UVA10020 - Minimal coverage

 

 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

 

2

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

1
-1 0
0 1
0 0

 

Sample Output

 

0

1
0 1
题目大意：给定N条线段，最少需要选择多少条线段可以覆盖区间[0,M]。
题解：贪心问题。对于输入的数据，我们可以过滤掉和区间[0,M]没有公共区间的线段，因为这些线段没有意义。之后按左端值对线段进行升序排序。假设s为需要覆盖区间的起点，如果区间1的左端点大于s，则说明问题无解，否则选择左端点小于等于s的右端点值最大的区间。选择区间[Li,Ri]之后，新的起点应该设置为Ri,并重复进行上述操作，直到遇到无法覆盖的情况或者已经覆盖完毕。我这题WA了三次，原因是每更新一次起点之后，忘记判断不能够再进行覆盖的情况，也就是说之后的所有点的左端点都在起点的右边，害得我一直循环，o(╯□╰)o。

#include<stdio.h>
#include<string.h>
#define MAXSN 100005
typedef struct
{
    long x;
    long y;
} NODE;
NODE s[MAXSN];
NODE f[MAXSN];
void qsort(long l,long r)
{
    long i,j,mid;
    NODE temp;
    i=l;
    j=r;
    mid=s[(l+r)>>1].x;
    while(i<=j)
    {
        while(s[i].x<mid) i++;
        while(s[j].x>mid) j--;
        if(i<=j)
        {
            temp=s[i];
            s[i]=s[j];
            s[j]=temp;
            i++;
            j--;

        }
    }
    if(i<r) qsort(i,r);
    if(j>l) qsort(l,j);
}
int main(void)
{
    long i,j,k,m,n,ans,a,b,r,max,flag;
    NODE tt;
    scanf("%ld",&n);
    while(n--)
    {
        scanf("%ld",&m);
        ans=0;
        r=0;
        k=0;
        while(scanf("%ld%ld",&a,&b)==2)
        {
            if(a==0&&b==0) break;
            if((a<=0&&b<=0)||(a>=m&&b>=m)) continue;
            s[ans].x=a;
            s[ans].y=b;
            ans++;
        }
        qsort(0,ans-1);
        if(s[0].x>0)
        {
            printf("0\n");
            if(n) printf("\n");
            continue;
        }
        i=0;
        max=-1;
        j=0;
        flag=0;
        while(i<ans)
        {
            while(s[i].x<=r&&i<ans)
            {
                if(s[i].y>max)
                {
                    tt=s[i];
                    max=s[i].y;

                }
                i++;
            }
            f[j++]=tt;
            r=max;
            if(r>=m) break;
            if(s[i].x>r) break;
        }
        if(r<m)
            printf("0\n");
        else
        {
            printf("%ld\n",j);
            for(i=0; i<j; i++)
                printf("%ld %ld\n",f[i].x,f[i].y);
        }
        if(n) printf("\n");

    }
    return 0;
}