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.
原題:點擊打開鏈接
/*
這種題最好畫畫圖找下規律,由於要兩人見面,所以地圖被劃分成4塊,而且有兩種情況,
所以要從4個角向中間dp出4個表,之後找到交點四周dp值最大的那個,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;
}