YJJ's Salesman

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1990    Accepted Submission(s): 741

 

Problem Description

YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109). YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109), he will only forward to (x+1,y), (x,y+1) or (x+1,y+1).
On the rectangle map from (0,0) to (109,109), there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109), only the one who was from (xk−1,yk−1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.

Input

The first line of the input contains an integer T (1≤T≤10),which is the number of test cases.

In each case, the first line of the input contains an integer N (1≤N≤105).The following N lines, the k-th line contains 3 integers, xk,yk,vk (0≤vk≤103), which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.

Output

The maximum of dollars YJJ can get.

Sample Input

1 3 1 1 1 1 2 2 3 3 1

Sample Output

3

Source

2018中國大學生程式設計競賽 - 網路選拔賽

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題意:有n個點,每個點有一個價值,必須從(x-1,y-1)走到(x,y)才能獲得(x,y)點的價值,當走到最後能獲得的最大價值是多少？

題解:首先對於1e9的x座標和y座標無法開陣列,於是先對x軸和y軸進行離散,很容易可以想到對於一個點的最大值就是找到y和x都小於這個點的最大值再加上這個點的價值,於是用線段樹為何0--p[i].y-1行的最大值,直接進行轉移即可。

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=1e6+7;

struct node
{
    int x,y,w;
    bool operator <(const node &a)const
    {
        return (x==a.x)?y>a.y:x<a.x;
    }
}p[maxn];

int tree[maxn],n;
void build(int rt=1,int l=0,int r=n-1)
{
    tree[rt]=0;
    if(l==r) return;
    int m=l+r>>1;
    build(rt<<1,l,m);
    build(rt<<1|1,m+1,r);
}
void update(int pos,int w,int l=0,int r=n-1,int rt=1)
{
    if(l==r)
    {
        tree[rt]=max(tree[rt],w);
        return;
    }
    int m=l+r>>1;
    if(pos<=m) update(pos,w,l,m,rt<<1);
    else update(pos,w,m+1,r,rt<<1|1);
    tree[rt]=max(tree[rt<<1],tree[rt<<1|1]);
}
int query(int L,int R,int l=0,int r=n-1,int rt=1)
{
    if(l>=L && r<=R)
        return tree[rt];
    int MAX=0;
    int m=l+r>>1;
    if(L<=m) MAX=max(MAX,query(L,R,l,m,rt<<1));
    if(R>m) MAX=max(MAX,query(L,R,m+1,r,rt<<1|1));
    return MAX;
}

int hax[maxn],hay[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].w);
            hax[i]=p[i].x;hay[i]=p[i].y;
        }
        sort(hax,hax+n);sort(hay,hay+n);
        int n1=unique(hax,hax+n)-hax;
        int n2=unique(hay,hay+n)-hay;
        for(int i=0;i<n;i++)
        {
            p[i].x=lower_bound(hax,hax+n1,p[i].x)-hax;
            p[i].y=lower_bound(hay,hay+n2,p[i].y)-hay;
        }
        sort(p,p+n);
        build();

        int ans=0;
        for(int i=0;i<n;i++)
        {
            int tmp;
            if(!p[i].y) tmp=p[i].w;
            else tmp=query(0,p[i].y-1)+p[i].w;
            update(p[i].y,tmp);
            ans=max(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}