http://poj.org/problem?id=2983

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

Unreliable
Reliable
这是一道最短路差分约束题，通俗一点讲就是给你一些表达试看是不是可靠，就是存在不存在矛盾。
这里的P A　Ｂ　Ｘ可以转化为Ａ－Ｂ＜＝Ｘ或Ｂ－Ａ＞＝Ｘ
Ｖ　Ａ　Ｂ，则可以转化为Ａ－Ｂ＜＝－１
对于《=的差分约束题，就要用到最短路，而判断信息的可靠与否就要判断图中是否存在负权回路即环。

代码：

#include<iostream>
#include<string.h>
#define  N 1005
#define  M 200005
#define  MAX 99999999
using namespace std;
struct edge{int s,e,w;
       }aa[M];
int dist[N];
int n,m,tot;
bool Bellman_ford()
{     bool flag;本来是进行n-1次松弛的，在这里我用了一些小技巧松弛了n次。
   for(int i=1;i<=n;++i)
   {         flag=true;
           for(int j=1;j<=tot;++j)
            {     int temp=dist[aa[j].s]+aa[j].w;
                 if(dist[aa[j].e]>temp)
                   { dist[aa[j].e]=temp;
                      flag=false;
                    }
             }
             if(flag) return true;如果在某一次dist中的值不改变的时候，则以后不会在改变，直接还回true即可。
    }
    return flag;而在进行第N次松弛时如果还可以松弛则说明图中存在回路，而此时flag为false。否者falg为true因此可以直接返回flag即可。

    }
    int main()
    {    char ch;
        while(cin>>n>>m)
        {       tot=0;
            memset(dist,0,sizeof(dist));
           for(int i=0;i<m;++i)
                    {      getchar();
                         cin>>ch;
                         int a,b,c;
                        cin>>a>>b;
                        if(ch=='P')
                        {   cin>>c;
                            aa[++tot].s=a;
                            aa[tot].e=b;
                            aa[tot].w=-c;
                            aa[++tot].s=b;
                            aa[tot].e=a;
                            aa[tot].w=c;
                        }
                        else{ aa[++tot].s=a;
                               aa[tot].e=b;
                               aa[tot].w=-1;
                             }
                             
                  }
                  if(Bellman_ford()) cout<<"Reliable"<<endl;
                   else cout<<"Unreliable"<<endl;
                        } return 0;
        }