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  • 双线程dp

    zznu 1625

    详解链接:http://blog.csdn.net/su20145104009/article/details/51277041

    下面是四维dp

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<cctype>
    #include<algorithm>
    #include <vector>
    #include <queue>
    
    using namespace std;
    #define INF 0x3f3f3f3f
    #define N 55
    
    int maps[N][N];
    int dp[N][N][N][N];
    
    int main()
    {
        int T, m, n;
        scanf("%d", &T);
    
        while(T--)
        {
            scanf("%d%d", &m, &n);
            for(int i=1; i<=m; i++)
            for(int j=1; j<=n; j++)
                scanf("%d", &maps[i][j]);
    
            memset(dp, 0, sizeof(dp));
    
            for(int i=1; i<=m; i++)
            for(int j=1; j<=n; j++)
            for(int k=i+1; k<=m; k++)
            {
                int l=i+j-k;
                if(l<0 || l>n) break;
                dp[i][j][k][l]=max(max(dp[i-1][j][k][l-1], dp[i-1][j][k-1][l]),
                                   max(dp[i][j-1][k-1][l], dp[i][j-1][k][l-1]))+maps[i][j]+maps[k][l];
            }
            printf("%d
    ", max(max(dp[m-1][n][m][n-1], dp[m-1][n][m-1][n]),
                                   max(dp[m][n-1][m-1][n], dp[m][n-1][m-1][n]))+maps[m][n]);
        }
        return 0;
    }

     如果n和m太大就要根据条件,降维为三维,这样回省很多的空间

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<cctype>
    #include<algorithm>
    #include <vector>
    #include <queue>
    
    using namespace std;
    #define INF 0x3f3f3f3f
    #define N 110
    #define max4(a, b, c, d) max(max(a, b), max(c, d))
    
    int a[N][N];
    int dp[210][N][N];
    
    int main()
    {
        int T, n, m;
        scanf("%d", &T);
    
        while(T--)
        {
            scanf("%d%d", &m, &n);
            for(int i=1; i<=m; 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];//走零步所得价值
            for(int k=1; k<=m+n-2; k++)
            for(int i=1; i<=k+1; i++)
            for(int j=1; j<=k+1; j++)
            {
                dp[k][i][j]=max4(dp[k-1][i][j], dp[k-1][i][j-1], dp[k-1][i-1][j], dp[k-1][i-1][j-1]);//走k步的状态可由k-1的4个状态得来
                if(i!=j)
                    dp[k][i][j]+=a[i][k+2-i]+a[j][k+2-j];//不同点
                else
                    dp[k][i][j]+=a[i][k+2-i];//同一个点
            }
            printf("%d
    ", dp[m+n-2][m][m]);
        }
        return 0;
    }
    下面代码对于有些题目有可能是不对的,用上面的三维的最好
    #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<cctype> #include<algorithm> #include <vector> #include <queue> using namespace std; #define INF 0x3f3f3f3f #define N 55 int maps[N][N]; int dp[110][N][N]; int main() { int T, m, n, step; scanf("%d", &T); while(T--) { scanf("%d%d", &m, &n); for(int i=1; i<=m; i++) for(int j=1; j<=n; j++) scanf("%d", &maps[i][j]); step=m+n-2; memset(dp, 0, sizeof(dp)); for(int i=1; i<=m; i++) for(int j=i+1; j<=m; j++) for(int k=1; k<step; k++) { if(k+2>=i&&k+2>=j) { dp[k][i][j]=max(max(dp[k-1][i-1][j], dp[k-1][i-1][j-1]), max(dp[k-1][i][j], dp[k-1][i][j-1]))+maps[i][k+2-i]+maps[j][k+2-j]; } } printf("%d ", max(max(dp[step-1][m][m-1], dp[step][m][m]), max(dp[step-1][m-1][m], dp[step-1][m-1][m-1]))+maps[m][n]); } return 0; }
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  • 原文地址:https://www.cnblogs.com/9968jie/p/5532638.html
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