Is the Information Reliable?（差分约束）

Is the Information Reliable?Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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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 withN 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 stationB, 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 integersN (0 < N ≤ 1000) and M (1 ≤ M ≤ 100000).The nextM 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 arrangeN defense stations satisfying all the M tips, otherwise output “Unreliable”.

Sample Input

3 4P 1 2 1P 2 3 1V 1 3P 1 3 15 5V 1 2V 2 3V 3 4V 4 5V 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）

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;#define inf 9999999999;struct node{    int u,v,w;} edge[200000];int dis[200000];int n,m,cnt;int Bellman_ford(){    int i,j;    int flag;    memset(dis,0,sizeof(dis));    for(i=0;i<n;i++)    {        flag = 0;        for(j=0;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=1;            }        }        if(!flag)            break;    }    for(j=0;j<cnt;j++)    {        if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)            return 1;    }    return 0;}int main(){    int i,j;    int u,v,w;    char str;    while(~scanf("%d %d",&n,&m))    {        cnt=0;        for(i=0; i<m; i++)        {            getchar();            scanf("%c",&str);            if(str=='P')            {                scanf("%d %d %d",&u,&v,&w);                edge[cnt].u=u;                edge[cnt].v=v;                edge[cnt].w=w;                cnt++;                edge[cnt].u=v;                edge[cnt].v=u;                edge[cnt].w=-w;                cnt++;            }            else            {                scanf("%d %d",&u,&v);                edge[cnt].u=v;                edge[cnt].v=u;                edge[cnt].w=-1;                cnt++;            }        }        if(Bellman_ford())            puts("Unreliable");        else            puts("Reliable");    }    return 0;}