ural 1310 Minimal Coverage

看题都没看懂，不得不说，中国人的思维和老外还是有蛮大区别的，

参考思路：http://www.knightzone.org/wordpress/archives/1414

1303. Minimal Coverage

Time limit: 1.0 second
Memory limit: 64 MB

Given set of line segments [L_i, R_i] with integer coordinates of their end points. Your task is to find the minimal subset of the given set which covers segment [0, M] completely (M is a positive integer).

Input

First line of the input contains an integer M (1 ≤ M ≤ 5000). Subsequent lines of input contain pairs of integers L_i and R_i (−50000 ≤ L_i < R_i ≤ 50000). Each pair of coordinates is placed on separate line. Numbers in the pair are separated with space. Last line of input data contains a pair of zeroes. The set contains at least one and at most 99999 segments.

Output

Your program should print in the first line of output the power of minimal subset of segments which covers segment [0, M]. The list of segments of covering subset must follow. Format of the list must be the same as described in input with exception that ending pair of zeroes should not be printed. Segments should be printed in increasing order of their left end point coordinate.

If there is no covering subset then print “No solution” to output.

Samples

input	output
1 -1 0 -5 -3 2 5 0 0	No solution
1 -1 0 0 1 0 0	1 0 1

Problem Source: II Collegiate Students Urals Programming Contest. Yekaterinburg, April 3-4, 1998

Tags: dynamic programming  (

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#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
     int r;
     int l;
}p[200000],q[200000];
bool cmp(node a,node b)
{
     return a.l<b.l;
}
int main()
{
     int m,i,j,x,y,n,flag;
     while(scanf("%d",&m)!=EOF)
     {
          i=0;
          flag=0;

          while(scanf("%d %d",&x,&y)!=EOF)
          {
               if(x==0&&y==0)
               break;
               if(x<m&&y>0)//与区间[0,m]没有交集的话就没必要保存了
               {
               p[i].l=x;
               p[i].r=y;
               i++;
               }
          }
          n=i;
          int temp=0,right=0,num=0,k;
          sort(p,p+n,cmp);//按区间左界的值排序
          for(i=0;i<n;i++)
          {
               k=-1;
               for(;i<n&&p[i].l<=right;i++)//找右界最大的区间
               {
                    if(p[i].r>temp)//
                    {
                         temp=p[i].r;
                         k=i;
                    }
               }
               if(k==-1)//两个区间没有交集跳出
               {
                    break;
               }
               q[num++]=p[k];//保存找到的区间
               if(temp>=m)//找到，跳出
               {
                    flag=1;
                 break;
               }
               right=temp;//右值要更新
               i--;//i要回退1
          }
          if(k!=-1&&flag)
          {
               printf("%d\n",num);
               for(i=0;i<num;i++)
               printf("%d %d\n",q[i].l,q[i].r);
          }
          else
          printf("No solution\n");
     }
     return 0;
}