Brownie Slicing G

P3017 [USACO11MAR] Brownie Slicing G

题目描述

Bessie has baked a rectangular brownie that can be thought of as an RxC grid (1 <= R <= 500; 1 <= C <= 500) of little brownie squares. The square at row i, column j contains N_ij (0 <= N_ij <= 4,000) chocolate chips.

Bessie wants to partition the brownie up into A*B chunks (1 <= A <= R; 1 <= B <= C): one for each of the A*B cows. The brownie is cut by first making A-1 horizontal cuts (always along integer

coordinates) to divide the brownie into A strips. Then cut each strip *independently* with B-1 vertical cuts, also on integer

boundaries. The other A*B-1 cows then each choose a brownie piece, leaving the last chunk for Bessie. Being greedy, they leave Bessie the brownie that has the least number of chocolate chips on it.

Determine the maximum number of chocolate chips Bessie can receive, assuming she cuts the brownies optimally.

As an example, consider a 5 row x 4 column brownie with chips

distributed like this:

1 2 2 1
3 1 1 1
2 0 1 3
1 1 1 1
1 1 1 1

Bessie must partition the brownie into 4 horizontal strips, each with two pieces. Bessie can cut the brownie like this:

1 2 | 2 1
---------
3 | 1 1 1
---------
2 0 1 | 3
---------
1 1 | 1 1
1 1 | 1 1

Thus, when the other greedy cows take their brownie piece, Bessie still gets 3 chocolate chips.

输入格式

* Line 1: Four space-separated integers: R, C, A, and B

* Lines 2…R+1: Line i+1 contains C space-separated integers: N_i1, …, N_iC

输出格式

* Line 1: A single integer: the maximum number of chocolate chips that Bessie guarantee on her brownie

输入输出样例 #1

输入 #1

5 4 4 2 
1 2 2 1 
3 1 1 1 
2 0 1 3 
1 1 1 1 
1 1 1 1

输出 #1

3

代码内容

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll r,c,a,b;
ll m[501][501],s[501][501];

bool find(ll x)
{
    ll now=0,num=0;
    for(ll i=1;i<=r;i++)
    {
        ll dis=0,sum=0;
        for(ll j=1;j<=c;j++)
            if(dis+(s[i][j]-s[i][j-1])-(s[now][j]-s[now][j-1])<x) dis+=(s[i][j]-s[i][j-1])-(s[now][j]-s[now][j-1]);
            else sum++,dis=0;
        if(sum>=b) now=i,num++;

    }
    if(num<a) return 0;
    return 1; 
}

int main()
{
    cin>>r>>c>>a>>b;
    for(ll i=1;i<=r;i++)
        for(ll j=1;j<=c;j++)
            cin>>m[i][j];

    for(ll i=1;i<=r;i++)
        for(ll j=1;j<=c;j++)
            s[i][j]=s[i-1][j]+s[i][j-1]+m[i][j]-s[i-1][j-1];

    ll h=0,t=s[r][c];
    ll ans=0;
    while(h<=t)
    {
        ll mid=(h+t)/2;
        if(find(mid))
        {
            h=mid+1;
            ans=mid;
        }
        else t=mid-1;
    }
    
    cout<<ans<<endl;
    return 0;
}