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 }

 

posted @ 2013-07-18 14:42  fangguo  阅读(160)  评论(0)  编辑  收藏  举报