poj 1364 差分约束

思路：设dis[i]为从0点到第i点的序列总和。那么对于A B gt  k 来讲意思是dis[B+A]-dis[A]>k; 对于A B lt k来讲就是dis[B+A]-dis[A]<k;将两个不等式都化为

dis[A]-dis[B+A]<=-k-1;  dis[A+B]-dis[A]<=k-1;那么就可以根据公式来建边了，用bellman_ford算法判断是否存在负圈就行了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define Maxn 1100
 #define inf 1<<30
 using namespace std;
 int dis[Maxn],vi[Maxn],n,index[Maxn],e;
 struct Edge{
     int from,to,val,next;
 }edge[Maxn*];
 void addedge(int from, int to, int val)
 {
     edge[e].from=from;
     edge[e].to=to;
     edge[e].val=val;
     edge[e].next=index[from];
     index[from]=e++;
 }
 void init()
 {
     int i;
     memset(index,-,sizeof(index));
     memset(vi,,sizeof(vi));
     for(i=;i<Maxn;i++)
         dis[i]=inf;
     e=;
 }
 int bellman_ford()
 {
     int i,j,temp,flag;
     for(i=;i<=n+;i++)
     {
         flag=;
         for(j=;j<e;j++)
         {
             temp=edge[j].from;
             //cout<<edge[j].from<<" "<<edge[j].to<<" "<<edge[j].val<<" "<<j<<" "<<e<<endl;
             if(dis[temp]+edge[j].val<dis[edge[j].to])
             {
                 dis[edge[j].to]=dis[temp]+edge[j].val;
                 flag=;
             }
         }
         if(flag)
         return ;
     }

     return ;
 }
 int main()
 {
     int i,j,a,b,m,k;
     char str[];
     while(scanf("%d",&n)!=EOF,n)
     {
         scanf("%d",&m);
         init();
         for(i=;i<=m;i++)
         {
             scanf("%d%d%s%d",&a,&b,&str,&k);
             if(str[]=='g')
                 addedge(a+b+,a,-k-);
             else
                 addedge(a,a+b+,--k);
         }
         if(bellman_ford())
             printf("lamentable kingdom\n");
         else
             printf("successful conspiracy\n");
     }
     return ;
 }