WOJ 1026

万恶的动态规划题. 其实这题已经属于DP里面比较基础的了. 由下面的公式来写程序:

DP[i][j]=board[i][j],i=0;

DP[i][j]=max(DP[i−1][j−1],DP[i−1][j],DP[i−1][j+1])+borad[i][j],i>=1

然后注意的一点就是, 写程序的时候棋盘最两边的那俩只能取上排的两个哦…不然数组要越界…
代码如下:

//// Created by Frank Yan on 2017/12/19.//#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <stdlib.h>#include <string>#include <vector>#include <queue>using namespace std;int problem26(){    int t;    cin>>t;    for(int k=0;k<t;k++){        int n;        cin>>n;        int board[n][n];        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                cin>>board[i][j];        int DP[n][n];        for(int j=0;j<n;j++)            DP[0][j] = board[0][j];        for(int i=1;i<n;i++)            for(int j=0;j<n;j++){                if(j!=0&&j!=n-1)                    DP[i][j] = max(max(DP[i-1][j],DP[i-1][j-1]),DP[i-1][j+1])+board[i][j];                else if(j==0) DP[i][j] =max(DP[i-1][j],DP[i-1][j+1])+board[i][j];                else DP[i][j] = max(DP[i-1][j],DP[i-1][j-1]);            }        //cout<<DP[1][1]<<endl;        int result = -1;        for(int i=0;i<n;i++)            if(DP[n-1][i]>result)                result = DP[n-1][i];        cout<<"Case "<<k+1<<":"<<endl;        cout<<result<<endl;        if(k!=t-1) cout<<endl;    }}