Codeforces Round #245 (Div. 1) B. Working out dp

题目链接：

http://codeforces.com/contest/429/problem/B

B. Working out

time limit per test2 seconds
memory limit per test256 megabytes #### 问题描述 > 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. #### 输入 > 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). #### 输出 > The output contains a single number — the maximum total gain possible. #### 样例 > **sample input** > 3 3 > 100 100 100 > 100 1 100 > 100 100 100 > > **sample output** > 800

题意

每个点有a[i][j]的物品（非负数），现在有一个人要从（1,1）到（n,m），且只能往下或者往右走，另一个人从(n,1)->(1,m)，且只能往上或往右走，现在要让这两个人的路线有且只有一个公共点（且公共点的物品谁都不能取），问如何规划使得两个人能够获得的物品总数最多。

题解

这题比较特殊的地方在公共点，从公共点(x,y)出发，你会发现我们把原问题划分成了四个简单的子问题：(1,1)->(x,y)，（n,1)->(x,y)，(n,m)->(x,y)，(1,m)->(x,y)，但是你发现枚举这个点是不够的，还需要枚举这个点的上下左右四个点（既如何安排这两个人进入x,y的入口，使得刚好能够只有（x,y）这一个公共点）

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e3+10;

int arr[maxn][maxn];
int dp[4][maxn][maxn];
int n,m;

void init(){
    clr(dp,0);
}

int main() {
    scf("%d%d",&n,&m);
    init();
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scf("%d",&arr[i][j]);
        }
    }
    //(1,1)->(n,m)
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            dp[0][i][j]=max(dp[0][i-1][j],dp[0][i][j-1])+arr[i][j];
        }
    }
    //(n,1)->(1,m)
    for(int i=n;i>=1;i--){
        for(int j=1;j<=m;j++){
            dp[1][i][j]=max(dp[1][i][j-1],dp[1][i+1][j])+arr[i][j];
        }
    }
    //(n,m)->(1,1)
    for(int i=n;i>=1;i--){
        for(int j=m;j>=1;j--){
            dp[2][i][j]=max(dp[2][i+1][j],dp[2][i][j+1])+arr[i][j];
        }
    }
    //(1,m)->(n,1)
    for(int i=1;i<=n;i++){
        for(int j=m;j>=1;j--){
            dp[3][i][j]=max(dp[3][i-1][j],dp[3][i][j+1])+arr[i][j];
        }
    }

    int ans=-1;
    //枚举相遇点
    for(int i=2;i<n;i++){
        for(int j=2;j<m;j++){
            ans=max(ans,dp[0][i-1][j]+dp[2][i+1][j]+dp[1][i][j-1]+dp[3][i][j+1]);
            ans=max(ans,dp[0][i][j-1]+dp[2][i][j+1]+dp[1][i+1][j]+dp[3][i-1][j]);
        }
    }

    printf("%d
",ans);
    return 0;
}

//end-----------------------------------------------------------------------