HDOJ 1051. Wooden Sticks 貪心 結構體排序

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)

 

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

 

          這是個貪心算法入門題，題目大意是給定$n$根木棍，以及每根木棍的長度$l$和重量$w$。現在要對它們進行加工，機器調整時間為1分鍾，如果加工完一根長$l$重$w$的木棍后，下一根長$l'$重$w'$的木棍滿足$l\leq l'$且$w\leq w'$，那么機器就可以繼續加工而不需要進入調整時間。否則的話，又需要1分鍾的調整時間才能繼續加工。求總共最小需要的調整時間。

          如果要求總共最小的時間，那么就要盡量把木棍分成幾個序列，使得每個序列中前一個根木棍和后一根木棍的$l$和$w$都相差很小，但又能滿足 $l\leq l'$且$w\leq w'$。瞬間想到應該用排序來做，但排序分主次，很遺憾我們並不能同時對兩個主元進行排序，必須先對$l$排序，然后對$w$貪心，或者對$w$排序對$l$貪心。

          基本思路是比如我對$l$排序，這樣整個序列就是滿足 $l\leq l'$的辣，然后遍歷這個序列，如果還滿足 $w\leq w'$那么太好了，把這個木棍的下標記下來，去找下一個滿足條件的木棍，這樣找下去就找到第一個子序列辣。再去找剩下的，發現我們不知道哪些是找過的，好，給木棍加一個vis狀態，0為未訪問1為訪問過噠，OK那么每次找子序列呢先找到一個未訪問過的木棍，把res加一，然后對於這個序列我貪心的去找下一根木棍並把它的vis記為1。全部木棍被標為1的時候也就找完辣！

#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\n", res);
}

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

 

By Black Storm（使用為知筆記）