poj1716——差分约束

poj1716——差分约束

Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13029		Accepted: 5516

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
题意：给定n个区间，求一个集合，每个区间都有两个数在集合里，求集合里的数的最小个数
思路：设s(x)为0-x中集合里的数的个数，则对区间[a,b]，有s(b)-s(a-1)>=2，0<=s(i+1)-s(i)<=1；移项得s(b)-s(a-1)>=2,s(i+1)-s(i)>=0,s(i)-s(i+1)>=-1，符合dist[v]>=dist[u]+e，以此条件建图并松弛，求最长路即可，改为求最短路WA，不知道为什么。。

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>



using namespace std;



const int maxn=10010;

const int INF=(1<<28);



int N;

int a,b;

struct Edge

{

    int u,v,w;

};Edge edge[maxn*10];int e;

int Min,Max;

int dist[maxn];



void add_edge(int u,int v,int w)

{

    edge[e++]={u,v,w};

}



bool relax(int cur)

{

    int tmp=dist[edge[cur].u]+edge[cur].w;

    if(tmp>dist[edge[cur].v]){

        dist[edge[cur].v]=tmp;

        return true;

    }

    return false;

}



int bellman_ford()

{

    memset(dist,0,sizeof(dist));

    for(int i=Min;i<Max+1;i++){

        bool flag=0;

        for(int j=0;j<e;j++){

            if(relax(j)) flag=1;

        }

        if(!flag) break;

    }

    return dist[Max+1]-dist[Min];

}



int main()

{

    while(cin>>N){

        e=0;

        Min=INF;Max=-INF;

        for(int i=0;i<N;i++){

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

            add_edge(a,b+1,2);

            if(a<Min) Min=a;

            if(b>Max) Max=b;

        }

        for(int i=Min;i<=Max;i++){

            add_edge(i,i+1,0);

            add_edge(i+1,i,-1);

        }

        cout<<bellman_ford()<<endl;

    }

    return 0;

}

View Code