题解:nState为状态数,state数组为可能的状态
代码:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; const int INF=0x3f3f3f3f;
typedef long long ll;
typedef unsigned long long ull;
#define fi first
#define se second
#define prN printf("\n")
#define SI(N) scanf("%d",&(N))
#define SII(N,M) scanf("%d%d",&(N),&(M))
#define SIII(N,M,K) scanf("%d%d%d",&(N),&(M),&(K))
#define cle(a,val) memset(a,(val),sizeof(a))
#define rep(i,b) for(int i=0;i<(b);i++)
#define Rep(i,a,b) for(int i=(a);i<=(b);i++) int n,m,dp[105][80][80];
int row[105];
int nState,state[80],num[80]; void init()
{
int k=1<<m;
nState=0;
rep(i,k)
{
if ((i&(i<<1))==0&&(i&(i<<2))==0)
{
state[nState]=i;
num[nState]=0;
int j=i;
while(j)
{
num[nState]+=j%2;
j/=2;
}
nState++;
}
}
} int main()
{
#ifndef ONLINE_JUDGE
freopen("C:\\Users\\Zmy\\Desktop\\in.txt","r",stdin);
// freopen("C:\\Users\\Zmy\\Desktop\\out.txt","w",stdout);
#endif // ONLINE_JUDGE char str[20]= {0}; while(cin>>n>>m)
{
init();
rep(i,n)
{
row[i]=0;
scanf("%s",str);
rep(j,m)
{
if (str[j]=='P')
{
row[i]+=1<<j;
}
}
} cle(dp,0); rep(j,nState)
{
if ((state[j]&row[0])!=state[j])
{
continue;
}
rep(k,nState)
{
dp[0][j][k]=num[j];
}
} if (n>1)
rep(j,nState)
{
if ((state[j]&row[1])!=state[j])
{
continue;
}
rep(k,nState)//这的k代表的是上上个
{
if ((state[j]&state[k])==0)
{
dp[1][j][k]=dp[0][k][0] + num[j];/**< 这是啥意思?? */
}
}
} Rep(i,2,n-1)
{
rep(j,nState)
{
if ((state[j]&row[i])!=state[j])
{
continue;
}
rep(k,nState)//这的k代表的是上上个
{
if ( state[j] & state[k] )
continue;
for (int h = 0; h < nState; h++)/**< i是这个,j是上1,k是上2,h是上3 */
{
if ( state[j] & state[h] )
continue;
if ( dp[i-1][k][h] > dp[i][j][k] )
dp[i][j][k] = dp[i-1][k][h];
}
dp[i][j][k] += num[j]; }
}
}
int maxa = 0;
for (int j = 0; j < nState; j++)
{
for (int k = 0; k < nState; k++)
if (maxa < dp[n-1][j][k])
maxa = dp[n-1][j][k];
} printf("%d\n", maxa); } return 0;
}