Rectangles

Rectangles

时间限制: 1 Sec 内存限制: 128 MB

---

题目描述

Given N (4 <= N <= 100) rectangles and the lengths of their sides ( integers in the range 1..1,000), write a program that finds the maximum K for which there is a sequence of K of the given rectangles that can “nest”, (i.e., some sequence P1, P2, …, Pk, such that P1 can completely fit into P2, P2 can completely fit into P3, etc.).

A rectangle fits inside another rectangle if one of its sides is strictly smaller than the other rectangle’s and the remaining side is no larger. If two rectangles are identical they are considered not to fit into each other. For example, a 2*1 rectangle fits in a 2*2 rectangle, but not in another 2*1 rectangle.

The list can be created from rectangles in any order and in either orientation.

输入

The first line of input gives a single integer, 1 ≤ T ≤10, the number of test cases. Then follow, for each test case：
* Line 1: a integer N , Given the number ofrectangles N<=100
* Lines 2..N+1: Each line contains two space-separated integers X Y, the sides of the respective rectangle. 1<= X , Y<=5000

输出

Output for each test case , a single line with a integer K , the length of the longest sequence of fitting rectangles.

样例输入

1
4
8 14
16 28
29 12
14 8

样例输出

2

题意概括

给出n个箱子的宽和高，宽和高都小于另一个箱子的时候这个箱子可以套进去那个箱子里面，当然又可以容忍一个边一样的时候，也是可以套进去的；

解题思路

按照所给边的最大边的值从小打到排序，之后从小开始遍历，使用一个变量c来记录当前套进去的筐子数，然后用两重循环暴力更新，最后找出最大值。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <queue>

using namespace std;

struct date {
    int a,b,c;
}line[550];
int cmp (date a,date b)
{
    if(a.a!=b.a){
        return a.a<b.a;
    }else {
        return a.b<b.b;
    }
}
int main ()
{
    int t,n,m,i,j;
    int vis[550];

    scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        int a,b;
        for(i=0;i<n;i++){
            scanf("%d %d",&a,&b);
            if(a>b){
                line[i].a=a;
                line[i].b=b;
            }else {
                line[i].b=a;
                line[i].a=b;
            }
            line[i].c=1;
        }
        sort(line,line+n,cmp);
        int maxx=0,k=0;
        for(i=1;i<n;i++){
            maxx=line[i].c;
            k=-1;
            for(j=i-1;j>=0;j--){
                if(vis[j]==1||line[j].a>line[i].a||line[j].b>line[i].b)
                    continue;
                if(line[j].a<line[i].a){
                    if(maxx<line[i].c+line[j].c){
                        k=j;
                        maxx=line[i].c+line[j].c;
                    }
                }else if(line[j].b<line[i].b){
                    if(maxx<line[i].c+line[j].c){
                        k=j;
                        maxx=line[i].c+line[j].c;
                    }
                }
            }
            if(k!=-1){
                //vis[k]=1;
                line[i].c=maxx;
            }
        }
        maxx=0;
        for(i=0;i<n;i++){
            //printf("%d==  %d  %d
",i+1,line[i].c,vis[i]);
            if(vis[i]==0&&line[i].c>maxx){
                maxx=line[i].c;
            }
        }
        printf("%d
",maxx);
    }
    return 0;
}