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。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #define Maxn 1100

 6 #define inf 1<<30

 7 using namespace std;

 8 int dis[Maxn],vi[Maxn],n,index[Maxn],e;

 9 struct Edge{

10     int from,to,val,next;

11 }edge[Maxn*1000];

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

13 {

14     edge[e].from=from;

15     edge[e].to=to;

16     edge[e].val=val;

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

18     index[from]=e++;

19 }

20 void init()

21 {

22     int i;

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

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

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

26         dis[i]=inf;

27     e=0;

28 }

29 int bellman_ford()

30 {

31     int i,j,temp,flag;

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

33     {

34         flag=1;

35         for(j=0;j<e;j++)

36         {

37             temp=edge[j].from;

38             //cout<<edge[j].from<<" "<<edge[j].to<<" "<<edge[j].val<<" "<<j<<" "<<e<<endl;

39             if(dis[temp]+edge[j].val<dis[edge[j].to])

40             {

41                 dis[edge[j].to]=dis[temp]+edge[j].val;

42                 flag=0;

43             }

44         }

45         if(flag)

46         return 1;

47     }

48 

49     return 0;

50 }

51 int main()

52 {

53     int i,j,a,b,m,k;

54     char str[10];

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

56     {

57         scanf("%d",&m);

58         init();

59         for(i=1;i<=m;i++)

60         {

61             scanf("%d%d%s%d",&a,&b,&str,&k);

62             if(str[0]=='g')

63                 addedge(a+b+1,a,-k-1);

64             else

65                 addedge(a,a+b+1,--k);

66         }

67         if(bellman_ford())

68             printf("lamentable kingdom\n");

69         else

70             printf("successful conspiracy\n");

71     }

72     return 0;

73 }