zoukankan      html  css  js  c++  java
  • POJ 1089 Intervals【合并n个区间/贪心】

    There is given the series of n closed intervals [ai; bi], where i=1,2,...,n. The sum of those intervals may be represented as a sum of closed pairwise non−intersecting intervals. The task is to find such representation with the minimal number of intervals. The intervals of this representation should be written in the output file in acceding order. We say that the intervals [a; b] and [c; d] are in ascending order if, and only if a <= b < c <= d. 
    Task 
    Write a program which: 
    reads from the std input the description of the series of intervals, 
    computes pairwise non−intersecting intervals satisfying the conditions given above,
    writes the computed intervals in ascending order into std output

    Input

    In the first line of input there is one integer n, 3 <= n <= 50000. This is the number of intervals. In the (i+1)−st line, 1 <= i <= n, there is a description of the interval [ai; bi] in the form of two integers ai and bi separated by a single space, which are respectively the beginning and the end of the interval,1 <= ai <= bi <= 1000000.

    Output

    The output should contain descriptions of all computed pairwise non−intersecting intervals. In each line should be written a description of one interval. It should be composed of two integers, separated by a single space, the beginning and the end of the interval respectively. The intervals should be written into the output in ascending order.

    Sample Input

    5
    5 6
    1 4
    10 10
    6 9
    8 10

    Sample Output

    1 4
    5 10
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <cmath>
    #include <list>
    #include <deque>
    #include <map>
    #include <set>
    using namespace std;
    #define ll long long
    const double PI = acos(-1.0);
    const int maxn = 101;
    const int INF = 0x3f3f3f3f;
    int dx[]={0,0,-1,1};
    int dy[]={-1,1,0,0};
    
    int n,m;
    int f[maxn],cnt=0,sum=0;
    struct node
    {
        int l,r;
    }a[maxn];
    bool cmp(node a,node b)
    {
        return a.l<b.l;
    }
    int main()
    {
        while(~scanf("%d",&n))
        {
            for(int i=0;i<n;i++)
                scanf("%d %d",&a[i].l,&a[i].r);
            sort(a,a+n,cmp);
            int x=a[0].l,y=a[0].r;
            for(int i=0;i<n;i++)
            {
                if(a[i].l>=x && a[i].l<=y)
                {
                    if(a[i].r>y)
                        y=a[i].r;
                }
                else
                {
                    printf("%d %d
    ",x,y);
                    x=a[i].l;
                    y=a[i].r;
                }
            }
            printf("%d %d
    ",x,y);
        }
    }
    /*
    
    */
  • 相关阅读:
    java中异常的处理
    java中异常的处理
    python入门(一)
    logstash urldecode filter 插件
    logstash urlencode解码
    go get下载第三方包问题的解决
    饿了么这样跳过Redis Cluster遇到的“坑”
    Installation Guide 安装指南
    python pip 切换到阿里云 镜像
    守得云开见月明:一次ASM存储高可用故障解决过程分析
  • 原文地址:https://www.cnblogs.com/Roni-i/p/9565229.html
Copyright © 2011-2022 走看看