hdu 1051 Wooden Sticks

Wooden Sticks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20782 Accepted Submission(s): 8400 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 2 1 3

题意：机器加工木头需要时间，放入第一个木头的时间是1，如果后面木头的长度和重量都大于等于前面那一个，那么时间不变，否则就要加一。

利用贪心对其进行排序，先对长度进行从小到大排序，如果长度相等，那么把重量也按照从小到大排序。

然后从前往后找，对找过的点进行标记，这样等所有的点都找完，那么所要用的时间也就求出来了。

AC代码：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct node { int length; int weight; }p[5001]; bool cmp(node x, node y) { if(x.length==y.length) return x.weight < y.weight; else return x.length < y.length; } int main() { int t; int vis[5001]; scanf("%d",&t); while(t--) { memset(vis, 0, sizeof(vis)); int n; scanf("%d",&n); for(int i = 0; i < n; i++) { scanf("%d%d",&p[i].length, &p[i].weight); } sort(p, p + n, cmp); int sum = 0, x, y; for(int i = 0; i < n; i++) { x = p[i].length, y = p[i].weight; if(vis[i]!=1) { sum++; for(int j = i + 1; j < n; j++) { if(vis[j]!=1&&p[j].length >= x&&p[j].weight >= y) { x = p[j].length; y = p[j].weight; vis[j] = 1; } } } } printf("%d\n",sum); } return 0; }