标签：acm   dp   

方格取数

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 5

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Problem Description

Input

Output

Sample Input

8
2 3 13
2 6 6
3 5 7
4 4 14
5 2 21
5 6 4
6 3 15
7 2 14
0 0 0

Sample Output

67

Author

// 二取方格数     多进程dp

我们令 dp[x1,y1,x2,y2]表示第一条路径走到了（x1,y1）,第二条路径走到了(x2,y2)是的最优解，可以很自然的得到状态转移方程： 

设p=max(sum[i1-1,j1,i2-1,j2],sum[i1-1,j1,i2,j2-1],sum[i1,j1-1,i2-1,j2],sum[i1,j1-1,i2,j2-1])
dp[x1][y1][x2][y2]= 0 (x1||y1||x2||y2==0)
                  = p+data[x1][y1]   (x1==x2&&y1==y2)
                  = p+data[x1][y1]+data[x2][y2] (x1!=x2||y1!=y2)

#include<cstdio>
#include<cstring>
int dp[15][15][15][15],p[15][15];
int main()
{
    int n,i1,j1,i2,j2,x,y,z;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(p,0,sizeof(p));
        while(1)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(x+y+z==0) break;
            p[x][y]=z;
        }
        for(i1=1;i1<=n;i1++)
            for(j1=1;j1<=n;j1++)
                for(i2=1;i2<=n;i2++)
                    for(j2=1;j2<=n;j2++)
                    {
                        if(dp[i1-1][j1][i2-1][j2]>dp[i1][j1][i2][j2])
                            dp[i1][j1][i2][j2]=dp[i1-1][j1][i2-1][j2];
                        if(dp[i1-1][j1][i2][j2-1]>dp[i1][j1][i2][j2])
                            dp[i1][j1][i2][j2]=dp[i1-1][j1][i2][j2-1];
                        if(dp[i1][j1-1][i2-1][j2]>dp[i1][j1][i2][j2])
                            dp[i1][j1][i2][j2]=dp[i1][j1-1][i2-1][j2];
                        if(dp[i1][j1-1][i2][j2-1]>dp[i1][j1][i2][j2])
                            dp[i1][j1][i2][j2]=dp[i1][j1-1][i2][j2-1];
                        dp[i1][j1][i2][j2]+=p[i1][j1];
                        if(i1!=i2||j1!=j2)
                            dp[i1][j1][i2][j2]+=p[i2][j2];
                    }
        printf("%d\n",dp[n][n][n][n]);
    }
    return 0;
}

方格取数

标签：acm   dp   

原文地址：http://blog.csdn.net/u012773338/article/details/38061189