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  • 图论--差分约束--POJ 2983--Is the Information Reliable?

    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 的条件可不可以全部满足。

    由P 可以得到 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,判断是否会有负环。初始dis要全部为0.

    #include<cstdio>
    #include<cstring>
    #include<queue>
    #include<algorithm>
    #define INF 1e9
    using namespace std;
    const int maxn=1000+10;
    const int maxm=100000*3;
    struct Edge
    {
        int from,to,dist;
        Edge(){}
        Edge(int f,int t,int d):from(f),to(t),dist(d){}
    };
     
    struct BellmanFord
    {
        int n,m;
        int head[maxn],next[maxm];
        Edge edges[maxm];
        int d[maxn];
        int cnt[maxn];
        bool inq[maxn];
     
        void init(int n)
        {
            this->n=n;
            m=0;
            memset(head,-1,sizeof(head));
        }
     
        void AddEdge(int from,int to,int dist)
        {
            edges[m]=Edge(from,to,dist);
            next[m]=head[from];
            head[from]=m++;
        }
     
        bool bellman_ford()
        {
            memset(inq,0,sizeof(inq));
            memset(cnt,0,sizeof(cnt));
            queue<int> Q;
            for(int i=0;i<n;i++) d[i]= i==0?0:INF;
            Q.push(0);
     
            while(!Q.empty())
            {
                int u=Q.front(); Q.pop();
                inq[u]=false;
                for(int i=head[u];i!=-1;i=next[i])
                {
                    Edge &e=edges[i];
                    if(d[e.to] > d[u]+e.dist)
                    {
                        d[e.to] = d[u]+e.dist;
                        if(!inq[e.to])
                        {
                            inq[e.to]=true;
                            Q.push(e.to);
                            if(++cnt[e.to]>n) return true;
                        }
                    }
                }
            }
            return false;
        }
    }BF;
     
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)==2)
        {
            BF.init(n+1);
            while(m--)
            {
                char s[10];
                int u,v,d;
                scanf("%s",s);
                if(s[0]=='P')
                {
                    scanf("%d%d%d",&u,&v,&d);
                    BF.AddEdge(u,v,d);
                    BF.AddEdge(v,u,-d);
                }
                else if(s[0]=='V')
                {
                    scanf("%d%d",&u,&v);
                    BF.AddEdge(v,u,-1);
                }
            }
            for(int i=1;i<=n;i++)
                BF.AddEdge(0,i,0);
            printf("%s
    ",BF.bellman_ford()?"Unreliable":"Reliable");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798580.html
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