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 2 1 3 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;
}