poj 2983 差分约束+SPFA

来源:互联网 发布:sql默认地址不详设置 编辑:程序博客网 时间:2024/03/29 01:52
Is the Information Reliable?
Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 9890 Accepted: 3067

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

Source

POJ Monthly--2006.08.27, Dagger
题意:第一行n,m,代表有n个点和m条情报,接下来m行,以P开头的后跟3个数字a、b、c,代表a在b的北方距离c处,以V开头的后跟俩数字a、b,代表a在b北方至少距离为1的地方,让你判断情报是否可信。
思路:由此可以建立不等式与方程,其中方程转化为等价的两个不等式,如第一组样例,可以变为不等式组:d2-d1<=-1,d1-d2<=1,d3-d2<=-1,d2-d3<=1,d3-d1<=-1,d1-d3<=1(重复的已去除)。这样问题就变成了不等式组有没有解的问题,有解则情报可信,没解则情报不可信,好吧,这里就是一个差分约束问题,什么?你不知道差分约束?那你自己百度一下吧。
下面是用SPFA写的程序,题目有坑,对于给你俩点编号相同的情况要特殊处理。
#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;struct node{    int a,b,c;    int next;}e[202005];int fst[1005];bool v[1005];int m,n,a,b,c,num;int cnt[1005];char p[5];int dist[1005];bool spfa()//spfa判断是否有解{    for(int i=0;i<=n;i++)dist[i]=1;//比0点到它的边的长度大即可    dist[0]=0;    queue<int>q;    q.push(0);    cnt[0]++;    v[0]=1;    while(!q.empty())    {        a=q.front();        q.pop();        for(int i=fst[a];i!=-1;i=e[i].next)        {            b=e[i].b;            c=e[i].c;            if(dist[b]>dist[a]+c)            {                dist[b]=dist[a]+c;                if(!v[b])                {                    q.push(b);                    v[b]=1;                    cnt[b]++;                    if(cnt[b]>n)return false;//说明有负环路,无解退出                }            }        }        v[a]=0;    }    return true;}int main(){    int ans;    while(scanf("%d%d",&n,&m)!=EOF)    {        num=0;        ans=0;        memset(fst,-1,sizeof(fst));        memset(cnt,0,sizeof(cnt));        memset(v,0,sizeof(v));        for(int i=0;i<m;i++)        {            scanf("%s",&p);            if(p[0]=='P')            {                scanf("%d%d%d",&a,&b,&c);                if(a==b&&c)ans=1;//自己连自己的点特殊处理                                e[num].next=fst[a];//等式变为俩不等式,即建立双向边,这里邻接表存储                fst[a]=num;                e[num].a=a;                e[num].b=b;                e[num++].c=-c;                e[num].next=fst[b];                fst[b]=num;                e[num].a=b;                e[num].b=a;                e[num++].c=c;            }            else            {                scanf("%d%d",&a,&b);                if(a==b)ans=1;//自己连自己的特殊处理                e[num].next=fst[a];//不等式只建立单向边                fst[a]=num;                e[num].a=a;                e[num].b=b;                e[num++].c=-1;            }        }        for(int i=1;i<=n;i++)        {            e[num].next=fst[0];            fst[0]=num;            e[num].a=0;            e[num].b=i;            e[num++].c=0;        }        if(ans)cout<<"Unreliable"<<endl;        else if(spfa())cout<<"Reliable"<<endl;        else cout<<"Unreliable"<<endl;    }    return 0;}