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  • Wooden Sticks

    Wooden Sticks 分享至QQ空间 去爱问答提问或回答

    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
    ", 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;
    }
    


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  • 原文地址:https://www.cnblogs.com/pangblog/p/3262948.html
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