Codefroces 429 B. Working out

Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.


Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].


There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.

If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.

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/*    这种题最好画画图找下规律,由于要两人见面,所以地图被划分成4块,而且有两种情况,    所以要从4个角向中间dp出4个表,之后找到交点四周dp值最大的那个,就是答案了.*/#include <iostream>#include <memory.h>#include <stdio.h>using namespace std;const int q=1005;long long int dp1[q][q],dp2[q][q],dp3[q][q],dp4[q][q];int main(){    memset(dp1,0,sizeof(dp1));    memset(dp2,0,sizeof(dp2));    memset(dp3,0,sizeof(dp3));    memset(dp4,0,sizeof(dp4));    int m,n,i,j;    long long int maxx=0;    cin>>m>>n;    for(i=1;i<=m;i++)        for(j=1;j<=n;j++)            {scanf("%lld",&dp1[i][j]);dp2[i][j]=dp3[i][j]=dp4[i][j]=dp1[i][j];}    for(i=1;i<=m;i++)    {        for(j=1;j<=n;j++)        {            dp1[i][j]+=max(dp1[i][j-1],dp1[i-1][j]);        }    }    for(i=1;i<=m;i++)    {        for(j=n;j>=1;j--)        {            dp2[i][j]+=max(dp2[i][j+1],dp2[i-1][j]);        }    }    for(i=m;i>=1;i--)    {        for(j=1;j<=n;j++)        {            dp3[i][j]+=max(dp3[i+1][j],dp3[i][j-1]);        }    }    for(i=m;i>=1;i--)    {        for(j=n;j>=1;j--)        {            dp4[i][j]+=max(dp4[i+1][j],dp4[i][j+1]);        }    }    for(i=2;i<m;i++)        {            for(j=2;j<n;j++)            {                maxx=max(maxx,dp1[i][j-1]+dp2[i-1][j]+dp3[i+1][j]+dp4[i][j+1]);                maxx=max(maxx,dp1[i-1][j]+dp2[i][j+1]+dp3[i][j-1]+dp4[i+1][j]);            }        }    cout<<maxx<<endl;    return 0;}