hduoj1677

Nested Dolls

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3666    Accepted Submission(s): 1105

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

4320 30 40 50 30 40420 30 10 10 30 20 40 50310 30 20 20 30 10410 10 20 30 40 50 39 51

 

Sample Output

1232
题解：和hduoj1057很类似，只不过等于的时候装不进去了
#include<iostream>#include<cstdio>#include<cmath>#include<vector>#include<algorithm>#define INF 10000000+10using namespace std;int dp[20010];struct No{    int l, w;}no[20010],na[20010];bool com(No a,No b){    if(a.l == b.l)    return a.w < b.w;    return a.l > b.l;}int main(){    int t;    cin >> t;    while(t--)    {        int n,i,j;        cin >> n;        for( i = 0;i < n;i++)        {            cin >> no[i].l >> no[i].w;             na[i].l = na[i].w = 11000;        }        sort(no,no+n,com);        int ans = 0;        for(i = 0;i < n;i++)        {         j = 0;         while(na[j].l <= no[i].l|| na[j].w <= no[i].w) j++;         na[j].l = no[i].l;         na[j].w = no[i].w;           }  for(i = 0;i < n;i++)  if(na[i].l != 11000)  ans++;        cout << ans << endl;    }    return 0;
}