hdu2686 多决策dp

来回可以看作从(1,1)出发到(n,n)的两条不相交的路径

//dp[k][x1][y1][x2][y2] = max(dp[k-1][x1-1][y1][x2-1][y2],dp[k-1][x1-1][y1][x2][y2-1],dp[k-1][x1][y1-1][x2-1][y2],dp[k-1][x1][y1-1][x2][y2-1])
//k为步数
//k与x1,y1有关，可以降到三维

//多决策dp
//dp[k][x1][y1][x2][y2] = max(dp[k-1][x1-1][y1][x2-1][y2],dp[k-1][x1-1][y1][x2][y2-1],dp[k-1][x1][y1-1][x2-1][y2],dp[k-1][x1][y1-1][x2][y2-1])
//k为步数
//k与x1,y1有关，可以降到三维
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>

using namespace std;
const int inf = (1<<31)-1;
const int MAXN = 3e1+10;

int dp[2*MAXN][MAXN][MAXN];
int a[MAXN][MAXN];
int t1[2],t2[2];

int main()
{
    int n;
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&a[i][j]);
            }
        }
        memset(dp,0,sizeof(dp));
       // dp[0][1][1] = a[1][1];
        for(int i=1;i<=2*n-3;i++){
            for(int j=1;j<=i+1&&j<=n-1;j++){//x1满足
                    if(i+2-j>n)continue;//y1满足
                for(int k=j+1;k<=i+1&&k<=n;k++){ 
                    if(i+2-k>n-1)continue; 
                    t1[0] = j;
                    t1[1] = j-1;
                    t2[0] = k;
                    t2[1] = k-1;
                    for(int l=0;l<2;l++){
                        for(int r=0;r<2;r++){
                            if(t1[l]==t2[r])continue;
                            dp[i][j][k] = max(dp[i][j][k],dp[i-1][t1[l]][t2[r]]);
                        }
                    }
                    dp[i][j][k] += a[j][i+2-j]+a[k][i+2-k];
                   /* cout<<i<<" "<<j<<" "<<k<<" "<<dp[i][j][k]<<endl;
                    cout<<"debug"<<endl;*/
                }
            }
        }
        dp[2*n-2][n][n] = dp[2*n-3][n-1][n] + a[n][n];
        cout<<dp[2*n-2][n][n]+a[1][1]<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}