Wooden Sticks

Wooden Sticks  

时间限制(普通/Java):1500MS/15000MS     运行内存限制:65536KByte
总提交: 31            测试通过: 12

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

题目上传者

crq

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;struct info {    int x;    int y;    bool tag;};int n;vector<info> v;bool cmp(const info& a, const info& b) {    return a.x < b.x;     if(a.x != b.x)     return a.y < b.y;}void work() {    int i, j;    int Size = v.size();    int cnt = 0;// store result;    for(i = 0; i < Size; i++) {        if(v[i].tag==false) {            cnt++;            for(j = i+1; j < Size; j++){                if(v[j].x >= v[i].x && v[j].y >= v[i].y && v[j].tag==false){                    v[i].x = v[j].x; v[i].y = v[j].y;                    v[j].tag = true;                }            }            v[i].tag = true;        }    }    printf("%d\n", cnt);}int main(){    int T, i;    info tmp;    scanf("%d", &T);    while(T--) {        scanf("%d", &n);        v.clear();        for(i = 0; i < n; i++) {            scanf("%d%d", &tmp.x, &tmp.y);            tmp.tag = false;            v.push_back(tmp);        }        sort(v.begin(), v.end(), cmp);        work();    }    return 0;}