return false; -     return true; the } * voidInit () {//Find all possible legal states, up to 60 kinds $top=0;Panax Notoginseng     intTotal=1<<m; -      for(intI=0; i<total;i++){ the         if(ok (i)) { +stk[top++]=i; A         } the     } + } - intCount_one (intx) {//Number of 1 in the binary of an integer x (for initialization) $     intans=0; $      while(x) { -ans+=x%2; -x>>=1; the     } -     returnans;Wuyi } the intMain () - { Wu      while(SCANF ("%d%d", &n,&m) = =2){ -         if(n==0&& m==0) About            Break; $            - init (); -          for(intI=0; i<n;i++){ -scanf"%s", Mp[i]); A           } +            the            for(intI=0; i<n;i++) {//Initialize first row -cnt[i]=0; $              for(intj=0; j<m;j++){ the                   if(mp[i][j]=='H'){ thecnt[i]+= (1<<j); the                   } the               } -           } in            thememset (dp,-1,sizeof(DP)); the            About            for(intI=0; i<top;i++){ thenum[i]=Count_one (Stk[i]); the             if((stk[i]&cnt[0])==0){ thedp[0][i][0]=Num[i]; +             } -           }  the           Bayi          /*for (int i=0;i<top;i++) { the if (stk[i]&cnt[1]) continue; the for (int j=0;j<top;j++) { - if (stk[j]&cnt[0]) continue; - if (stk[i]&stk[j]) continue; the Dp[1][i][j]=max (Dp[1][i][j],dp[0][j][0]+num[i]); the               } the           } the           *///this way, the second line is initialized, not initialized, and not even wrong . -            the            for(intI=1; i<n;i++){ the                for(intt=0; t<top;t++){ the                   if(Stk[t]&cnt[i])Continue;94                    for(intj=0; j<top;j++){ the                       if(Stk[j]&stk[t])Continue; the                        for(intk=0; k<top;k++){ the                           if(Stk[k]&stk[t])Continue;98                           if(dp[i-1][j][k]==-1)Continue; AboutDp[i][t][j]=max (dp[i][t][j],dp[i-1][j][k]+num[t]); -                       }101                   }102               }103           }104           intans=0; the            for(intI=0; i<n;i++){106                for(intj=0; j<top;j++){107                    for(intk=0; k<top;k++){108ans=Max (ans,dp[i][j][k]);109                   } the               }111           } theprintf"%d\n", ans);113     } the     return 0; the}

View Code

POJ 1185 Artillery positions (state compression DP)