hdu 1051 Wooden Sticks(贪心)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15640    Accepted Submission(s): 6429

Problem Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 

(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup. 

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

 

Output

The output should contain the minimum setup time in minutes, one per line.

 

Sample Input

3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1

 

Sample Output

213

 

Source

Asia 2001, Taejon (South Korea)

 

题意：给你n个点（x,y），当xi>=xj&&yi>=yj是可以归为一个集合，求最终可以归为的集合数最少是多少。

题解：贪心。排序后，对于当前a[i].每次都找 没被用过&&离他最近&&符合条件的a[j]归为一个集合，再从a[j]开始找。

#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>#define N 5005#define INF 1e8using namespace std;int n;struct point {    int x,y;} a[N];bool cmp(point a,point b) {    if(a.x==b.x)return a.y>b.y;    return a.x>b.x;}bool vis[N];bool is(point a,point b) {    return a.x>=b.x&&a.y>=b.y;}int main() {    //freopen("test.in","r",stdin);    int t;    cin>>t;    while(t--) {        scanf("%d",&n);        for(int i=0; i<n; i++) {            scanf("%d%d",&a[i].x,&a[i].y);        }        sort(a,a+n,cmp);        int ans=n;        memset(vis,0,sizeof vis);        int cnt=0;        while(cnt<n) {            while(vis[cnt]&&cnt<n)cnt++;            vis[cnt]=1;            int i=cnt;            for(int j=i+1; j<n; j++) {                if(!vis[j]&&is(a[i],a[j])) {                    i=j;                    ans--;                    vis[j]=1;                }            }        }        printf("%d\n",ans);    }    return 0;}