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  • 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;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/algorithms/p/2606861.html
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