【POJ 2983】Is the Information Reliable?（差分约束系统）

【POJ 2983】Is the Information Reliable?（差分约束系统）

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

Source

POJ Monthly--2006.08.27, Dagger

建立差分约束系统进行求解

存在两种关系

P a b x 表示a在b北边x光年 等价与Pb - Pa = x

想要表示等于 就要转换成 Pb - Pa >= x Pb - Pa <= x

即为 Pb-Pa >= x Pa - Pb >= -x

V a b 表示a在b北边至少一光年 即为Pb - Pa >= 1

用三个公式建立差分约束系统即可 由于可能是多个不连通图 就需要用一个超级源点把他们都链接起来

如果跑最短的过程中没有负环 即说明是合法的关系图 否则Unreliable

代码如下:

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const double eps = 1e-8;

int dcmp(double x)
{
	return x < -eps? -1: x > eps;
}

struct Edge
{
	int v,w,next;
};

Edge eg[233333];
int head[1010];
int dis[1010];
int cnt[1010];
bool vis[1010];
int n,m,tp;

void Add(int u,int v,int w)
{
	eg[tp].v = v;
	eg[tp].w = w;
	eg[tp].next = head[u];
	head[u] = tp++;
}

bool SPFA()
{
	memset(dis,-INF,sizeof(dis));
	memset(cnt,0,sizeof(cnt));
	memset(vis,0,sizeof(vis));
	vis[0] = 1;
	dis[0] = 0;
	cnt[0]++;
	queue <int> q;
	q.push(0);
	int u,v,w;

	while(!q.empty())
	{
		u = q.front();
		vis[u] = 0;
		q.pop();
		for(int i = head[u]; i != -1; i = eg[i].next)
		{
			v = eg[i].v;
			w = eg[i].w;
			if(dis[v] < dis[u]+w)
			{
				dis[v] = dis[u]+w;
				cnt[v]++;
				if(cnt[v] > n) return false;
				if(!vis[v])
				{
					q.push(v);
					vis[v] = 1;
				}
			}
		}
	}
	return true;
}

int main()
{
	int u,v,w;
	char opt[3];
	while(~scanf("%d%d",&n,&m))
	{
		memset(head,-1,sizeof(head));
		tp = 0;
		while(m--)
		{
			scanf("%s",opt);

			if(opt[0] == 'P')
			{
				scanf("%d%d%d",&u,&v,&w);
				Add(u,v,w);
				Add(v,u,-w);
			}
			else
			{
				scanf("%d%d",&u,&v);
				Add(u,v,1);
			}

		}
		for(int i = 1; i <= n; ++i)
		{
			Add(0,i,0);
		}

		puts(SPFA()? "Reliable": "Unreliable");
	}

	return 0;
}