OJ-Problems-Source/POJ/1185.cpp

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2016-08-03 15:50:31 +08:00
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define _BITMOVE(x) (1<<x)
#define _BITMAX(x) _BITMOVE(x)
#define _BITMAXVALUE(x) (_BITMAX(x)-1)
#define MAXLINE 105
#define MAXSTATUS _BITMAX(12)
#define ESTIMATED_MAXSTATUS 129+1
typedef unsigned long STATE;
STATE map[MAXLINE];
int dp[MAXLINE][ESTIMATED_MAXSTATUS][ESTIMATED_MAXSTATUS];
STATE availableStatus[ESTIMATED_MAXSTATUS];
int countOfOne[ESTIMATED_MAXSTATUS];
char _tmpbuff[8];
inline bool isGoodState(STATE x)
{
if(x&(x<<1)) return false;
if(x&(x<<2)) return false;
return true;
}
inline STATE PlaceOnMap(int line,int stateid)
{
return map[line]&availableStatus[stateid];
}
inline STATE CheckBetweenLines(int line1,int line2)
{
return availableStatus[line1]&availableStatus[line2];
}
inline int getCountOfOneByState(STATE x)
{
int c=0;
while(x>0)
{
if(x&1) ++c;
x=x>>1;
}
return c;
}
int solve(int n,int m)
{
memset(dp,-1,sizeof(int)*MAXLINE*ESTIMATED_MAXSTATUS*ESTIMATED_MAXSTATUS);
memset(map,0,sizeof(STATE)*MAXLINE);
for(int i=0;i<n;i++)
{
gets(_tmpbuff);
for(int j=0;j<m;j++)
{
int tmp=(_tmpbuff[j]=='P')?0:1;
map[i]=(map[i]<<1)|tmp;
}
}
if(n==1&&m==1)
{
if(map[0]>0) printf("0\n");
else printf("1\n");
return 0;
}
int statusCount=-1;
for(STATE ci=0;ci<(STATE)_BITMAX(m);ci++)
{
if(isGoodState(ci))
{
availableStatus[++statusCount]=ci;
countOfOne[statusCount]=getCountOfOneByState(ci);
}
}
++statusCount;
/// Initialize the Line 0 with all available status.
for(int ci=0;ci<statusCount;ci++)
{
if(!PlaceOnMap(0,ci))
{
dp[0][ci][0]=countOfOne[ci];
}
}
/// Main DP
for(int i=1;i<n;i++)
{
for(int j=0;j<statusCount;j++)
{
if(PlaceOnMap(i,j)) continue;
/// Last Line
for(int k=0;k<statusCount;k++)
{
if(CheckBetweenLines(j,k)) continue;
/// Last Last Line
for(int x=0;x<statusCount;x++)
{
if(CheckBetweenLines(k,x)) continue;
if(CheckBetweenLines(j,x)) continue;
if(dp[i-1][k][x]==-1) continue;
dp[i][j][k]=max(dp[i][j][k],dp[i-1][k][x]+countOfOne[j]);
}
}
}
}
/// Calculate the answer
int ans=-1;
for(int ci=0;ci<statusCount;ci++)
{
for(int cj=0;cj<statusCount;cj++)
{
///此处是max运算,因为结果要求最大值
ans=max(ans,dp[n-1][ci][cj]);
}
}
printf("%d\n",ans);
return 0;
}
int main()
{
/**
/// A Program To Estimate the max value of statusCount
/// The answer is 129
int statusCount=-1;
for(STATE ci=0;ci<MAXSTATUS;ci++)
{
if(isGoodState(ci))
{
availableStatus[++statusCount]=ci;
countOfOne[statusCount]=getCountOfOneByState(ci);
}
}
++statusCount;
printf("%d\n",statusCount);
*/
int n,m;
/// %*c重要
while(scanf("%d %d%*c",&n,&m)==2)
{
solve(n,m);
}
return 0;
}