poj 1716 Integer Intervals (差分约束 或 贪心)

Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12192		Accepted: 5145

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

4
3 6
2 4
0 2
4 7

Sample Output

4

Source

CEOI 1997

参考: http://blog.csdn.net/lyy289065406/article/details/6648679

贪心算法的思想我没证明，是参考别人的 ，不过不知道代码为什么没有过..所以就不贴出来了

差分约束的算法我自己也想了一下，其实并不难，不知道为什么老是错..可能poj的数据比较大吧

所以代码我也是参考别人的，自己的改了一段时间还没改好就没改了，贴一下：

 //Accepted    324K    297MS    C++    1088B    2013-12-01 22:11:55
 /*

     题意:
         给出n个区间[ai,bi]，求在每个区间内最少有两个数的一组数

     差分约束：
         S[ai - 1] <= S[bi] - 2
         S[i] <= S[i - 1] + 1
         S[i - 1] <= S[i] 

     自己看着建吧.. 

 */
 #include<iostream>
 #include<queue>
 #include<string.h>
 #define N 10005
 #define inf 0x7ffffff
 using namespace std;
 struct node{
     int s,e;
 }edge[N];
 int d[N];
 int n,up,down;
 void bellman_ford()
 {
     memset(d,,sizeof(d));
     int flag=;
     while(flag){
         flag=;
         for(int i=;i<n;i++)
             if(d[edge[i].s]>d[edge[i].e]-){
                 d[edge[i].s]=d[edge[i].e]-;
                 flag=;
             }
         for(int i=down;i<up;i++)
             if(d[i+]>d[i]+){
                 d[i+]=d[i]+;
                 flag=;
             }
         for(int i=up-;i>=down;i--)
             if(d[i]>d[i+]){
                 d[i]=d[i+];
                 flag=;
             }
     }
 }
 int main(void)
 {
     while(cin>>n)
     {
         up=;
         down=n;
         for(int i=;i<n;i++){
             int a,b;
             cin>>a>>b;
             edge[i].s=a;
             edge[i].e=b+;
             if(down>a) down=a;
             if(up<b+) up=b+;
         }
         bellman_ford();
         cout<<d[up]-d[down]<<endl;
     }
     return ;
 }