题意：给定一个n*m的矩阵，可以在矩阵内将A个元素改成0，现在要在矩阵内选出B个不相交的宽为m的矩阵，使得这3个矩阵的和最大
这个可以将所有行压成一个点然后变成序列上的问题，修改可以预处理，显然对每行来说，应优先修改最小的元素，那么就可以预处理出a[i][j]，代表第i行修改j次后的权值。显然修改的效果会越来越小，所以f(j)=a[i][j]是一个上凸函数。。
然后设d[p][i][j]为到第i行，选了p个矩形(第p个矩形以第i行结尾)，修改了j次的最大值
那么d[p][i][j]=max{max{d[p][i-1][k]+a[i][j-k]},max{d[p-1][v][k]+a[i][j-k]}}
=max{max(d[p][i-1][k],d[p-1][v][k])+a[i][j-k]}
那么设g[p][i][k]=max(d[p][i][k],d[p-1][v][k])
d[p][i][j]=max{g[p][i][k]+a[i][j-k]}
复杂度是O(TnABm)，显然会T
而由于a的性质比较特殊，所以这个决策具有单调性
对于2个决策点k<v<j，如果g[p][i][k]+a[i][j-k]<g[p][i][v]+a[i][j-v]，由于a[i][x]为上凸函数，所以随着j增加，v一直比k优，因此可以用单调队列维护决策，用二分找到决策区间就可以了。。
/*
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1LL<<(x))
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 105
#define nm 10005
#define pi 3.1415926535897931
using namespace std;
const ll inf=1e9+7;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
int q[nm],qh,qt,b[NM],n,tot,m,_p,v[nm],f[nm];
ll a[NM][3005],g[NM][nm],d[NM][nm],ans;
int main(){
//freopen("data.in","r",stdin);
int _=read();while(_--){
n=read();tot=read();m=read();_p=read();
inc(i,1,n){
a[i][0]=0;
inc(j,1,tot)b[j]=read(),a[i][0]+=b[j];
sort(b+1,b+tot+1);
inc(j,1,tot)a[i][j]=a[i][j-1]+max(0,-b[j]);
}
//inc(i,1,n){inc(j,0,tot)printf("%lld ",a[i][j]);putchar('\n');}
mem(g);ans=0;
inc(i,1,n)inc(j,0,m)d[i][j]=-inf;
inc(p,1,_p){
inc(i,1,n){
q[qh=qt=1]=0;mem(f);mem(v);
d[i][0]=max(d[i][0],g[i-1][0]+a[i][0]);
g[i][0]=max(g[i][0],d[i][0]);
inc(j,1,m){
f[j]=max(f[j],f[j-1]);
while(qh<=qt&&f[j]!=q[qh])qh++;
int s=j;
while(qh<=qt){
s=q[qt]+tot+1;
for(int x=j,y=min(m,q[qt]+tot);x<=y;)
if(g[i-1][q[qt]]+a[i][mid-q[qt]]<=g[i-1][j]+a[i][mid-j])s=mid,y=mid-1;else x=mid+1;
if(s<=v[q[qt]])qt--;else break;
}
if(s<=m)v[j]=s,f[s]=j,q[++qt]=j;
while(qh<=qt&&f[j]!=q[qh])qh++;
//printf("%d %d:%d\n",i,j,q[qh]);
d[i][j]=max(d[i][j],g[i-1][q[qh]]+a[i][j-q[qh]]);
g[i][j]=max(g[i][j],d[i][j]);
}
}
//inc(i,1,n){inc(j,0,m)printf("%lld ",d[i][j]);putchar('\n');}putchar('\n');
inc(i,1,n)inc(j,0,m)g[i][j]=max(g[i][j],g[i-1][j]);
}
inc(j,0,m)ans=max(ans,g[n][j]);
printf("%lld\n",ans);
}
return 0;
}
Problem E. Find The Submatrix
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 43 Accepted Submission(s): 15
Problem Description
Little Q is searching for the submatrix with maximum sum in a matrix of n rows and m columns. The standard algorithm is too hard for him to understand, so he (and you) only considers those submatrixes with exactly m columns.
It is much easier now. But Little Q always thinks the answer is too small. So he decides to reset no more than A cells' value to 0, and choose no more than B disjoint submatrixes to achieve the maximum sum. Two submatrix are considered disjoint only if they do not share any common cell.
Please write a program to help Little Q find the maximum sum. Note that he can choose nothing so the answer is always non-negative.
Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there are 4 integers n,m,A,B(1≤n≤100,1≤m≤3000,0≤A≤10000,1≤B≤3).
Each of the following n lines contains m integers, the j-th number on the i-th of these lines is wi,j(|wi,j|≤109), denoting the value of each cell.
Output
For each test case, print a single line containing an integer, denoting the maximum sum.
Sample Input
2 5 1 0 1 3 -1 5 -1 -2 5 1 1 1 3 -1 5 -1 -2
Sample Output
7 8
Source
2018 Multi-University Training Contest 3
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