典型的狀態(tài)壓縮題，只是要定義dp[i][j][k]，

#include<bits/stdc++.h>
using namespace std;

int n,m,state[1<<10],tot=0;

int dp[110][66][66],sum[1<<10],mp[110];

int getsum(int x){
??? int cnt=0;
??? while(x){
??????? if(x&1) cnt++;
??????? x>>=1;
??? }
??? return cnt;
}

bool check(int x)
{
??? if(x&(x>>1)) return 0;
??? if(x&(x>>2)) return 0;
??? return 1;
}

int main(){
??? memset(dp,0,sizeof(dp));
??? scanf("%d%d",&n,&m);
??? for(int i=0;i<(1<<m);i++){
??????? if(check(i))
??????? {
??????????? state[++tot]=i;
??????????? sum[tot]=getsum(i);
??????? }
??? }
??? for(int i=1;i<=n;i++) for(int j=0;j<m;j++)
??? {
??????? char ch=getchar();
??????? while(ch!='P' && ch!='H') ch=getchar();
??????? if(ch=='H') mp[i]+=(1<<(m-1-j));
??? }
??? //處理第一行
??? for(int i=1;i<=tot;i++){
??????? if(state[i]&mp[1]) continue;
??????? dp[1][i][0]=sum[i];
??? }
??? //處理第二行
??? for(int i=1;i<=tot;i++){
??????? if(mp[2]&state[i]) continue;
??????? for(int j=1;j<=tot;j++){
??????????? if(mp[1]&state[j]) continue;
??????????? if(state[i]&state[j]) continue;
??????????? dp[2][i][j]=max(dp[2][i][j],dp[1][j][0]+sum[i]);
??????? }
??? }
?? ?
??? for(int i=3;i<=n;i++){
??????? for(int k1=1;k1<=tot;k1++){
??????????? if(state[k1]&mp[i]) continue;
??????????? for(int k2=1;k2<=tot;k2++){
??????????????? if(state[k2]&mp[i-1]) continue;
??????????????? if(state[k2]&state[k1]) continue;
??????????????? for(int k3=1;k3<=tot;k3++){
??????????????????? if(state[k3]&mp[i-2]) continue;
??????????????????? if(state[k3]&state[k2]) continue;
??????????????????? if(state[k3]&state[k1]) continue;
??????????????????? dp[i][k1][k2]=max(dp[i][k1][k2],dp[i-1][k2][k3]+sum[k1]);
??????????????? }
??????????? }
??????? }
??? }
??? int ans=0;
??? for(int i=1;i<=tot;i++){
??????? for(int j=1;j<=tot;j++){
??????????? ans=max(ans,dp[n][i][j]);
??????? }
??? }
??? printf("%d",ans);
??? return 0;
}

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i指所在的行號(hào)，j指當(dāng)前行的狀態(tài)，k指上一行的狀態(tài)

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轉(zhuǎn)載于:https://www.cnblogs.com/plysc/p/10300301.html

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