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 "); 72 else printf("Reliable "); 73 } 74 return 0; 75 }