题目来源：http://noi.openjudge.cn/ch0406/718/
718: Integer Intervals
总时间限制: 1000ms 内存限制: 65536kB
描述
An integer interval [a,b], a < b, is a set of all consecutiveintegers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set containingat least two different integers from each interval.
输入
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, bseparated by a single space, 0 <= a < b <= 10000. They are thebeginning and the end of an interval.
输出
Output the minimal number of elements in a set containing atleast two different integers from each interval.
样例输入
4
3 6
2 4
0 2
4 7
样例输出
4
来源
CEOI 1997
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思路
题意：求最小的整数集合，使得每个区间中都有至少两个数在该集合中。
思路：将区间按末端升序排列。用s1,s2表示在每个区间取的两个数(s1<s2). s1,s2初始化为第一个区间的最后两个数。之后循环其余(n-1)个集合，如果该集合的始端小于等于s1, 则不用改变s1,s2，结果数不变；若该集合的始端大于s1小于等于s2, 则s1=s2, s2变为当前集合的末端，结果数+1；若该集合的始端大于s2,则s1,s2为当前集合的最后两个数，结果数+2。
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代码
#include<iostream>
#include<fstream>
#include<algorithm>
using namespace std;
struct interval {
int a,b;
interval(){}
interval(int aa, int bb): a(aa),b(bb){}
bool operator< (const interval &bint)
{
return b < bint.b;
}
};
const int NMAX = 10005;
interval vec[NMAX] = {};
int main()
{
#ifndef ONLINE_JUDGE
ifstream fin ("0406_718.txt");
int n,i,a,b,s1,s2,ans=0;
fin >> n;
for (i=0; i<n; i++)
{
fin >> a >> b;
interval myint(a,b);
vec[i] = myint;
}
fin.close();
sort(vec, vec+n);
s1 = vec[0].b-1;
s2 = vec[0].b;
ans = 2;
for (i=1; i<n; i++)
{
interval myint = vec[i];
if (myint.a <= s1)
{
continue;
}
else if (myint.a<=s2)
{
ans++;
s1 = s2;
s2 = myint.b;
}
else
{
ans += 2;
s1 = myint.b-1;
s2 = myint.b;
}
}
cout << ans;
return 0;
#endif
#ifdef ONLINE_JUDGE
int n,i,a,b,s1,s2,ans=0;
cin >> n;
for (i=0; i<n; i++)
{
cin >> a >> b;
interval myint(a,b);
vec[i] = myint;
}
sort(vec, vec+n);
s1 = vec[0].b-1;
s2 = vec[0].b;
ans = 2;
for (i=1; i<n; i++)
{
interval myint = vec[i];
if (myint.a <= s1)
{
continue;
}
else if (myint.a<=s2)
{
ans++;
s1 = s2;
s2 = myint.b;
}
else
{
ans += 2;
s1 = myint.b-1;
s2 = myint.b;
}
}
cout << ans;
return 0;
#endif
}