Poj 1201 intervals

Difference Constraint

The meaning of the question is simple, but it is difficult to establish constraints (Edge building ). You can initialize inf to obtain the minimum value and then output-Dis [maxn]. You can also initialize-inf to find the maximum and output dis [maxn].

When the maximum value is obtained:

Minn is the smallest and maxn is the largest.

Enter u, v Len build constraints for U-> V = Len, and finally between Minn and maxn also create I-> I-1 =-1, I-1-> I = 0.

Finally find minn-1 ~ Maxn.

The minimum value is that the edge is not changed, and the distance is changed to negative.

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6#define LL long longusing namespace std;int n,maxn,minn,m;struct lx{    int v,len;};vector<lx>g[51001];queue<int>q;void SPFA(){    bool vis[51001];    int dis[51001];    for(int i=0;i<=maxn;i++)        vis[i]=0,dis[i]=-INF;    vis[minn-1]=1,dis[minn-1]=0;    q.push(minn-1);    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int j=0;j<g[u].size();j++)        {            int v=g[u][j].v;            int len=g[u][j].len;//            printf("v=%d,u=%d:%d>%d+%d=\n",v,u,dis[v],dis[u],len);//                system("pause");            if(dis[v]<dis[u]+len)            {                dis[v]=dis[u]+len;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    printf("%d\n",dis[maxn]);//    for(int i=minn-1;i<=maxn;i++)//        printf("%d : %d ==\n",i,dis[i]);}int main(){    while(scanf("%d",&m)!=EOF)    {        minn=INF,maxn=0;        for(int i=0;i<51001;i++)            g[i].clear();        int u,v,len;        lx now;        while(m--)        {            scanf("%d%d%d",&u,&v,&len);            maxn=max(maxn,v);            minn=min(minn,u);            now.v=v,now.len=len;            g[u-1].push_back(now);        }        for(int i=minn;i<=maxn;i++)        {            now.v=i-1,now.len=-1;            g[i].push_back(now);            now.v=i,now.len=0;            g[i-1].push_back(now);        }//        for(int i=minn-1;i<=maxn;i++)//        {//            printf("%d-> ",i);//            for(int j=0;j<g[i].size();j++)//                printf("%d:%d  ",g[i][j].v,g[i][j].len);//            printf("\n");//        }        while(!q.empty())q.pop();        SPFA();    }}

Poj 1201 intervals