poj 3169 Layout（差分约束求最短路）

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5425		Accepted: 2590

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.

题意：有n头牛。有ml个关系：每个关系A,B,C表示B距离A不能超过C。有md个关系：每个关系A,B,C表示B距离A至少为C。为牛1和牛n的距离是最大是多少，可能是任意的，也可能是无解的。

思路：对于每个关系B-A<=C,建一条边A->B，权值为C；对于每个关系B-A>=C，建一条边B->A，权值为-C。用spfa求最短路，若最后求得dis[n]=INF,则距离是任意的，若图中有负权回路，则无解。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <map>


using namespace std;

const int INF=100000000;
const int maxn=1005;

struct node
{
    int v,w,next;
}edge[20005];
int head[maxn],dis[maxn],cnt[maxn];
bool vis[maxn];
int num,n,ml,md;
void init()
{
    for(int i=1;i<=n;i++)
    head[i]=-1;
    num=0;
}
void add(int u,int v,int w)
{
    edge[num].v=v;
    edge[num].w=w;
    edge[num].next=head[u];
    head[u]=num++;
}
int spfa()
{
    queue<int>Q;
    for(int i=1;i<=n;i++)
    dis[i]=INF;
    dis[1]=0;
    memset(vis,false,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    vis[1]=true;
    Q.push(1);
    cnt[1]++;
    while(!Q.empty())
    {
        int x=Q.front();
        Q.pop();
        vis[x]=false;
        for(int i=head[x];i!=-1;i=edge[i].next)
        {
            if(dis[edge[i].v]>dis[x]+edge[i].w)
            {
                dis[edge[i].v]=dis[x]+edge[i].w;
                if(!vis[edge[i].v])
                {
                    vis[edge[i].v]=true;
                    if(++cnt[edge[i].v]>n) return -1;
                    Q.push(edge[i].v);
                }
            }
        }
    }
    if(dis[n]==INF) return -2;
    return dis[n];
}
int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&n,&ml,&md)!=EOF)
    {
        init();
        while(ml--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
        }
        while(md--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(b,a,-c);
        }
        printf("%d\n",spfa());
    }
    return 0;
}