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:
    最全的C#图片处理类ImageHelper.cs
    基于Asp.net C#实现HTML转图片(网页快照)
    WebSiteThumbnail 直接根据html生成图片
    C#里面如何判断一个Object是否是某种类型
    对datatable里面的表进行排序
    spark中RDD的transformation&action
    腾讯大数据之TDW计算引擎解析——Shuffle
    什么是RDD?
    java.io.IOException: No space left on device 错误
    win7 audio repeater 虚拟声卡 屏幕录像专家
  • 原文地址:https://www.cnblogs.com/Roni-i/p/9565229.html
Copyright © 2011-2022 走看看