1、http://acm.hdu.edu.cn/showproblem.php?pid=1677
2、題目大意：
有n個盒子，知道寬和高，如果第一個盒子的寬和高是w1,h1,第二個的寬和高是w2,h2,如果w1<w2 && h1<h2 ,那麼第二個盒子就能裝下第一個盒子，求最終剩下多少個盒子
3、思路分析：
這道題目和裝娃娃的題目有點像，首先得排序，這個很關鍵，按照w從大到小排序，再按照h從小到大排序，一開始還以爲只要都按照從大到小就可以，但是對於w相同的盒子此排序就不對了，排序後對h做最長上升子序列，輸出的最大長度即可
4、題目：
Nested Dolls
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2275 Accepted Submission(s): 653
Problem Description
Dilworth is the world’s most prominent collector of Russian nested dolls: he literally has thousands of them! You know, the wooden hollow dolls of different sizes of which the smallest doll is contained in the second smallest, and
this doll is in turn contained in the next one and so forth. One day he wonders if there is another way of nesting them so he will end up with fewer nested dolls? After all, that would make his collection even more magnificent! He unpacks each nested doll
and measures the width and height of each contained doll. A doll with width w1 and height h1 will fit in another doll of width w2 and height h2 if and only if w1 < w2 and h1 < h2. Can you help him calculate the smallest number of nested dolls possible to assemble
from his massive list of measurements?
Input
On the first line of input is a single positive integer 1 <= t <= 20 specifying the number of test cases to follow. Each test case begins with a positive integer 1 <= m <= 20000 on a line of itself telling the number of dolls in the
test case. Next follow 2m positive integers w1, h1,w2, h2, . . . ,wm, hm, where wi is the width and hi is the height of doll number i. 1 <= wi, hi <= 10000 for all i.
Output
For each test case there should be one line of output containing the minimum number of nested dolls possible.
Sample Input
4
3
20 30 40 50 30 40
4
20 30 10 10 30 20 40 50
3
10 30 20 20 30 10
4
10 10 20 30 40 50 39 51
Sample Output
1
2
3
2
Source
Recommend
lcy
5、Ac代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 20005
struct node
{
int h;
int w;
}a[N];
int dp[N];
int stack[N];
int cmp(node a,node b)
{
if(a.w==b.w)
return a.h<b.h;
return a.w>b.w;
}
int lis(int n)
{
memset(dp,0,sizeof(dp));
memset(stack,0,sizeof(stack));
int top=0;
stack[top]=-99999999;
int maxx=-1;
for(int i=1; i<=n; i++)
{
if(a[i].h>=stack[top])
{
stack[++top]=a[i].h;
dp[i]=top;
}
else
{
int l=1,r=top;
while(l<=r)
{
int mid=(l+r)>>1;
if(a[i].h>=stack[mid])
{
l=mid+1;
}
else
r=mid-1;
}
stack[l]=a[i].h;
dp[i]=l;
}
if(dp[i]>maxx)
maxx=dp[i];
}
return maxx;
}
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a[i].w,&a[i].h);
}
sort(a+1,a+n+1,cmp);
int ans=lis(n);
printf("%d\n",ans);
}
return 0;
}