贪心算法 WOODEN STICKS 贪心算法 WOODEN STICKS 实例代码
人气:0Problem 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
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 5000;
struct node
{
int l;
int w;
int flag;
}sticks[5000];
int cmp(const void *p,const void *q)
{
struct node *a = (struct node *)p;
struct node *b = (struct node *)q;
if(a->l > b->l) return 1;
else if(a->l < b->l) return -1;
else return a->w > b->w ? 1 : -1;
}
int main()
{
int t,n,cnt,cl,cw;
int i,j;
scanf("%d",&t);
while(t--)
{
memset(sticks,0,sizeof(sticks));
scanf("%d",&n);
for( i = 0; i < n; i++)
scanf("%d %d",&sticks[i].l,&sticks[i].w);
qsort(sticks,n,sizeof(sticks[0]),cmp);
sticks[0].flag = 1;
cl = sticks[0].l;
cw = sticks[0].w;
cnt = 1;
for( j = 1; j < n; j++)
{
for( i = j; i < n; i++)
{
if(!sticks[i].flag&&sticks[i].l>=cl&&sticks[i].w>=cw)
{
cl = sticks[i].l;
cw = sticks[i].w;
sticks[i].flag = 1;
}
}
i = 1;
while(sticks[i].flag)
i++;
j = i;
if(j == n) break;
cl = sticks[j].l;
cw = sticks[j].w;
sticks[j].flag = 1;
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}
题意：
我们要处理一些木棍，第一根的时间是1分钟，另外的，在长度为l，重为w的木棍后面的那根的长度为l', 重量w'，只要l <=l' 并且w <= w'，就不需要时间，否则需要1分钟，求如何安排处理木棍的顺序，才能使花的时间最少。
思路：
贪心算法。先把这些木棍按长度和重量都从小到大的顺序排列。cl和cw是第一根的长度和重量，依次比较后面的是不是比当前的cl，cw大，是的话把标志flag设为1，并跟新cl,cw。比较完后，再从前往后扫描，找到第一个标志位为0的，作为是第二批的最小的一根，计数器加一。把它的长度和重量作为当前的cl，cw，再循环往后比较。直到所有的都处理了。
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