Problem 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 5Sample Output
Unreliable
Reliable
题意：n 个防御站排成一条线，有 m 个条件共 2 种形式：
- P A B X：表示 A 在 B 的北面 X 光年处
- V A B：表示 A 在 B 的北面至少 1 光年处
问防御系统是否能满足这 m 个条件
思路：
由题意可得：
- P A B X：B+X=A
- V A B：B+1<=A
假设在最北方存在一个超级源点 0 号，那么设 dis[i] 为 0 号站离 i 号防御站的距离，那么：
- P A B X：dis[A]+X=dis[B]，那么：dis[B]<=dis[A]+X 且 dis[B]>=dis[A]+X
dis[B]<=dis[A]+X：A 到 B 权值为 X 的边
dis[B]>=dis[A]+X：B 到 A 的权值 - X 的边 - V A B：dis[A]<=dis[B]-1，据此可以添加：B 到 A 的权值为 -1 的边
由于边的权值存在负数，因此可以使用 SPFA 算法来判断负环是否存在从而判断该差分约束系统是否有解，将所有点的最短距离都设为 INF，即 dis[i]=INF，再将超级源点 0 号点设为 dis[0]=0，然后求 0 号点到其他所有点的最短距离并判断负环是否存在即可
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
#define LL long long
const int MOD=10007;
const int N=1000000+5;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;
struct Edge{
int from;
int to;
int w;
Edge(){}
Edge(int from,int to,int w):from(from),to(to),w(w){}
}edge[N];
int head[N],next[N],num;
int dis[N];
int outque[N];
bool vis[N];
void init(){
num=0;
memset(head,-1,sizeof(head));
}
void add(int from,int to,int w){
edge[num]=Edge(from,to,w);
next[num]=head[from];
head[from]=num++;
}
int SPFA(int s,int n){
int res=0;
memset(vis,false,sizeof(vis));
memset(outque,0,sizeof(outque));
for(int i=1;i<n;i++)
dis[i]=INF;
dis[s]=0;
vis[s]=true;
queue<int> Q;
Q.push(s);
while(!Q.empty()){
int x=Q.front();
Q.pop();
vis[x]=false;
outque[x]++;
if(outque[x]>n-1){
res=-1;
break;
}
for(int i=head[x];i!=-1;i=next[i]){
Edge &e=edge[i];
if(dis[e.to]>dis[x]+e.w){
dis[e.to]=dis[x]+e.w;
if(!vis[e.to]){
vis[e.to]=true;
Q.push(e.to);
}
}
}
}
return res;
}
int main(){
int n,m;
while(scanf("%d%d%d",&n,&m)!=EOF){
init();
for(int i=0;i<m;i++){
char ch[10];
scanf("%s",ch);
if(ch[0]=='P'){
int A,B,X;
scanf("%d%d%d",&A,&B,&X);
add(A,B,X);
add(B,A,-X);
}
else{
int A,B;
scanf("%d%d",&A,&B);
add(B,A,-1);
}
}
for(int i=1;i<=n;i++)//虚拟节点到各点的边
add(0,i,0);
int res=SPFA(0,n+1);
if(res==-1)
printf("Unreliable\n");
else
printf("Reliable\n");
}
return 0;
}