## poj 2983 差分约束+SPFA

2019年02月28日 算法 ⁄ 共 2938字 ⁄ 字号 评论关闭
Is the Information Reliable?
 Time Limit: 3000MS Memory Limit: 131072K Total 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 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

```#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;
}
```
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