hdu 3458 Rectangles Too!

转自：http://blog.csdn.net/gotoac/article/details/7564668

非常感谢原问的作者，确实写的很好，代码风格都好看，没有一点啰嗦的地方

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3458

题目大意:给定一串矩形列,求最长子矩形列.

题目思路:
(1)
dp[cur]=max(dp[i],...,dp[j])(rct[i]~rct[j]均小于rct[cur],dp[]存放以rct[]结尾的最长子矩阵列),求max(dp[i],...,dp[j])可以用线段树.
(2)
对于一维的线段树我们只能保证x或y满足条件,所以我们把某一个rct加到线段树前必须保证该rct之后的x1(或y1)一定大于它的x2(或y2).可以把x1(或y1)升序排序,扫过的rct加到优先队列pq中,队首的矩形x2(或y2)最小,这样只要队首矩形的x2(或y2)小于rct[cur].x1(或y1)就以意味着当前位置起后面的矩形的x1(或y1)均大于队首矩形的x2(或y2),这样就可以把队首矩形加到线段树中了.

#include <stdlib.h>#include <string.h>#include <stdio.h>#include <ctype.h>#include <math.h>#include <time.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <string>#include <iostream>#include <algorithm>using namespace std;#define ull unsigned __int64#define ll __int64//#define ull unsigned long long//#define ll long long#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define middle (l+r)>>1#define MOD 1000000007#define esp (1e-8)const int INF=0x3F3F3F3F;const double DINF=10000.00;//const double pi=acos(-1.0);const int N=100010;int n,m;int X[N<<1],mmax[N<<4],dp[N];struct rct{int x1,y1,x2,y2;void write(){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);}bool operator < (const rct& p) const {return y1 < p.y1;}}a[N];struct node{int pos,id;node(){}node(int _pos,int _id) : pos(_pos),id(_id){}bool operator < (const node& p) const {return a[id].y2 > a[p.id].y2;}}nd;void PushUp(int rt){mmax[rt]=max(mmax[rt<<1],mmax[rt<<1|1]);}void Update(int l,int r,int rt,int p,int c){if(l==r){mmax[rt]=max(mmax[rt],c);return;}int mid=middle;if(p<=mid) Update(lson,p,c);else Update(rson,p,c);PushUp(rt);}int Query(int l,int r,int rt,int L,int R){if(L<=l && r<=R) return mmax[rt];int mid=middle,ret=0;if(L<=mid) ret=max(ret,Query(lson,L,R));if(mid<R) ret=max(ret,Query(rson,L,R));return ret;}int bs(int key,int size,int A[]){int l=0,r=size-1,mid;while(l<=r){mid=middle;if(key>A[mid]) l=mid+1;else if(key<A[mid]) r=mid-1;else return mid;}return -1;}int main(){//freopen("1.in","r",stdin);//freopen("1.out","w",stdout);int i,j,k,ret;//int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++)while(~scanf("%d",&n)){if(!n) break;for(i=m=0;i<n;i++){a[i].write(),dp[i]=1;X[m++]=a[i].x1;X[m++]=a[i].x2;}sort(X,X+m);for(i=k=1;i<m;i++) if(X[i]!=X[i-1]) X[k++]=X[i];sort(a,a+n);priority_queue<node>pq;memset(mmax,0,sizeof(mmax));for(i=ret=0;i<n;i++){while(!pq.empty()){nd=pq.top();if(a[nd.id].y2<a[i].y1){pq.pop();Update(0,k,1,nd.pos+1,dp[nd.id]);}else break;}dp[i]=Query(0,k,1,0,bs(a[i].x1,k,X))+1;pq.push(node(bs(a[i].x2,k,X),i));ret=max(ret,dp[i]);}printf("%d\n",ret);}return 0;}