poj 1716

Integer Intervals

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11866 Accepted: 5007

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

Source

CEOI 1997

0 1 2 3 4 5 6 7-----           (0 to 2)    -----       (2 to 4)       -------   (3 to 6)        ------- (4 to 7)

数据如上，要找的就是一个集合，集合与数据的每个集合的交集至少有2个元素，求集合最少的数的个数。

贪心：先按照结尾数排序，第一个集合找末尾2个数，第二个集合，现根据前面判断已经有几个数在交集中了，如果够2个就跳到第三个集合，如果不够就从末尾找，往前补够2个，第三及以后的类推。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int a,b;}p[10005];int n,ans;bool cmp(node x,node y){    return x.b<y.b;}bool v[10005];int main(){    while(scanf("%d",&n)!=EOF){    for(int i=0;i<n;i++)scanf("%d%d",&p[i].a,&p[i].b);    sort(p,p+n,cmp);    memset(v,0,sizeof(v));    v[p[0].b-1]=1;    v[p[0].b]=1;    ans=2;    for(int i=1;i<n;i++)    {        int num=2;        for(int j=p[i].a;j<=p[i].b;j++)        {            if(v[j]==1)num--;        }        if(num>0)        while(num)        {            v[p[i].b-num+1]=1;            ans++;            num--;        }    }    cout<<ans<<endl;}    return 0;}