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poj 1716 Integer Intervals

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. 
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

43 62 40 24 7

Sample Output

4

以前做的是选一个点,这个题是选两个点,不过做法都一样,都是选区间最后两个点
#include<iostream>#include<cstdio>#include<map>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<vector>#include<set>using namespace std;const int maxn=10000+10;struct node{          int x,y;}a[maxn];bool vis[maxn];bool cmp(node xx,node yy){          return xx.y<yy.y;}int main(){          int n;          while(scanf("%d",&n)!=EOF)          {                    for (int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y);                    sort(a+1,a+n+1,cmp);                    memset(vis,0,sizeof(vis));                    vis[a[1].y]=1;                    vis[a[1].y-1]=1;                    int ans=2;                    for (int i=2;i<=n;i++)                    {                              int m=0;                              for (int j=a[i].x;j<=a[i].y;j++)                              {                                        if (vis[j]) m++;                                        if (m==2) break;                              }                              if (m==0)                              {                                        vis[a[i].y]=1;                                        vis[a[i].y-1]=1;                                        ans+=2;                              }                              if (m==1)                              {                                        vis[a[i].y]=1;                                        ans++;                              }                    }                    printf("%d\n",ans);          }          return 0;}