POJ 1716 Integer Intervals（差分约束）

Integer Intervals

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 14689

 

Accepted: 6226

Description

An integer interval [a,b], a < b, is a set of all consecutive integersbeginning 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.

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 asingle space, 0 <= a < b <= 10000. They are the beginning and the endof an interval.

Output

Out put the minimal number of elements in a set containing at least two differentintegers from each interval.

SampleInput

4

3 6

2 4

0 2

4 7

Sample Output

4

差分约束总结：传送门

题意：给出数轴上的n个闭合int型区间。现在要在数轴上任意取一堆元素，构成一个元素集合V，要求给出的每个区间和元素集合V的交集至少有两个不同的元素，求集合V最小的元素个数。

思路：（1）贪心

           （2）差分约束：

对于差分约束，我们可以这样考虑，令sum[x]为[0,x]内所在集合V中元素个数，

那么我们就能很容易得到：对于每个区间【a,b】,sum[b+1]-sum[a]>=2。

但千万不能忘了隐含的约束条件：0<=sum[i+1]-sum[i]<=1.

综上所述：不等数组统一化为：sum[b+1]-sum[a]>= 2         建边：add(a,b+1,2)

                                                  sum[i+1]-sum[i]>= 0            建边：add(i,i+1,0)

                                                  sum[i]-sum[i+1]>= -1           建边：add(i+1,i,-1)

建边完毕后跑个最长路即可。

#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <algorithm>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<(b);++i)typedef long long ll;const int maxn= 10005;const int mod = 10007;const int INF = 0x3f3f3f3f;const double eps = 1e-6;#define rush() int T;scanf("%d",&T);while(T--)int dis[maxn];int vis[maxn];int num[maxn];int head[maxn];int n,cnt;struct node{    int v,w,next;} e[maxn*3];void add(int u,int v,int w){    e[cnt].v=v;    e[cnt].w=w;    e[cnt].next=head[u];    head[u]=cnt++;}void init(){    cnt=0;    mst(head,-1);}int spfa(int s,int t){    mst(vis,0);    mst(num,0);    for(int i=0;i<=t;i++)    {        dis[i]=-INF;    }    queue<int>q;    vis[s]=1;    dis[s]=0;    num[s]++;    q.push(s);    while(q.size())    {        int now=q.front();        q.pop();        vis[now]=0;        for(int i=head[now]; i!=-1; i=e[i].next)        {            int v=e[i].v;            int w=e[i].w;            if(dis[v]<dis[now]+w)            {                dis[v]=dis[now]+w;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    return dis[t];}int main(){    int a,b;    while(~scanf("%d",&n))    {        init();        int Max=0;        for(int i=0;i<n;i++)        {            scanf("%d%d",&a,&b);            add(a,b+1,2);            Max=max(Max,b+1);        }        for(int i=0;i<=Max;i++)        {            add(i,i+1,0);            add(i+1,i,-1);        }        int ans=spfa(0,Max);        printf("%d\n",ans);    }    return 0;}