uva 10020

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

題意：

有n個閉區間[Li,Ri],給你一個整數，要你求出最少區間覆蓋[Li,Ri].

思路：

參考算法競賽與入門經典（紫書）P233，首先進行預處理，在區間[Li,Ri]外面的首先要去掉，預處理完成後在相互包含的情況下，小區間顯然不能考慮。再把各個區間按照

Li從小排到大，如果li>0肯定就是無解，如果Li<=0進行搜索，讓Ri儘可能的大。

代碼：

#include<cstdio>
#include<algorithm>
int L[50001][2];
using namespace std;
int T,M;
int f,n;
int pos,maxn,sum,posmax;
struct node
{
    int l;
    int r;
};
node arr[100001],brr[100001];
bool cmp(node a,node b)
{
    return a.l<b.l;
}
void solve()
{
    sort(arr,arr+n,cmp);
    if(arr[0].l>0)
        printf("0\n");
    else
    {
        pos=0;maxn=0;sum=1;
        while(pos+1<n&&arr[pos+1].l<=0)
        {
            pos++;
            if(arr[pos].r>arr[maxn].r)
                maxn=pos;

        }
        pos=maxn; brr[1].l=arr[pos].l;brr[1].r=arr[pos].r;
        while(pos<n&&brr[sum].r<=M)
        {
            f=1;
            posmax=maxn;
            while(pos+1<n&&arr[pos+1].l<=arr[maxn].r)
            {
                f=0;
                pos++;
                if(arr[pos].r>arr[posmax].r) posmax=pos;
            }
            if(f)   break;
            maxn=posmax;
            sum++;
            brr[sum].l=arr[maxn].l;
            brr[sum].r=arr[maxn].r;
        }
        if(brr[sum].r>=M)
        {
            printf("%d\n",sum);
            for(int i=1;i<=sum;i++)
                printf("%d %d\n",brr[i].l,brr[i].r);

        }
        else printf("0\n");
    }

}
void init()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&M);
         n=0;
        while(scanf("%d%d",&arr[n].l,&arr[n].r))
        {
            if(arr[n].l==0&&arr[n].r==0)
                break;
                n++;
        }
        solve();
    }
}
int main()
{

    init();
   // solve();
    return 0;
}