Minimal coverage-区间覆盖问题

最新推荐文章于 2019-02-21 12:56:46 发布

lele_pipi

最新推荐文章于 2019-02-21 12:56:46 发布

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分类专栏：贪心高效算法设计文章标签：高效算法设计贪心区间覆盖问题

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Minimal coverage

The Problem

Given several segments of line (int the X axis) with coordinates [L_i,R_i]. 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 "L_i R_i"(|L_i|, |R_i|<=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 (L_i), 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 在别人的博客里发现几组数据

Sample Output

题意:数轴上有多个闭区间[Li,Ri],选择尽量少的区间覆盖一条指定线段[0,M];

[分析]:把各个区间按照Li从小到大排序,如果区间i的终点<线段起点,直接pass,否则如果线段的起点处于区间内,则找出区间Ri最大的那个,然后把线段的起点更新为Ri最大的那个,继续查找下一个区间,直到找到Ri>m,结束查找.

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int a,b;
}ss[100010];
int cmp(node A,node B)  //将区间按左端点从小到大排列,如果左端点相同,按右端点从小到大排列
{
    if(A.a==B.a)
        return A.b<B.b;
    return A.a<B.a;
}
int main()
{
    int n,m,l,s,flag,cnt,i,f,t,aa[100010][2],k,g;
    scanf("%d",&n);
    while(n--)
    {
        k=0;flag=0;l=0;
        scanf("%d",&m);
        while(scanf("%d%d",&ss[l].a,&ss[l].b)!=EOF&&(ss[l].a!=0||ss[l].b!=0))
        {
            l++;
        }
        sort(ss,ss+l,cmp);
        f=0;s=0;cnt=1;
        for(i=0;i<l;i++)
        {
            if(s>ss[i].b)   //如果区间的右端点比线段的起点小,直接不用考虑
            {
                continue;
            }
            else if(s>=ss[i].a) //如果线段的起点处于区间内,并且该区间的右端点比上一个满足条件的区间的右端点大,即覆盖区域更往后,则将该区间的左右端点记录下来
            {
                if(f<=ss[i].b)
                {
                   f=ss[i].b;
                   g=ss[i].a;
                }
            }
            else if(s<ss[i].a)  //当区间左端点>线段起点,这时候要更新线段起点为之前区间覆盖最远的地方
            {
                if(f<ss[i].a)   //如果之前区间覆盖最远的地方<当前区间的左端点,说明中间中断了,不可能将线段覆盖完,flag=1,直接跳出循环
                {
                    flag=1;
                    break;
                }
                else
                {
                    aa[k][0]=g;
                    aa[k][1]=f;
                    k++;
                    s=f;
                    i--;
                    cnt++;
                }

            }
            if(f>=m)    //当把线段覆盖完了,就不用考虑后面的区间,提前结束循环
                break;
        }
        if(f<m||flag==1)
            printf("0\n");
        else
        {
            printf("%d\n",cnt);
            aa[k][0]=g;aa[k][1]=f;k++;  //最后一组符合条件的区间未加入aa数组中
            for(i=0;i<k;i++)
            {
                printf("%d %d\n",aa[i][0],aa[i][1]);
            }
        }
        if(n)
        printf("\n");
    }

}