poj2761 Feed the dogs
2015-11-21 20:32
393 查看
Feed the dogs
Description
Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to
n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want
to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Input
The first line contains n and m, indicates the number of dogs and the number of feedings.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
Output
Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.
Sample Input
Sample Output
Source
POJ Monthly--2006.02.26,zgl & twb
Treap模板题
这道题是支持离线的,所以我们可以将区间按左端点从小到大排序,每次相应加点和删点在求出Treap中第k小的数。
Treap模板写挂了...整整调了3个小时才找出错...最后发现是在删点时没有将节点数减1...QAQ
Time Limit: 6000MS | Memory Limit: 65536K | |
Total Submissions: 17095 | Accepted: 5349 |
Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to
n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want
to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Input
The first line contains n and m, indicates the number of dogs and the number of feedings.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
Output
Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.
Sample Input
7 2 1 5 2 6 3 7 4 1 5 3 2 7 1
Sample Output
3 2
Source
POJ Monthly--2006.02.26,zgl & twb
Treap模板题
这道题是支持离线的,所以我们可以将区间按左端点从小到大排序,每次相应加点和删点在求出Treap中第k小的数。
Treap模板写挂了...整整调了3个小时才找出错...最后发现是在删点时没有将节点数减1...QAQ
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<cstdlib> #define F(i,j,n) for(int i=(j);i<=(n);i++) #define D(i,j,n) for(int i=(j);i>=(n);i--) #define LL long long #define pa pair<int,int> #define MAXN 100005 using namespace std; int n,m,root=0,tot=0,b[MAXN],ans[MAXN]; struct tree_type { int l,r,s,rnd,v,w; }t[MAXN]; struct query_type { int l,r,x,num; }a[MAXN]; int read() { int ret=0,flag=1; char ch=getchar(); while (ch<'0'||ch>'9') { if (ch=='-') flag=-1; ch=getchar(); } while (ch>='0'&&ch<='9') { ret=ret*10+ch-'0'; ch=getchar(); } return ret*flag; } void rturn(int &k) { int tmp=t[k].l; t[k].l=t[tmp].r; t[tmp].r=k; t[tmp].s=t[k].s; t[k].s=t[t[k].l].s+t[t[k].r].s+t[k].w; k=tmp; } void lturn(int &k) { int tmp=t[k].r; t[k].r=t[tmp].l; t[tmp].l=k; t[tmp].s=t[k].s; t[k].s=t[t[k].l].s+t[t[k].r].s+t[k].w; k=tmp; } void ins(int &k,int x) { if (!k) { k=++tot; t[k].v=x; t[k].s=t[k].w=1; t[k].l=t[k].r=0; t[k].rnd=rand(); return; } t[k].s++; if (t[k].v==x) t[k].w++; else if (t[k].v>x) { ins(t[k].l,x); if (t[t[k].l].rnd<t[k].rnd) rturn(k); } else { ins(t[k].r,x); if (t[t[k].r].rnd<t[k].rnd) lturn(k); } } void del(int &k,int x) { if (t[k].v==x) { if (t[k].w>1) t[k].w--,t[k].s--; else if (!t[k].l||!t[k].r) k=t[k].l+t[k].r; else if (t[t[k].l].rnd<t[t[k].r].rnd) { rturn(k); del(k,x); } else { lturn(k); del(k,x); } return; } t[k].s--; if (x<t[k].v) del(t[k].l,x); else del(t[k].r,x); } int getans(int k,int x) { int ln=t[t[k].l].s; if (ln<x&&ln+t[k].w>=x) return t[k].v; else if (x<=ln) return getans(t[k].l,x); else return getans(t[k].r,x-ln-t[k].w); } bool cmp(query_type x,query_type y) { return x.l==y.l?x.r<y.r:x.l<y.l; } int main() { t[0].s=0; a[0].l=1;a[0].r=0; n=read();m=read(); F(i,1,n) b[i]=read(); F(i,1,m) { a[i].l=read(); a[i].r=read(); a[i].x=read(); a[i].num=i; } sort(a+1,a+m+1,cmp); F(i,1,m) { if (a[i].l>a[i-1].r) { root=0; F(j,a[i].l,a[i].r) ins(root,b[j]); } else { F(j,a[i-1].l,a[i].l-1) del(root,b[j]); F(j,a[i-1].r+1,a[i].r) ins(root,b[j]); } ans[a[i].num]=getans(root,a[i].x); } F(i,1,m) printf("%d\n",ans[i]); }
相关文章推荐
- LintCode-剑指Offer-(245)子树
- 南大软院大神养成计划--HTML和CSS基础课程(二)
- 在VS中让JS文件显示智能感知的一个犀利方法
- 2 jquery选择器
- jquery each跳出循环的问题
- html中字体英文
- Bootstrap CSS概览
- 南大软院大神养成计划--js
- Angular.js
- jsp本质,jsp静态导入与动态导入区别,jsp九个内置对象,jsp传值
- bootstrap框架
- 从最大似然到EM算法浅解
- 同样版本的jstl,都是jstl1.2版本,有个有问题,另一个没有问题
- Deep Learning(深度学习)学习笔记整理系列之(八)
- fedora22 安装fcitx 输入法
- Deep Learning(深度学习)学习笔记整理系列之(七)
- Deep Learning(深度学习)学习笔记整理系列之(六)
- 怎样成长为一个优秀的web前端工程师
- Deep Learning(深度学习)学习笔记整理系列之(五)
- 今天在win7下安装Fedora22