POJ 3167 Layout（差分约束）

题面

Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.

Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.

Input

Line 1: Three space-separated integers: N, ML, and MD.

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

Output

Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.

Sample Input

4 2 1

1 3 10

2 4 20

2 3 3

Sample Output

27

Hint

Explanation of the sample:

There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.

题解

翻译一下：有N头牛，ML个关系1，MD个关系2

对于关系1 a b c 而言 指 a牛和b牛之间的距离不能够超过c

对于关系2 a b d 而言 指 a牛和b牛之间的距离至少为d

求出1和N的最短距离，如果无解输出-1，距离可以无限大输出-2

题解：

差分约束

关系1而言直接建边

关系2而言 Xb-Xa>=d 变为 Xa-Xa<=-d

建边

然后SPFA求解

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 100000000
#define MAX 1100
#define MAXL 50000
inline int read()
{
	  register int x=0,t=1;
	  register char ch=getchar();
	  while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	  if(ch=='-'){t=-1;ch=getchar();}
	  while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	  return x*t;
}
struct Line
{
	  int v,next,w;
}e[MAXL];
int h[MAX],cnt=1;
int dis[MAX];
int Count[MAX];
bool vis[MAX];
int N,Ma,Mb;
inline void Add(int u,int v,int w)
{
	  e[cnt]=(Line){v,h[u],w};
	  h[u]=cnt++;
}
bool SPFA()
{
	  for(int i=1;i<=N;++i)dis[i]=INF;
	  for(int i=1;i<=N;++i)Count[i]=0;
	  for(int i=1;i<=N;++i)vis[i]=false;
	  dis[1]=0;
	  queue<int> Q;
	  while(!Q.empty())Q.pop();
	  Q.push(1);
	  while(!Q.empty())
	  {
	  	     int u=Q.front();Q.pop();
	  	     vis[u]=false;
	  	     if(Count[u]>=N)return false;
	  	     for(int i=h[u];i;i=e[i].next)
	  	     {
	  	     	     int v=e[i].v,w=e[i].w;
	  	     	     if(dis[v]>dis[u]+w)
	  	     	     {
	  	     	     	   dis[v]=dis[u]+w;
	  	     	     	   if(!vis[v])
	  	     	     	   {
	  	     	     	   	       	vis[v]=true;
	  	     	     	   	       	Q.push(v);
	  	     	     	   	       	Count[v]++;
	  	     	     	   }
	  	     	     }
	  	     }
	  }
	  return true;
}
int main()
{
	  N=read();Ma=read();Mb=read();
	  for(int i=1;i<=Ma;++i)
	  {
	  	     int a=read(),b=read(),c=read();
	  	     Add(a,b,c);
	  }
	  for(int i=1;i<=Mb;++i)
	  {
	  	     int a=read(),b=read(),c=read();
	  	     Add(b,a,-c);
	  }
	  if(!SPFA())//存在负环
	  	     printf("%d\n",-1);
	  else
	  	     if(dis[N]==INF)//可以无限大
			 	     printf("%d\n",-2);
			 else
			         printf("%d\n",dis[N]);
	  return 0;
}