P4363 [九省联考2018]一双木棋chess

思路

容易发现只能在轮廓线的拐点处落子，所以棋盘的状态可以用一个n+m长度的二进制数表示
转移就是10变成01

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
// #define int long long
using namespace std;
int A[20][20],B[20][20],n,m;
//map<int,int> M,vis;
int M[(1<<21)],vis[(1<<21)];
void print(int x){
    for(int i=n+m-1;i>=0;i--)
        printf("%d",(x>>i)&1);
    putchar('\n');
}
int dfs(int state,int which){//0 feifei 1 niuniu
//	print(state);
//	printf("which:%d\n",which);
//	getchar();
    if(vis[state]){
//		printf("req=%d\n",M[state]);
        return M[state];
    }
    vis[state]=true;
    int x=m,y=0;
    int mid;
    if(!which)
        mid=-0x3f3f3f3f3f3f3f3fLL;
    else
        mid=0x3f3f3f3f3f3f3f3fLL;
    for(int i=0;i<n+m;i++){
        if((i+1)<n+m&&((state>>i)&1)==0&&((state>>(i+1))&1)==1){
//			printf("y=%d x=%d\n",y+1,x);
            int to=state^(3<<i);
            if(!which)
                mid=max(mid,dfs(to,which^1)+A[y+1][x]);
            else
                mid=min(mid,dfs(to,which^1)-B[y+1][x]);
//			printf("mid=%d\n",mid);
        }
        if((state>>i)&1)
            y++;
        else
            x--;
    }
    M[state]=mid;
//	printf("req=%d\n",M[state]);
    return mid;
}
signed main(){
//	freopen("test.in","r",stdin);
    scanf("%d %d",&n,&m);
//	int t=clock();
    int st=(((1<<(n))-1)<<m),end=(1<<n)-1;
//	print(st);
//	print(end);
    vis[end]=true;
    M[end]=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&A[i][j]);
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&B[i][j]);
        }
    }
    printf("%d\n",dfs(st,0));
//	printf("time=%lf\n",1.0*(clock()-t)/CLOCKS_PER_SEC);
    return 0;
}