hdu1051（贪心）

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)

简单排序，先排长度，长度相等再按照重量排，然后跟区间相交的思路差不多，参照“今年寒假不AC”这道题，用l,w记录每一次的值 用一个vis数组用2来标记每一个开头，1标记能被开头包含的组，0表示没有被包含过，最后vis中值为2的个数就是最少的时间数！

#include<bits/stdc++.h>using namespace std;int vis[5050];struct wood{int l;int w;}wooden[5005];bool cmp(wood a,wood b){return a.l==b.l?a.w<b.w:a.l<b.l;}int main(){int n,i,j;int t;int l=0,w=0;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&wooden[i].l,&wooden[i].w);}     sort(wooden,wooden+n,cmp);memset(vis,0,sizeof(vis));for(i=0;i<n;i++){if(vis[i]==0) vis[i]=2;else continue;w=wooden[i].w;for(j=i+1;j<n;j++){if(wooden[j].w>=w&&vis[j]==0){w=wooden[j].w;vis[j]=1;} }}int sum=0;for(i=0;i<n;i++)if(vis[i]==2) sum++;cout<<sum<<endl;}}