Is the Information Reliable?（差分约束）

Description

The galaxy war between the Empire Draco and the Commonwealth of Zibu broke out 3 years ago. Draco established a line of defense called Grot. Grot is a straight line with N defense stations. Because of the cooperation of the stations, Zibu’s Marine Glory cannot march any further but stay outside the line.

A mystery Information Group X benefits form selling information to both sides of the war. Today you the administrator of Zibu’s Intelligence Department got a piece of information about Grot’s defense stations’ arrangement from Information Group X. Your task is to determine whether the information is reliable.

The information consists of M tips. Each tip is either precise or vague.

Precise tip is in the form of P A B X, means defense station A is X light-years north of defense station B.

Vague tip is in the form of V A B, means defense station A is in the north of defense station B, at least 1 light-year, but the precise distance is unknown.

Input

There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 1000) and M (1 ≤ M ≤ 100000).The next M line each describe a tip, either in precise form or vague form.

Output

Output one line for each test case in the input. Output “Reliable” if It is possible to arrange N defense stations satisfying all the M tips, otherwise output “Unreliable”.

Sample Input

3 4

P 1 2 1

P 2 3 1

V 1 3

P 1 3 1

5 5

V 1 2

V 2 3

V 3 4

V 4 5

V 3 5

Sample Output UnreliableReliable

题意：有N个车站，给出一些点的精确信息和模糊信息，精确信息给出两点的位置和距离，模糊信息给出两点的位置，但距离大于等于一。试确定是否所有的信息满足条件；

　思路：对于精确信息，可以得出两个差分条件，b-a = c;可以化为b-a >= c && a - b <= -c;(因为是精确信息，故要建立双向边)

对于模糊信息，只能得出一个差分条件，可以化为 b - a <= 1;所以a <= b-1;说明b到a有一条长度为-1的边；（模糊信息，建立单向边）

WA了N次，最终不知道为什么。看到别人的博客里说Bellman_ford判负环比SPFA简单，它不用考虑不连通的情况，而SPFA要考虑是否连通，保证从源点开始，能到达各个顶点，这样才能保证差分约束里的各个不等式成立。因为要是源点到达不了某个顶点的话(即图是不连通的)，那么从该顶点就无法入队，导致从该顶点出发的所有不等式，都没有得到检查，因此要添加一个超级源点，而Bellman_ford不需要添加源点，每个顶点都能被松弛n-1次。

 1 #include<stdio.h>

 2 #include<string.h>

 3 const int maxn = 210000;

 4 const int INF = 0x3f3f3f3f;

 5 struct node

 6 {

 7     int u,v,w;

 8 }edge[maxn];

 9 int dis[maxn];

10 int n,m,cnt;

11 //普通的Bellman_ford算法

12 bool Bellman_ford()

13 {

14     bool flag;

15     for(int i = 1; i <= n; i++)

16         dis[i] = INF;

17     for(int i = 1; i <= n; i++)

18     {

19         flag = false;

20         for(int j = 0; j < cnt; j++)

21         {

22             if(dis[edge[j].v] > dis[edge[j].u] + edge[j].w)

23             {

24                 dis[edge[j].v] = dis[edge[j].u] + edge[j].w;

25                 flag = true;

26             }

27         }

28         if( !flag )

29             break;

30     }

31     for(int j = 0; j < cnt; j++)

32     {

33         if(dis[edge[j].v] > dis[edge[j].u] + edge[j].w)

34             return true;

35     }

36     return false;

37 }

38 

39 int main()

40 {

41     int u,v,w;

42     char ch;

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

44     {

45         cnt = 0;

46         for(int i = 0; i < m; i++)

47         {

48             getchar();

49             scanf("%c",&ch);

50             if(ch == 'P')

51             {

52                 scanf("%d %d %d",&u,&v,&w);//双向边

53                 edge[cnt].u = u;

54                 edge[cnt].v = v;

55                 edge[cnt++].w = w;

56 

57                 edge[cnt].u = v;

58                 edge[cnt].v = u;

59                 edge[cnt++].w = -w;

60 

61             }

62             else

63             {

64                 scanf("%d %d",&u,&v);//单向边

65                 edge[cnt].u = v;

66                 edge[cnt].v = u;

67                 edge[cnt++].w = -1;

68             }

69         }

70         if(Bellman_ford())

71             printf("Unreliable\n");

72         else printf("Reliable\n");

73     }

74     return 0;

75 }

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