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  • Codeforces Round #FF (Div. 2) D. DZY Loves Modification 优先队列

    D. DZY Loves Modification
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.

    Each modification is one of the following:

    1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
    2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.

    DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.

    Input

    The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).

    Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.

    Output

    Output a single integer — the maximum possible total pleasure value DZY could get.

    Examples
    input
    2 2 2 2
    1 3
    2 4
    output
    11
    input
    2 2 5 2
    1 3
    2 4
    output
    11
    Note

    For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:


    1 1
    0 0

    For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:


    -3 -3
    -2 -2

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<bitset>
    #include<set>
    #include<map>
    #include<time.h>
    using namespace std;
    #define LL long long
    #define bug(x)  cout<<"bug"<<x<<endl;
    const int N=1e3+10,M=1e6+10,inf=1e9+10;
    const LL INF=1e18+10,mod=1e9+7;
    const double eps=(1e-8),pi=(4*atan(1.0));
    
    LL a[N][N];
    priority_queue<LL>col,row;
    LL r[M],c[M];
    int main()
    {
        int n,m,k,p;
        scanf("%d%d%d%d",&n,&m,&k,&p);
        for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%lld",&a[i][j]);
        for(int i=1;i<=n;i++)
        {
            int r=0;
            for(int j=1;j<=m;j++)
            r+=a[i][j];
            row.push(r);
        }
        for(int j=1;j<=m;j++)
        {
            int c=0;
            for(int i=1;i<=n;i++)
            c+=a[i][j];
            col.push(c);
        }
        for(int i=1;i<=k;i++)
        {
            LL x=row.top();
            row.pop();
            r[i]=r[i-1]+x;
            x-=m*p;
            row.push(x);
        }
        for(int i=1;i<=k;i++)
        {
            LL x=col.top();
            col.pop();
            c[i]=c[i-1]+x;
            x-=n*p;
            col.push(x);
        }
        LL ans=-INF;
        for(int i=0;i<=k;i++)
            ans=max(ans,r[i]+c[k-i]-1LL*p*i*(k-i));
        printf("%lld
    ",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/7528196.html
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