2019 Multi-University Training Contest 6 Snowy Smile （最大子段和）

Snowy Smile

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2997    Accepted Submission(s): 954

 

Problem Description

There are n pirate chests buried in Byteland, labeled by 1,2,…,n. The i-th chest's location is (xi,yi), and its value is wi, wi can be negative since the pirate can add some poisonous gases into the chest. When you open the i-th pirate chest, you will get wi value.

You want to make money from these pirate chests. You can select a rectangle, the sides of which are all paralleled to the axes, and then all the chests inside it or on its border will be opened. Note that you must open all the chests within that range regardless of their values are positive or negative. But you can choose a rectangle with nothing in it to get a zero sum.

Please write a program to find the best rectangle with maximum total value.

 

 

Input

The first line of the input contains an integer T(1≤T≤100), denoting the number of test cases.

In each test case, there is one integer n(1≤n≤2000) in the first line, denoting the number of pirate chests.

For the next n lines, each line contains three integers xi,yi,wi(−109≤xi,yi,wi≤109), denoting each pirate chest.

It is guaranteed that ∑n≤10000.

 

 

Output

For each test case, print a single line containing an integer, denoting the maximum total value.

 

 

Sample Input

2 4 1 1 50 2 1 50 1 2 50 2 2 -500 2 -1 1 5 -1 1 1

 

 

Sample Output

100 6

 

 

Source

2019 Multi-University Training Contest 6

 

 

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题目大意：给一个矩阵，求一个子矩阵的和最大。

解题思路：将x离散化后，我们可以枚举左右边界。然后对y轴求一个最大子段和就可以了。

刚开始的时候pushup函数写错了。。。。。。

#include
#define LL long long
using namespace std;
#define pb(x) push_back(x)
#define sca(x) scanf("%d",&x)
#define mp(x,y) make_pair(x,y)

const int N = 2000+5;
struct poi
{
    int x,y,w;
} poi[2005];
vectorX,Y;
typedef pairpii;
vectorb[N];


void pre_work(int n)
{
    for(int i=0; i>1;
    if(pos<=m) upd(rt<<1,l,m,pos,val);
    else upd(rt<<1|1,m+1,r,pos,val);
    push_up(rt);
}

int main()
{
    int _t;
    cin>>_t;
    while(_t--)
    {
        int n;
        sca(n);
        X.clear();
        Y.clear();
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d",&poi[i].x,&poi[i].y,&poi[i].w);
            X.pb(poi[i].x);
            Y.pb(poi[i].y);
        }
        pre_work(n);
        int len1=X.size();
        int len2=Y.size();
        LL ans=0;

        for(int i=1; i<=len1; i++)
        {
            memset(t,0,sizeof(t));
            for(int j=i; j<=len1; j++)
            {
                for(int k=0; k