Is the Information Reliable?--差分约束系统

Is the Information Reliable?

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 13099 Accepted: 4129

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 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

题目链接：http://poj.org/problem?id=2983

差分约束系统！找了我一天的bug！最后找出来差点没有抽我自己一巴掌，b-a>=1，所以，a-b<=-1，所以connect(b,a,-1)，不是connect(a,b,-1)，天呐，卡了我一天！！！，脑子太白痴了。

差分约束系统详见我的上一篇博客吧，最近我都快让差分给约束了。

这个题的题意是说

给一段信息，有关于N个点，M条信息。其中每一条信息：

1、P A B X，意思是A站点在B站点北X光年

2、V A B，意思是A站点在B站点北，但不知道确切距离，最少一光年。

对于给出的这段信息，求是否是真是的消息，因为错误的消息有矛盾的信息。

懂差分约束系统的看我代码就懂了，我还有一个思路，我想用如果是V添加的话就用拓扑排序排一遍，其他的用精确的来计算一遍，然后不知道行不行，我一会儿去试试。

渣渣的代码：

#include <cstdio>#include <cstring>#include <iostream>#include <queue>#define inf 0x3f3f3f3fusing namespace std;struct node{    int v,w,pre;}edge[1000100];int dis[1000010];bool vis[10000];int fax[100000];//判断负环int p[100000],nEdge;int n,m;void Init(){

    memset(dis,inf,sizeof(dis));    memset(vis,false,sizeof(vis));    memset(fax,0,sizeof(fax));    memset(p,-1,sizeof(p));    nEdge=0;}void connect(int u,int v,int w){    nEdge++;    edge[nEdge].v=v;    edge[nEdge].w=w;    edge[nEdge].pre=p[u];    p[u]=nEdge;}bool spfa(){    queue<int >Q;    dis[0]=0;//超级源点，也可以把所有的点放进队列    Q.push(0);    vis[0]=true;    while(!Q.empty()){        int t=Q.front();        Q.pop();        vis[t]=false;        for(int i=p[t];i!=-1;i=edge[i].pre){            int v=edge[i].v;            int w=edge[i].w;            if(dis[v]>dis[t]+w){                dis[v]=dis[t]+w;                if(!vis[v]){                    fax[v]++;                    Q.push(v);                    vis[v]=true;                    if(fax[v]>n){//有负环，说明不合格                        return false;                    }                }            }        }    }    return true;}int main(){    while(~scanf("%d%d",&n,&m)){        Init();        for(int i=0;i<=n;i++){            connect(0,i,0);//超级源点的初始化        }        char s[10];        for(int i=0;i<m;i++)        {            scanf("%s",s);            if(s[0]=='P'){                int a,b,c;                scanf("%d%d%d",&a,&b,&c);                connect(a,b,c);                connect(b,a,-c);            }            else if(s[0]=='V'){                int a,b;                scanf("%d%d",&a,&b);                connect(b,a,-1);//b到a！！！            }        }        bool h=spfa();        if(h){            printf("Reliable\n");        }        else{            printf("Unreliable\n");        }    }    return 0;}