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

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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posted on 2013-11-29 21:48 straw_berry 阅读(257) 评论(0) 编辑收藏举报