hdu1051Wooden Sticks(贪心)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15585    Accepted Submission(s): 6413

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
题目大意：有 n个长度为 L[i],重量为w[i]的木棒需要加工,每次第一个加工的木棒花费一分钟，若后面的木棒长度和重量都大于等于第一个加工的木棒，
则不耗费任何时间。问加工 n 个木棒最少需要多少木棒？
解题思路：最近搞了几个贪心，发现多个变量决定一个物体时，在排序时，这几个变量都要考虑，否则就一直wrong,看来看问题不能太片面啊；一看是两个参数长度和重量决定一个木棒，又要求最少时间，所以可以现按长度从小到大排序，当长度相等时，重量小的靠前。这样第一步搞定，再看题中的例子说
(1,4), (3,5), (4,9), (2,1), (5,2).这样排序后所用的时间是最短，且是2分钟。咋一看，感觉不对啊，（2，1）（5，2）这两个明明不符合条件，怎么可能两分钟，应该是三分钟啊。其实这是分两步进行的，第一次先把（1，4）（3，5）（4，9）加工完，需要一分钟，第二次在把（2，1）（5，2）弄完，需要一分钟。所以共需2分钟。
所以可知在排完序后应对其进行扫描，把第一次可以加工的扫描并标记，在弄剩余的并标记，直到所有都被加工完。能扫描几次就需要几分钟。
代码：
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct wooden{int x,y;int flag;}d[5010];int cmp(wooden a,wooden b){  if(a.x!=b.x)return a.x<b.x;  else    return a.y<b.y;}int main(){int c;int n;int i,j,k,time;scanf("%d",&c);while(c--){memset(d,0,sizeof(d));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&d[i].x,&d[i].y);d[i].flag=0;}sort(d,d+n,cmp);time=0; for(i=0;i<n;i++){if(!d[i].flag){d[i].flag=1;for(j=i+1;j<n;j++){if(!d[j].flag&&d[j].x>=d[i].x&&d[j].y>=d[i].y){d[j].flag=1;d[i].x=d[j].x;d[i].y=d[j].y;}} time++;}}printf("%d\n",time);}return 0;}