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
",ans);
    }
    return 0;
}