cf 429B
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题目链接 CODEFORCES 429B
这道题是看了题解做出来的,不多说,学习吧
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.
The first line of the input contains two integers n and m (3 ≤ n, m ≤ 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 ≤ a[i][j] ≤ 105).
The output contains a single number — the maximum total gain possible.
3 3100 100 100100 1 100100 100 100
800
Iahub will choose exercises a[1][1] → a[1][2] → a[2][2] → a[3][2] → a[3][3]. Iahubina will choose exercises a[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].
解题思路:先预处理出从四个顶点出发到任一点的最大值,用dp即可。然后 枚举相遇点,相遇点不可能在边缘,因为那样交点就肯定不止一个,对于每个交点有两种方式走,
求出两种方式最大值就可。
代码
#include<stdio.h>#include<string.h>#include<algorithm>#define zuida 1000using namespace std;int i,j,k,l,maxn,n,m,a[zuida+10][zuida+10];long long dp1[zuida+10][zuida+10],dp2[zuida+10][zuida+10],dp3[zuida+10][zuida+10],dp4[zuida+10][zuida+10];int main(void){ scanf("%d%d",&n,&m); for(i=1;i<=n;i++) for(j=1;j<=m;j++) scanf("%d",&a[i][j]); memset(dp1,0,sizeof(dp1)); memset(dp2,0,sizeof(dp2)); memset(dp3,0,sizeof(dp3)); memset(dp4,0,sizeof(dp4)); for(i=1;i<=n;i++) for(j=1;j<=m;j++) dp1[i][j]=max(dp1[i][j-1],dp1[i-1][j])+a[i][j]; for(i=1;i<=n;i++) for(j=m;j>=1;j--) dp2[i][j]=max(dp2[i][j+1],dp2[i-1][j])+a[i][j]; for(i=n;i>=1;i--) for(j=m;j>=1;j--) dp3[i][j]=max(dp3[i][j+1],dp3[i+1][j])+a[i][j]; for(i=n;i>=1;i--) for(j=1;j<=m;j++) dp4[i][j]=max(dp4[i][j-1],dp4[i+1][j])+a[i][j]; maxn=0; int temp1,temp2; for(i=2;i<n;i++) for(j=2;j<m;j++) { temp1=max(dp1[i][j-1]+dp3[i][j+1]+dp4[i+1][j]+dp2[i-1][j],dp1[i-1][j]+dp3[i+1][j]+dp4[i][j-1]+dp2[i][j+1]); maxn=max(temp1,maxn); } printf("%d",maxn); return 0;}
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