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。
#include<stdio.h>
#include<string.h>
const int maxn = ;
const int INF = 0x3f3f3f3f;
struct node
{
int u,v,w;
}edge[maxn];
int dis[maxn];
int n,m,cnt;
//普通的Bellman_ford算法
bool Bellman_ford()
{
bool flag;
for(int i = ; i <= n; i++)
dis[i] = INF;
for(int i = ; i <= n; i++)
{
flag = false;
for(int j = ; j < cnt; j++)
{
if(dis[edge[j].v] > dis[edge[j].u] + edge[j].w)
{
dis[edge[j].v] = dis[edge[j].u] + edge[j].w;
flag = true;
}
}
if( !flag )
break;
}
for(int j = ; j < cnt; j++)
{
if(dis[edge[j].v] > dis[edge[j].u] + edge[j].w)
return true;
}
return false;
} int main()
{
int u,v,w;
char ch;
while(scanf("%d %d",&n,&m)!= EOF)
{
cnt = ;
for(int i = ; i < m; i++)
{
getchar();
scanf("%c",&ch);
if(ch == 'P')
{
scanf("%d %d %d",&u,&v,&w);//双向边
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt++].w = w; edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt++].w = -w; }
else
{
scanf("%d %d",&u,&v);//单向边
edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt++].w = -;
}
}
if(Bellman_ford())
printf("Unreliable\n");
else printf("Reliable\n");
}
return ;
}