hdu 4268 Alice and Bob

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4268

题目大意:
Alice和Bob各有N(N<=100,000)张卡,卡有两个属性(w,h),如果A卡的这两个属性均不小于B卡的,则A卡可以覆盖B卡,求Alice最多可以盖Bob几张卡.

题目思路:
如果Alice的某张卡的h大于Bob的某些卡的h,那么可以知道Alice的这张卡应该盖这些卡中w值最大的且w值小于Alice的这张卡的w值(设为W).
假设Alice不盖可以盖的最大w值的那张卡,盖了一张小的(设为W‘),那么如果Alice有一张卡属性为h'和w'(h'>h,W'<w'<W),那么容易发现,如果之前盖了W,那么就可以多盖一张卡了.
通俗的来说就是要盖就盖能盖的最大卡.

代码:

#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 ll long long#define ls rt<<1#define rs ls|1#define lson l,mid,ls#define rson mid+1,r,rs#define middle l+r>>1#define INF 0x3F3F3F3F#define esp (1e-10)#define MOD 1000000007#define type int//const double pi=acos(-1.0);const int M=100000+5;#define clr(x,c) memset(x,c,sizeof(x))type min(type x,type y){return x<y? x:y;}type max(type x,type y){return x>y? x:y;}void swap(type& x,type& y){type t=x;x=y;y=t;}int T,cas=0;int n,m,ww[M<<1];struct node{    int h,w;    void read(){        scanf("%d%d",&h,&w);        ww[++m]=w;    }    bool operator < (const node &p) const {        return h==p.h? w<p.w:h<p.h;    }};node a[M],b[M];int bs(int key){    int l=1,r=m,mid;    while(l<=r){        mid=middle;        if(key>ww[mid]) l=mid+1;        else if(key<ww[mid]) r=mid-1;        else return mid;    }return -1;}int mmax[M<<3],cnt[M<<1];void update(int l,int r,int rt,int p,int c){    if(l==r){        mmax[rt]= (cnt[l]+=c)? l:0;        return;    }    int mid=middle;    if(p<=mid) update(lson,p,c);    else update(rson,p,c);    mmax[rt]=max(mmax[ls],mmax[rs]);}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(R<=mid) return query(lson,L,R);    else if(mid<L) return query(rson,L,R);    else return max(query(lson,L,mid),query(rson,mid+1,R));}void run(){    int i,j,k,ret=0;    scanf("%d",&n),m=0;    for(i=1;i<=n;i++) a[i].read();    for(i=1;i<=n;i++) b[i].read();    sort(ww+1,ww+m+1);    //m=unique(ww+1,ww+m+1)-(ww+1);    for(i=2,k=1;i<=m;i++) if(ww[i]!=ww[i-1]) ww[++k]=ww[i];    m=k;    sort(a+1,a+n+1),sort(b+1,b+n+1);    clr(mmax,0),clr(cnt,0);    for(i=j=1;i<=n;i++){        while(j<=n && b[j].h<=a[i].h){            update(1,m,1,bs(b[j].w),1);            j++;        }        k=query(1,m,1,0,bs(a[i].w));        if(k>0) ret++,update(1,m,1,k,-1);    }    printf("%d\n",ret);}int main(){    //freopen("1.in","r",stdin);    //freopen("1.out","w",stdout);    //preSof();    //run();    //while(~scanf("%d",&n)) run();    for(scanf("%d",&T),cas=1;cas<=T;cas++) run();    //system("pause");    return 0;}