zoukankan      html  css  js  c++  java
  • UVA 10020

    原题:

    The Problem

    Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M].

    The Input

    The first line is the number of test cases, followed by a blank line.

    Each test case in the input should contains an integer M(1<=M<=5000), followed by pairs "Li Ri"(|Li|, |Ri|<=50000, i<=100000), each on a separate line. Each test case of input is terminated by pair "0 0".

    Each test case will be separated by a single line.

    The Output

    For each test case, in the first line of output your programm should print the minimal number of line segments which can cover segment [0,M]. In the following lines, the coordinates of segments, sorted by their left end (Li), should be printed in the same format as in the input. Pair "0 0" should not be printed. If [0,M] can not be covered by given line segments, your programm should print "0"(without quotes).

    Print a blank line between the outputs for two consecutive test cases.

    Sample Input

    2
    
    1
    -1 0
    -5 -3
    2 5
    0 0
    
    1
    -1 0
    0 1
    0 0
    

    Sample Output

    0
    
    1
    0 1


    分析:经典的贪心题,记得贪心一点就能做对了~

    代码:
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    
    struct interval
    {
        int x, y;
        
    }itvl[100006], max_end[100006];
    
    bool cmp(const interval &a, const interval &b) 
    {
        return a.x < b.x;
    }
    
    
    int main()
    {
        int kase;
        cin >> kase;
        getchar();
        while (kase--){
            int ok = 0;
            int m,n=0;
            int cnt = 0;
            cin >> m;
            while (scanf("%d%d", &itvl[n].x, &itvl[n].y) && (itvl[n].x != 0 || itvl[n].y != 0)){
                ++n;
            }
            sort(itvl, itvl + n,cmp);
    
    
            int beg = 0, end = m;
            while (beg < end){
                max_end[cnt].y = 0;
                for (int i = 0; i < n; i++){
                    if (itvl[i].x <= beg){
                        if (itvl[i].y>max_end[cnt].y)    {
                            max_end[cnt] = itvl[i];
                            ok = 1;
                        }
                    }
                    else{
                        break;
                    }
                }
                if (ok == 0)        break;
                beg = max_end[cnt].y;
                cnt++;
            }
    
            //                        输出
    
            if (ok == 1){
                cout << cnt << endl;
                for (int i = 0; i < cnt; i++){
                    cout << max_end[i].x << ' ' << max_end[i].y << endl;
                }
            }
            else{
                cout <<'0'<< endl;
            }
            if (kase != 0)            cout << endl;
        
        }
        return 0;
    }
  • 相关阅读:
    02-17 位图验证码(一般处理程序)+AJAX
    02-18 报表
    SQLite 函数大全
    SQLite中的时间日期函数(转)
    DES,AeS加解密,MD5,SHA加密
    suspendlayout
    AES--高级数据加密标准
    C#中Validating和Validated事件
    Net操作Excel(终极方法NPOI)
    decimal,float和double的区别
  • 原文地址:https://www.cnblogs.com/shawn-ji/p/4711787.html
Copyright © 2011-2022 走看看