Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15564 Accepted Submission(s): 6405
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
Source
这是个贪心算法入门题，题目大意是给定$n$根木棍，以及每根木棍的长度$l$和重量$w$。现在要对它们进行加工，机器调整时间为1分钟，如果加工完一根长$l$重$w$的木棍后，下一根长$l'$重$w'$的木棍满足$l\leq l'$且$w\leq w'$，那么机器就可以继续加工而不需要进入调整时间。否则的话，又需要1分钟的调整时间才能继续加工。求总共最小需要的调整时间。
如果要求总共最小的时间，那么就要尽量把木棍分成几个序列，使得每个序列中前一个根木棍和后一根木棍的$l$和$w$都相差很小，但又能满足$l\leq l'$且$w\leq w'$。瞬间想到应该用排序来做，但排序分主次，很遗憾我们并不能同时对两个主元进行排序，必须先对$l$排序，然后对$w$贪心，或者对$w$排序对$l$贪心。
基本思路是比如我对$l$排序，这样整个序列就是满足$l\leq l'$的辣，然后遍历这个序列，如果还满足$w\leq w'$那么太好了，把这个木棍的下标记下来，去找下一个满足条件的木棍，这样找下去就找到第一个子序列辣。再去找剩下的，发现我们不知道哪些是找过的，好，给木棍加一个vis状态，0为未访问1为访问过哒，OK那么每次找子序列呢先找到一个未访问过的木棍，把res加一，然后对于这个序列我贪心的去找下一根木棍并把它的vis记为1。全部木棍被标为1的时候也就找完辣！
#include <stdio.h>
#include <algorithm>
struct st {
int l, w, vis;
bool operator<(const st&c)const {
return l==c.l?w<c.w:l<c.l;
}
}stick[];
int n;
void read() {
scanf("%d", &n);
for(int i=; i<n; i++) {
scanf("%d%d", &stick[i].l, &stick[i].w);
stick[i].vis = ;
}
}
void find(int i) {
int k = i;
for(int j=i+; j<n; j++)
if(!stick[j].vis)
if(stick[k].w<=stick[j].w) {
stick[j].vis = ;
k=j;
}
}
void work() {
int res = ;
std::sort(stick, stick+n);
for(int i=; i<n; i++) {
if(!stick[i].vis) {
stick[i].vis = ;
++res;
find(i);
}
}
printf("%d\n", res);
}
int main() {
int T, n;
scanf("%d", &T);
while(T--) {
read();
work();
}
return ;
}
