1001 of greedy strategy

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

    

#include<cstdio>#include<iostream>#include<string.h>#include<algorithm>using namespace std;struct wood{    int l,w;};bool comp(wood a, wood b){if (a.l != b.l)return a.l < b.l;    return a.w < b.w;}int main(){wood a[5001];int b[5001] = { 0 };int t, n, i, j,k,count,max;scanf("%d", &t);for (j = 0;j < t;j++){scanf("%d", &n);for (i = 0;i < n;i++)scanf("%d%d", &a[i].l, &a[i].w);sort(a, a + n, comp); count = 0;memset(b, 0, sizeof(b));for (i = 0;i <n;i++){k = i;if(b[i]==0){max=a[i].w;b[i] = 1;count++;while (k<=n-1){k++;if (a[k].w >=max&&b[k]==0) {b[k] = 1;max=a[k].w; }}             }}printf("%d\n",count);}}

 题目要求：若长度和重量都小于上一个生产的木棍则可以同时生产。要求计算出最短的生产时间。

      思路：1对长度进行一次升序，若长度相同则以重量升序，需要结构体。从高向低或从低向高进行判断。以从低向高为例，若比上一个造的长度长则可以同时造，每造一次时间加一，最后得到的便是最短时间。（注：本质为贪心算法。）

      细节：一定要和造的上一个比较而不要与开始造的比较，因为这错了好多遍。