UVA 10020 Minimal coverage

最新推荐文章于 2019-11-06 23:48:58 发布

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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

Sample Output

一开始理解错题意，WA了几次之后才发现这是一道贪心题目。给出区间【a，b】，把各个区间按照a从小到大排序，然后判断a是否满足条件，如果满足条件，找出满足条件中b的最大值的那组数据，然后将这组数据的b设为新的起点a，直到找出a>m的那组区间的前一个，即这些区间就是所求的。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;

struct sa
{
    int x;
    int y;
}data[100007],data1[1000007];

int cmp(const sa a1,const sa b1)
{
    return a1.x<b1.x;
}

int main()
{
    int textcase;
    scanf("%d",&textcase);
    while(textcase--)
    {
        memset(data,0,sizeof(data));
        memset(data1,0,sizeof(data1));
        int m,i=0,j,k;
        scanf("%d",&m);
        int left,right;
        while(1)
        {
            scanf("%d%d",&left,&right);
            if(right==0&&left==0)break;
                data[i].x=left;
                data[i].y=right;
                i++;

        }
        sort(data,data+i,cmp);
        int max_left=0,flag,flag1,max,ans=0,tot=0;
        while(1)
        {
            if(max_left>=m)break;
            flag=max=0;
            for(j=0;j<i;j++)//找出符合条件的区间右面的值的最大的那个值
            {
                if(data[j].x<=max_left)
                {
                    if(data[j].y>max)
                    {
                        flag1=j;
                        max=data[j].y;
                        flag=1;
                    }
                }
            }
            if(flag)
            {
                ans++;
                max_left=max;//替换最大值
                data1[tot++]=data[flag1];
            }
            else break;
        }
        if(flag)
        {
            printf("%d\n",ans);
            for(j=0;j<tot;j++)
            printf("%d %d\n",data1[j].x,data1[j].y);
        }
        else printf("0\n");
        if(textcase)printf("\n");
    }
    return 0;
}