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.
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
4 3 6 2 4 0 2 4 7
Sample Output
4
找一个数组,使数组里至少有两个整数属于间隔,每个间隔里的两个数都在这个数组
分析:
贪心,先按照b从小到大排序,在将数组里的数跟后面的间隔前两个比较
代码:
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
int x;
int y;
}b[100001];
bool cmp(node a,node b)
{
return a.y<b.y;
}
int main()
{
int n,j,i,m[10001];
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d%d",&b[i].x,&b[i].y);
sort(b,b+n,cmp);
m[0]=b[0].y-1;
m[1]=b[0].y;
j=0;
for(i=1;i<n;i++)
{
if(b[i].x<=m[j]&&m[j+1]<=b[i].y)
{
continue;
}
if(m[j+1]>=b[i].x&&m[j]<b[i].x)
{
m[j+2]=b[i].y;
j=j+1;
}
if(m[j+1]<b[i].x)
{
m[j+2]=b[i].y-1;
m[j+3]=b[i].y;
j=j+2;
}
}
printf("%d\n",j+2);
}
感受:挺简单的,就是题意一开始没有读懂。。。。