POJ 1716 Integer Intervals (差分约束系统)

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


令d[k]为从区间[0, k]中选取的元素的个数，根据题意建立差分约束系统：d[bi+1]-d[ai]>=2, 0=<d[i+1]-d[i]<=1，然后用spfa求解。

View Code

#include <stdio.h>

#include <string.h>

#include <queue>

using namespace std;

#define MIN(a,b) ((a)<(b)?(a):(b))

#define MAX(a,b) ((a)>(b)?(a):(b))

#define N 10010

#define M 40010

#define INF 0x3fffffff

int n,s,t;

int d[N];

int inq[N];

int first[N],next[M],v[M],w[M];

int e;

void init()

{

    e=0;

    memset(first,-1,sizeof(first));

    s=N;

    t=-N;

}

void insert(int a,int b,int c)

{

    v[e]=b;

    w[e]=c;

    next[e]=first[a];

    first[a]=e++;

}

void spfa()

{

    queue<int> q;

    int a,b;

    memset(inq,0,sizeof(int)*(t+5));

    for(int i=s;i<=t;i++)   d[i]=INF;

    d[s]=0;

    inq[s]=1;

    q.push(s);

    while(!q.empty())

    {

        a=q.front(),q.pop();

        inq[a]=0;

        for(int i=first[a];i!=-1;i=next[i])

        {

            b=v[i];

            if(d[b]>d[a]+w[i])

            {

                d[b]=d[a]+w[i];

                if(inq[b]==0)   inq[b]=1,q.push(b);

            }

        }

    }

    printf("%d\n",-d[t]);

}

int main()

{

    int i,a,b;

    while(~scanf("%d",&n))

    {

        init();

        for(i=0;i<n;i++)

        {

            scanf("%d%d",&a,&b);

            s=MIN(s,a);

            t=MAX(t,b+1);

            insert(a,b+1,-2);

        }

        for(i=s;i<t;i++)    insert(i,i+1,0),insert(i+1,i,1);

        spfa();

    }

    return 0;

}