HDU 1051 Wooden Sticks （贪心入门）

Wooden Sticks

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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

213

   

     分析：长度不同按长度从小到大排，长度相同，按重量从小到大排。

AC代码：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;struct sgm{    int l,w;} a[5005];bool cmp(sgm A,sgm B){    if(A.l!=B.l)        return A.l<B.l;//按长度排    else        return A.w<B.w;//按重量排}int main(){    int n,i,j,m;    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        for(i=0; i<m; i++)            scanf("%d%d",&a[i].l,&a[i].w);        sort(a,a+m,cmp);        int book[5005]={0};        int x,s=0;        for(i=0; i<m; i++)        {            if(book[i]==0)            {                x=a[i].w;                for(j=i+1; j<m; j++)                {                    if(a[j].w>=x&&book[j]==0)                    {                        x=a[j].w;                        book[j]=1;                    }                }                s++;            }        }        printf("%d\n",s);    }    return 0;}