Artillery positions
Time Limit: 2000MS |
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Memory Limit: 65536K |
Total Submissions: 25647 |
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Accepted: 9892 |
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 (denoted by "P"), such as. A maximum of one artillery unit can be arranged on each Goping terrain (no artillery units can be deployed on the mountain); an artillery unit is shown in the black area of the attack range on the map:
If you deploy an artillery unit on a plain that is identified in gray on the map, the black grid in the figure represents the area it can attack: two grids along the horizontal, two in the vertical. Other white meshes on the graph are not attacked. It can be seen from the diagram that the artillery's attack range is unaffected by the topography.
Now, the generals plan how to deploy artillery units, in the premise of preventing accidental injury (to ensure that no two artillery units can attack each other, that is, any artillery unit is not within the range of other artillery units), the maximum number of troops in the entire map area can be placed in the Army of artillery units.
Input
The first line contains two positive integers separated by a space, representing N and M, respectively;
In the next n rows, each row contains a continuous m-character (' P ' or ' H ') with no spaces in the middle. Represents the data for each row in the map sequentially. N <= 100;m <= 10.
Output
Only one row, containing an integer k, indicates the maximum number of artillery units that can be placed.
Sample Input
5 4PHPPPPHHPPPPPHPPPHHP
Sample Output
6
1#include <map>2 3#include <cstdio>4 5#include <iostream>6 7#include <string.h>8 9#include <algorithm>Ten One using namespacestd; A - Const intmaxn= -; - the intN,m,cnt,ans; - - Charch; - + inta[maxn+1]; - + intvald[maxn+1][1<<Ten+1]; A at intvaldcount[maxn+1]; - - intdp[maxn+1][maxn+Ten][maxn+Ten]; - - //map<int,int> ma[maxn+10]; - in BOOLJudgeintx) { - return(! (x& (x<<1)) &&! (x& (x<<2))); to } + - intBitcount (unsignedintN) the { *Unsignedintc =0 ; $ for(c =0; N ++c)Panax Notoginseng { -n &= (N-1) ; the } + returnC; A } the + intMain () - { $scanf"%d%d",&n,&m); $ for(intI=1; i<=n;i++){ - for(intj=1; j<=m;j++){ -scanf"%c",&ch); the if(ch=='H'){ -A[i]|= (1<< (J-1));Wuyi } the } - } Wu for(intI=1; i<=n;i++){ -Cnt=0; About for(intj=0; J<= (1<<M)-1); J + +){ $ if((! (J&a[i])) &&judge (J)) { -vald[i][cnt]=J; - //ma[i][j]=cnt; -cnt++; A } + } thevaldcount[i]=cnt-1; - } $ for(intI=0; i<=valdcount[1];i++){ the intx=vald[1][i]; the for(intj=0; J<= (1<< (M)-1); J + +) dp[1][i][j]=Bitcount (x); the } the for(intI=0; i<=valdcount[2];i++){ - intx=vald[2][i]; in intq=Bitcount (x); the for(intj=0; j<=valdcount[1];j++) { the inty=vald[1][j]; About if(! (X&y)) dp[2][i][j]=max (dp[2][i][j],q+Bitcount (y)); the } the } the for(intI=3; i<=n;i++){ + for(intj=0; j<=valdcount[i];j++){ - intx=Vald[i][j]; the intq=Bitcount (x);Bayi for(intk=0; k<=valdcount[i-1];k++){ the inty=vald[i-1][k]; the for(intH=0; h<=valdcount[i-2];h++){ - intz=vald[i-2][h]; - if(! (x&y) &&! (y&z) &&! (x&z)) { theDp[i][j][k]=max (dp[i][j][k],dp[i-1][k][h]+q); the } the } the } - } the } the for(intI=0; i<=valdcount[n];i++){ the intx=Vald[n][i];94 for(intj=0; j<=valdcount[n-1];j++) { theans=Max (ans,dp[n][i][j]); the } the }98printf"%d\n", ans); About return 0; -}
poj1185-Artillery positions (state compression DP)