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

 

怎么写都感觉不对,后来参考下网上的解题报告。感觉挺好的。

-2e9这个没看明白为什么是这个数,不过喜欢这个代码。下面是参考了网上的结果

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct z{
    int x;
    int y;
};
int comp(const void *a,const void *b) {
    struct z *c=(struct z*)a;
    struct z *d=(struct z*)b;
    if(c->x==d->x)
        return c->y-d->y;
    return c->x-d->x;
}
int main() {
    int t,a,b,m,n,i;
    struct z g[5001];
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d%d",&g[i].x,&g[i].y);
        qsort(g+1,n,sizeof(g[0]),comp);
        a=b=0;
        while(b!=n){
            m=-2e9;
            for(i=1;i<=n;i++) {
                if(g[i].y>=m) {
                    m=g[i].y;
                    g[i].y=-2e9-1;
                    b++;
                }
            }
            a++;
        }
        printf("%d\n",a);
    }
    return 0;
}

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