poj 1716 差分约束

水水的。

给几个不等式：dis[b]-dis[a]>=2;  0<=dis[i+1]-dis[i]<=1;

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#define inf 1000000000

#define Maxn 10110

#define Maxm 160000

using namespace std;

int index[Maxn],dis[Maxn],vi[Maxn],e,n,Que[2000100];

struct Edge{

    int to,next,val;

}edge[Maxm];

void init()

{

    memset(vi,0,sizeof(vi));

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

    for(int i=0;i<Maxn;i++)

        dis[i]=-inf;

    e=0;

}

void addedge(int from,int to,int val)

{

    edge[e].to=to;

    edge[e].val=val;

    edge[e].next=index[from];

    index[from]=e++;

}

int spfa(int u,int ans)

{

    int i,j,temp,now,head,rear;

    head=rear=0;

    dis[u]=0;

    Que[head++]=u;

    while(head!=rear)

    {

        temp=Que[rear++];

        //cout<<temp<<endl;

        vi[temp]=0;

        for(i=index[temp];i!=-1;i=edge[i].next)

        {

            now=edge[i].to;

            if(dis[temp]+edge[i].val>dis[now])

            {

                dis[now]=dis[temp]+edge[i].val;

                if(!vi[now])

                    Que[head++]=now;

                vi[now]=1;

            }

        }

    }

    return dis[ans];

}

int main()

{

    int m,i,j,a,b,c;

    while(scanf("%d",&n)!=EOF)

    {

    init();

    int maxn=0;

    int minn=inf;

    for(i=1;i<=n;i++)

    {

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

        maxn=max(maxn,b+1);

        minn=min(minn,a);

        addedge(a,b+1,2);

    }

    for(i=minn;i<=maxn;i++)

    {

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

        addedge(i,i+1,0);

    }

    int ans;

    ans=spfa(minn,maxn);

    printf("%d\n",ans);

    }

    return 0;

}