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,0,sizeof(d));
    int flag=1;
    while(flag){
        flag=0;
        for(int i=0;i<n;i++)
            if(d[edge[i].s]>d[edge[i].e]-2){
                d[edge[i].s]=d[edge[i].e]-2;
                flag=1; 
            }
        for(int i=down;i<up;i++)
            if(d[i+1]>d[i]+1){
                d[i+1]=d[i]+1;
                flag=1;
            }
        for(int i=up-1;i>=down;i--)
            if(d[i]>d[i+1]){
                d[i]=d[i+1];
                flag=1;
            }
    }
}
int main(void)
{
    while(cin>>n)
    {
        up=0;
        down=n;
        for(int i=0;i<n;i++){
            int a,b;
            cin>>a>>b;
            edge[i].s=a;
            edge[i].e=b+1;
            if(down>a) down=a;
            if(up<b+1) up=b+1;
        }
        bellman_ford();
        cout<<d[up]-d[down]<<endl;
    }
    return 0;
}