poj Layout（差分约束）

 Layout

Description

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.

ps：一看就是差分约束，但是由于差分约束只懂了点皮毛，所以没有找到关键

分析一下，

对于每个好感描述（i，j，k），i<=j，体现到距离上的要求就是：

D[j]-D[i]<=k

对于每个反感描述（i，j，k），i<=j，体现到距离上的要求就是：

D[j]-D[i]>=k

还有个条件，对于每个i<=j，D[j]>=D[i]

那么我们可以将三个式子这样写

D[j]-D[i]<=k

D[i]-D[j]<=-k

D[i]-D[j]<=0

这样我们就找到了条件可以直接松弛了

还有就是，当出现负环的时候输出-1，当d[n]为无穷大时，表明1和n的距离可以为任意大，那么就输出-2

详细证明 http://blog.youkuaiyun.com/zhang20072844/article/details/7788672

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

#define maxn 1000+10
#define maxv 20000+10
const int inf=0x3f3f3f3f;

struct node
{
    int v,w,next;
} G[maxv];
int first[maxn],d[maxn],vis[maxn];
bool inq[maxn];
int n,ml,md,len;

bool spfa()
{
    int i,st;
    for(i=1; i<=n; i++)
    {
        inq[i]=0;
        d[i]=inf;
    }
    d[1]=0,inq[1]=1;
    queue<int>q;
    q.push(1);
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        inq[st]=0;
        for(i=first[st]; i!=-1; i=G[i].next)
        {
            int v=G[i].v,w=G[i].w;
            if(d[v]>d[st]+w)
            {
                d[v]=d[st]+w;
                if(++vis[v]>n)
                    return true;
                if(!inq[v])
                {
                    inq[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return false;
}

void add_egde(int u,int v,int w)
{
    G[len].v=v,G[len].w=w;
    G[len].next=first[u];
    first[u]=len++;
}

int main()
{
    while(~scanf("%d%d%d",&n,&ml,&md))
    {
        memset(first,-1,sizeof(first));
        memset(vis,0,sizeof(vis));
        len=1;
        int i,u,v,w;
        for(i=1; i<=ml; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add_egde(u,v,w);
        }
        for(i=1; i<=md; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add_egde(v,u,-w);
        }
        if(spfa())
            printf("-1\n");
        else if(d[n]==inf)
            printf("-2\n");
        else
            printf("%d\n",d[n]);
    }
    return 0;
}