poj2983--Is the Information Reliable?(差分约束)

Is the Information Reliable?
Time Limit: 3000MS   Memory Limit: 131072K
Total Submissions: 11125   Accepted: 3492

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
Reliabl

给出了 P a  b  w 表示 b在a以北w公里, V a  b 表示 b在a北边,最少1公里,问所有 的条件可不能够所有满足。

由P 能够得到 b - a = w 也就是b - a <= w  &&  a - b <= w ,由 V a  b 得到 b - a >= 1 也就是 a - b <= -1 ;建图,使用最短路,推断是否会有负环。初始dis要所有为0.

 

#include <cstdio>
#include <cstring>
#include <algorithm>
struct node
{
    int u , v , w ;
} p[300000];
int dis[2000] , cnt ;
void add(int u,int v,int w)
{
    p[cnt].u = u ;
    p[cnt].v = v ;
    p[cnt++].w = w ;
}
int main()
{
    int i , j , n , m , u , v , w ;
    char str[10] ;
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        cnt = 0 ;
        while(m--)
        {
            scanf("%s", str);
            if(str[0] == 'P')
            {
                scanf("%d %d %d", &u, &v, &w);
                add(u,v,-w);
                add(v,u,w);
            }
            else
            {
                scanf("%d %d", &u, &v);
                add(u,v,-1);
            }
        }
        memset(dis,0,sizeof(dis));
        for(i = 1 ; i <= n ; i++)
            for(j = 0 ; j < cnt ; j++)
                if( dis[ p[j].v ] >dis[ p[j].u ] + p[j].w )
                    dis[ p[j].v ] = dis[ p[j].u ] + p[j].w ;
        for(j = 0 ; j < cnt ; j++)
            if( dis[ p[j].v ] > dis[ p[j].u ] + p[j].w )
                break;
        if(j < cnt)
            printf("Unreliable
");
        else
            printf("Reliable
");
    }
}


 

 

 

原文地址:https://www.cnblogs.com/gcczhongduan/p/3996027.html