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  • 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
    Reliable

    Solution:

      还是差分约束判断可行性。。。

      我们依然罗列出约束条件(可行性我直接当做求最小值来做):

        1、$s[v]-s[u]=c,s[u]-s[v]=c$,所以建边$w[u,v]=w[v,u]=c$(注意是双向的,因为等于是个对于两边都成立的约束条件

        2、$s[v]-s[u]>0$,即$s[v]-s[u]geq 1$,所以建边$w[u,v]=1$

      记得新建一个源点,连向各点$w[1,i]=0,;iin 1 ightarrow n$,以保证图联通。

      然后跑一遍最长路,判断一下有无环输出就好了。

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<queue>
    #pragma GCC optimize(2)
    #define il inline
    #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
    using namespace std;
    const int N=300005,M=1005,inf=23333333;
    int h[M],to[N],net[N],w[N],cnt,dis[M],tot[M],n,m;
    bool vis[M];
    
    il void add(int u,int v,int c){to[++cnt]=v,net[cnt]=h[u],h[u]=cnt,w[cnt]=c;}
    
    il void gi(int &a){
        a=0;char x=getchar();bool f=0;
        while((x<'0'||x>'9')&&x!='-')x=getchar();
        if(x=='-')x=getchar(),f=1;
        while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    }
    
    il bool spfa(){
        queue<int>q;
        dis[0]=0;q.push(0);tot[0]++;
        while(!q.empty()){
            int u=q.front();q.pop();vis[u]=0;
            for(int i=h[u];i;i=net[i])
                if(dis[to[i]]<dis[u]+w[i]){
                    dis[to[i]]=dis[u]+w[i];
                    if(!vis[to[i]]){
                        if(++tot[to[i]]>n)return 0;
                        q.push(to[i]),vis[to[i]]=1;
                    }
                }
        }
        return 1;
    }
    
    int main(){
        while(~scanf("%d%d",&n,&m)){
            char s[5];
            cnt=0;
            memset(dis,-0x3f,sizeof(int)*(n+3));
            memset(vis,0,sizeof(int)*(n+3));
            memset(tot,0,sizeof(int)*(n+3));
            memset(h,0,sizeof(h));
            int u,v,c;
            while(m--){
                scanf("%s",s);
                gi(u),gi(v);
                if(s[0]=='P'){
                    gi(c);
                    add(v,u,-c);add(u,v,c);
                }
                else add(u,v,1);
            }
            For(i,1,n) add(0,i,0);
            if(spfa())puts("Reliable");
            else puts("Unreliable");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/five20/p/9177055.html
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