hdu 三部曲1 Is the Information Reliable? 差分约束 bellman_ford算法

Problem 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

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差分约束问题  建图很重要。

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#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cctype>
#include<queue>
#include<stack>
using namespace std;
struct edge
{
    int v,u,len;
}e[200011];
bool change;
int dis[1021];
int e_num,n,m;
char c;
int x,y,w;
void add(int x,int y,int len)
{
    e[e_num].v=x;
    e[e_num].u=y;
    e[e_num].len=len;
    e_num++;
}
bool bellman_ford()
{
    for(int i=0;i<n;i++)
    {
        change=false;
       for(int j=0;j<e_num;j++)
        {
          if(dis[e[j].u]>dis[e[j].v]+e[j].len)
          {
              dis[e[j].u]=dis[e[j].v]+e[j].len;
              change=true;
          }
        }
    }
    return !change;
}
int main()
{
    while(scanf("%d %d",&n,&m)==2)
    {
        int i,j,k;
        e_num=0;
        memset(dis,0,sizeof(dis));
        for(i=0;i<m;i++)
        {
          scanf("
%c %d %d",&c,&x,&y);
          //满足x-y>=w&&y-x<=-w
          if(c=='P')
           {
            scanf("%d",&w);
            add(y,x,w);
            add(x,y,-w);
           }
          else
          {
            //满足x-y<=-1
            add(x,y,-1);
          }
        }
    if(bellman_ford())//如果存在负环，则互相矛盾，不可信
      printf("Reliable
");
     else
      printf("Unreliable
");
  }
     return 0;
}

重在找关系（约束条件），建图