Is the Information Reliable? -POJ2983差分约束

Time Limit: 3000MS	Memory Limit: 131072K

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

Source

POJ Monthly–2006.08.27, Dagger

题意：有n个防御塔和m个情报，判断这些情报是不是可信，情报有两种：（1）：P u v w 表示u在v的w光年处 。（2）：V u v 表示ｕ在ｖ北的一光年之外（包括一光年）

思路：Dis[i]表示i在源点北面Dis[i]光年，所以对于（1）：P uv w 可以表示为Dis[u]-Dis[v]=w,所以Dis[u]-Dis[v]>=w且Dis[v]-Dis[u]>=-w,对于(2): V u v 可以表示为Dis[u]-Dis[v]>=1,所以建立图，跑最长路，判断是不是有矛盾即正环，我用的

SPFA，要判断自环。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <string>
#include <vector>
#include <stack>
#include <iostream>
#include <algorithm>

using namespace std;

const int INF = 0x3f3f3f3f;

const int MaxN = 1100;

const int MaxM = 100100;

typedef struct node
{
    int v,w,next;
}Line;

Line Li[MaxM*2];

int Head[MaxN],top;

int Dis[MaxN],Du[MaxN];

bool vis[MaxN];

int n,m;

void AddEdge(int u,int v,int w)
{
    Li[top].v = v ; Li[top].w = w;

    Li[top].next = Head[u];

    Head[u] = top++;
}

bool SPFA()
{
    queue<int>Q;

    for(int i=1;i<=n;i++)//相当于有一个超级源点
    {
        Dis[i] = 0;

        vis[i]=true;

        Du[i]=1;

        Q.push(i);
    }

    while(!Q.empty())
    {
        int u = Q.front();

        Q.pop();

        if(Du[u]>n)//有矛盾
        {
            return false;
        }
        for(int i = Head[u];i!=-1;i = Li[i].next)
        {
            int v = Li[i].v;

            if(Dis[v]<Dis[u]+Li[i].w)
            {
                Dis[v] = Dis[u]+Li[i].w;

                if(!vis[v])
                {
                    Q.push(v);
                    vis[v] = true;
                    Du[v]++;
                }
            }
        }
        vis[u]=false;
    }
    return true;
}

int main()
{
    char Op[5];

    int u,v,w;

    while(~scanf("%d %d",&n,&m))
    {
        memset(Head,-1,sizeof(Head));

        top = 0;

        bool flag=false;

        for(int i=1;i<=m;i++)
        {
            scanf("%s",Op);

            if(Op[0]=='P')
            {
                scanf("%d %d %d",&u,&v,&w);

                if(u==v&&w!=0)//自环
                {
                    flag=true;
                }

                AddEdge(u,v,w);

                AddEdge(v,u,-w);
            }
            else
            {
                scanf("%d %d",&u,&v);

                if(u==v)
                {
                    flag=true;
                }

                AddEdge(u,v,1);
            }
        }
        if(flag||!SPFA())
        {
            printf("Unreliable\n");
        }
        else
        {
            printf("Reliable\n");
        }
    }
    return 0;
}