HDU 3607 Traversal （线段树）

Traversal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 899    Accepted Submission(s): 320

Problem Description

Xiao Ming is travelling, and today he came to the West Lake, to see people playing a game, this game is like this, lake placed n-box, from 1 to n label. These boxes are floating in the water, there are some gold inside each box. Then participants from coast to jump the other side, each box have its’s height, you can only jump from lower height to higher height, and from a small label to a big label, you can skip some of the middle of the box. Suppose the minimum height is this side, the other side has the maximum height. Xiao Ming would like to jump how to get the most gold? He now needs your help.

 

Input

There are multiple test cases. Each test case contains one integer N , representing the number of boxes . The following N lines each line contains two integers hi and gi , indicate the height and the number of gold of the ith box.
1 < = N < 100 001
0 < hi < 100 000 001
0 < gi < = 10000

 

Output

For each test case you should output a single line, containing the number of maximum gold XiaoMing can get.

 

Sample Input

41 12 22 35 11 1 10000

 

Sample Output

510000

 

Source

2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

 

Recommend

lcy

 

解题思路：类似我博客的HDU2227,每进来一个数，算出它的前面的数的最大的值（用线段树），然后加上它本身，就是它的最大值，插入线段树，然后循环

 

 

#include <iostream>#include <cstdio>#include <map>#include <algorithm>using namespace std;const int maxn=100010;struct point{int h,w;}data[maxn];struct node{int l,r,c,maxc;}a[maxn*4];int n;map <int,int> mp;void build(int l,int r,int k){a[k].l=l;a[k].r=r;a[k].c=0,a[k].maxc=0;if(l<r){int mid=(l+r)/2;build(l,mid,2*k);build(mid+1,r,2*k+1);}}void insert(int l,int r,int c,int k){if(l<=a[k].l && a[k].r<=r){a[k].c=max(c,a[k].c);a[k].maxc=a[k].c;}else{int mid=(a[k].l+a[k].r)/2;if(l>=mid+1) insert(l,r,c,2*k+1);else if(r<=mid) insert(l,r,c,2*k);else{insert(l,mid,c,2*k);insert(mid+1,r,c,2*k+1);}a[k].maxc=max(a[2*k].maxc,a[2*k+1].maxc);}}//query maxc in l,rint query(int l,int r,int k){if(l<=a[k].l && a[k].r<=r){return a[k].maxc;}else{int mid=(a[k].l+a[k].r)/2;if(l>=mid+1) return query(l,r,2*k+1);else if(r<=mid) return query(l,r,2*k);else{ return max(query(l,mid,2*k),query(mid+1,r,2*k+1));}}}void initial(){mp.clear();build(0,n,1);}void input(){int cnt=1;map <int,int>::iterator it;for(int i=1;i<=n;i++){scanf("%d%d",&data[i].h,&data[i].w);mp[data[i].h]=0;}for(it=mp.begin();it!=mp.end();it++) it->second=cnt++;}void computing(){int ans=0,tmp,pos;for(int i=1;i<=n;i++){pos=mp[data[i].h];tmp=query(0,pos-1,1)+data[i].w;if(ans<tmp) ans=tmp;insert(pos,pos,tmp,1);}cout<<ans<<endl;}int main(){while(scanf("%d",&n)!=EOF){initial();input();computing();}return 0;}