HDOJ 1051. Wooden Sticks 贪心 结构体排序

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

2 1 3

 

Source

Asia 2001, Taejon (South Korea)

 

来源： <http://acm.hdu.edu.cn/showproblem.php?pid=1051>

 

        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$lleq l'$且$wleq w'$，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$lleq l'$且$wleq w'$。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$lleq l'$的辣，然后遍历这个序列，如果还满足$wleq w'$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[5001];

int n;
void read() {
    scanf("%d", &n);
    for(int i=0; i<n; i++) {
        scanf("%d%d", &stick[i].l, &stick[i].w);
        stick[i].vis = 0;
    }
}

void find(int i) {
    int k = i;
    for(int j=i+1; j<n; j++)
        if(!stick[j].vis)
            if(stick[k].w<=stick[j].w) {
                stick[j].vis = 1;
                k=j;
            }
}

void work() {
    int res = 0;
    std::sort(stick, stick+n);
    for(int i=0; i<n; i++) {
        if(!stick[i].vis) {
            stick[i].vis = 1;
            ++res;
            find(i);
        }
    }
    printf("%d
", res);
}

int main() {
    int T, n;
    scanf("%d", &T);
    while(T--) {
        read();
        work();
    }
    return 0;
}

 

By Black Storm（使用为知笔记）