[Pressure DP] POJ1185 Artillery Positions

Chinese question test instructions no longer repeat

For this "P" in the middle, we just need to consider the previous

It becomes just a matter of consideration:

That is, two lines before the pressure.

Specifically similar to HDOJ's 4539: see Hdoj 4539

It's just a different judgment of coexistence.

1 //#include <bits/stdc++.h>2#include <cstdio>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7typedefLong LongLL;8 9 intdp[2][1050][1050], mp[ the][ the];Ten intpos[1050], D; One intMain () A { -     intD=0; -      for(intI=0;i< (1<<Ten); i++) the         if(! (i& (i<<2)) &&! (i& (i<<1))) -Pos[d++]=i;//preprocess a row of eligible -     intN, M; -      while(~SCANF ("%d%d", &n, &m)) +     { -         intans=0; +Memset (MP,0,sizeof(MP)); A          for(intI=1; i<=n;i++) at              for(intj=0; j<m;j++) -             { -                 CharC; -Cin>>C; -Mp[i][j]= (c=='P'); -             } inMemset (DP,0,sizeof(DP)); -          for(intI=1; i<=n;i++)//Enumerate n rows to              for(intj=0; J<d && pos[j]< (1&LT;&LT;M); j + +) +             { -                 intsum=0; the                  for(intk=0; k<m;k++)//for the current (i,k position) whether the row (in front of the K-grid) satisfies *                     if(pos[j]&1<<k) $sum+=Mp[i][k];Panax Notoginseng                  for(intk=0; K<d && pos[k]< (1&LT;&LT;M); k++)//enumerate the state of a i-1 row to see if it can coexist with line I -                     if(! (pos[j]&Pos[k])) the                     { +                         inttmp=0; A                          for(intL=0; L<d && pos[l]< (1&LT;&LT;M); l++)//enumerate the state of i-2 rows to see if they can coexist with line I, line i-1 the                             if(! (pos[j]&Pos[l])) +Tmp=max (TMP, dp[1-i&1][k][l]); -dp[i&1][j][k]=tmp+sum; $Ans=max (ans, dp[i&1][j][k]); $                     } -             } -printf"%d\n", ans); the     } -     return 0;Wuyi}

POJ 1185

[Pressure DP] POJ1185 Artillery Positions