Wooden Sticks

题目连接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40692#problem/E

描述

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). 

 

输入

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. 

 

输出

The output should contain the minimum setup time in minutes, one per line. 

 

样例输入

 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 

 

样例输出

 213 

 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct wood{    int l,w,v;}a[10000];bool cmp(wood x,wood y){    if(x.l!=y.l)    return x.l < y.l;    return x.w < y.w;}int main(){    int t,n,m,i,j,sum;    scanf("%d",&t);    while(t--)    {        sum=0;        scanf("%d",&n);        memset(a,0,sizeof(a));        for(i=0;i<n;i++)        {            scanf("%d %d",&a[i].l,&a[i].w);        }        sort(a,a+n,cmp);        for(i=0;i<n;i++)        {            if(a[i].v) continue;            int d=a[i].w;            sum++;            for(j=i;j<n;j++)            {                if(a[j].w >= d && !a[j].v)                {                    d=a[j].w;                    a[j].v=1;                }            }        }        printf("%d\n",sum);    }    return 0;}