POJ 1716 Integer Intervals（差分约束）

题意：给你一些区间，问你确定一个集合，使得这个集合和这些区间至少有两个数字相同，求这个集合最小元素个数

思路：令d[i]为0到i区间的不同元素个数，那么有d[b]-d[a-1]>=2，d[i+1]>=d[i]，d[i]>=d[i+1]-1,d[i]>=d[0]，然后建边跑最长路求最小值就可以了，这里区间从0开始计数，我就都++了，如果不++的话那么要变成d[b+1]-d[a]>=2，这样跑了980MS...而我这样居然只跑了500MS..

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define inf -1e9
const int maxn = 10005;
vector<pair<int,int> >e[maxn];
int d[maxn],cnt[maxn],inq[maxn],n;
int spfa(int num)
{
    memset(cnt,0,sizeof(cnt));
	memset(inq,0,sizeof(inq));
	for(int i = 0;i<=num;i++)
		d[i]=inf;
	inq[0]=1;
	queue<int>q;
	d[0]=0;
	q.push(0);
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		inq[u]=0;
		for(int i  =0;i<e[u].size();i++)
		{
			int v = e[u][i].first;
			if(d[v]<d[u]+e[u][i].second)
			{
				d[v]=d[u]+e[u][i].second;
				if(!inq[v])
				{
					inq[v]=1;
					q.push(v);
					if(++cnt[v]>n)
						return false;
				}
			}
		}
	}
	return 1;
}
int main()
{
   int num = 0;
   while(scanf("%d",&n)!=EOF)
   {
	   for(int i = 0;i<=num;i++)
		   e[i].clear();
	   num = 0;
       for(int i = 1;i<=n;i++)
	   {
		   int u,v;
		   scanf("%d%d",&u,&v);
		   u++,v++;
		   num = max(num,v);
		   e[u-1].push_back(make_pair(v,2));
	   }
	   for(int i = 1;i<=num;i++)
	   {
		   e[0].push_back(make_pair(i,0));
		   e[i].push_back(make_pair(i+1,0));
		   e[i+1].push_back(make_pair(i,-1));
	   }
	   spfa(num);
	   printf("%d\n",d[num]);
   }
}

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