BOJ 394

Problem I. The Longest Sequence of
Rectangles
Description
A rectangle is specified by a pair of coordinates (x1 , y1) and (x2 , y2)
indicating its lower-left and upper-right corners (x1 <= x2 and y1 <= y2). For
a pair of rectangles, A = ((X_A1, Y_A1), (X_A2, Y_A2)) and B = ((X_B1, Y_B1),
(X_B2, Y_B2)), we define A <= B if X_A2<X_B1 and Y_A2<Y_B1. Given a
number of rectangles on the plane, you are asked to find the length L of the
longest sequence of rectangles (A1, A2, ..., AL) such that A1 <= A2 <= ...
<= AL.
Input
The input contains multiple test cases. The first line of the input is an integer
T, indicating the number of test cases. Each test case begins with a line
containing a single integer n (1 <= n <= 100000), indicating the number of
rectangles. Each of the next n lines contains four integers x1, y1, x2, y2 (-
1000000 <= x1 < x2 <= 1000000, -1000000 <= y1 < y2 <= 1000000),
indicating the lower left and upper right corners of a rectangle.
Output
For each input test case, output a single integer indicating the length of the
longest sequence.
Sample Input
2
3
1 5 2 8
3 -1 5 4
10 10 20 20
2
2 1 4 5
6 5 8 10
Sample Output
2
1

解题思路：（此题的原型就是上一篇日志nlogn最长上升子序列(LIS) ）显然，最终的答案序列中矩形的左下角和右上角点坐标一定是按x严格递增的。这就给了我们将二维问题降至一维的思路。将每个矩形视为两个点（左下角和右上   角）的组合。将这2N 个点按照x坐标排序。这样可以将问题从二维空间降至一维空间：维护数组y[m] ，表示长度为m矩形序列的最后一个矩形y坐标最小是多少。依次遍历这排序后的2N 个点。若当前点为矩形的左下角，二分查找该点的y坐标在F数组中的位置。若当前点为矩形的右上角，则用其左下角的值更新F 数组。之所以不可以立即更新F 数组，是因为要求矩形序列中后一个矩形的左下角严格在前一个的右上方。

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

const int MV=200010;

struct point{
    int x,y,flag,id;
    bool operator<(const point& t)const{
        if(x!=t.x)return x<t.x;
        return flag < t.flag;
    }
}p[MV];

int y[MV], pos[MV];

int main(){
    int n, T;
    scanf("%d", &T);
    while(T--){
    	scanf("%d", &n);
        if(n==0)break;
        for( int i=0;i<n;i++){
            scanf("%d%d",&p[2*i].x,&p[2*i].y);
            p[2*i].flag=0;

            scanf("%d%d",&p[2*i+1].x,&p[2*i+1].y);
            p[2*i+1].flag=1;

            p[2*i].id=p[2*i+1].id=i;
        }
        n*=2;
        sort(p,p+n);

        int cnt=0;

        for(int i=0;i<n;i++){
            if(p[i].flag == 0){
                    pos[p[i].id] = lower_bound(y+1, y+1+cnt, p[i].y) - y ;
            } else {
                int tp = pos[p[i].id];
                if(tp > cnt)y[++cnt] = p[i].y;
                else y[tp] = min(y[tp], p[i].y);
            }
        }

        printf("%d\n",cnt);
    }
    return 0;
}