POJ 3169 Layout（差分约束+SPFA）

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4641		Accepted: 2240

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.

Source

USACO 2005 December Gold

 

 

 

/*
POJ 3169 Layout

差分约束+SPFA
*/
//队列实现SPFA，需要有负环回路判断
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXN=1010;
const int MAXE=20020;
const int INF=0x3f3f3f3f;
int head[MAXN];//每个结点的头指针
int vis[MAXN];//在队列标志
int cnt[MAXN];//每个点的入队列次数
int que[MAXN];//SPFA循环指针
int dist[MAXN];

struct Edge
{
    int to;
    int v;
    int next;
}edge[MAXE];
int tol;
void add(int a,int b,int v)//加边
{
    edge[tol].to=b;
    edge[tol].v=v;
    edge[tol].next=head[a];
    head[a]=tol++;
}
bool SPFA(int start,int n)
{
    int front=0,rear=0;
    for(int v=1;v<=n;v++)//初始化
    {
        if(v==start)
        {
            que[rear++]=v;
            vis[v]=true;
            cnt[v]=1;
            dist[v]=0;
        }
        else
        {
            vis[v]=false;
            cnt[v]=0;
            dist[v]=INF;
        }
    }
    while(front!=rear)
    {
        int u=que[front++];
        vis[u]=false;
        if(front>=MAXN)front=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dist[v]>dist[u]+edge[i].v)
            {
                dist[v]=dist[u]+edge[i].v;
                if(!vis[v])
                {
                    vis[v]=true;
                    que[rear++]=v;
                    if(rear>=MAXN)rear=0;
                    if(++cnt[v]>n) return false;
                    //cnt[i]为入队列次数，用来判断是否存在负环回来
                    //这条好像放在这个if外面也可以？？
                }
            }
        }
    }
    return true;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    int ML,MD;
    int a,b,c;
    while(scanf("%d%d%d",&n,&ML,&MD)!=EOF)
    {
        tol=0;//加边计数，这个不要忘
        memset(head,-1,sizeof(head));
        while(ML--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a>b)swap(a,b);//注意加边顺序
            add(a,b,c);
            //大-小<=c ,有向边(小，大）：c
        }
        while(MD--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a<b)swap(a,b);
            add(a,b,-c);
            //大-小>=c,小-大<=-c,有向边(大，小):-c
        }
        if(!SPFA(1,n)) printf("-1\n");//无解
        else if(dist[n]==INF) printf("-2\n");
        else printf("%d\n",dist[n]);
    }
    return 0;
}

代码2：

/*
POJ 3169
差分约束+BellmenFord
*/
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=1010;
const int MAXE=20020;

int dist[MAXN];

int edge[MAXE][3];//存边
int tol;//边的总数

bool bellman(int start,int n)
{
    for(int i=1;i<=n;i++)dist[i]=INF;
    dist[start]=0;
    for(int i=1;i<n;i++)//最多做n-1次
    {
        bool flag=false;//优化，没有更新就跳出
        for(int j=0;j<tol;j++)
        {
            int u=edge[j][0];
            int v=edge[j][1];
            if(dist[v]>dist[u]+edge[j][2])
            {
                dist[v]=dist[u]+edge[j][2];
                flag=true;
            }
        }
        if(!flag)break;
    }
    for(int j=0;j<tol;j++)
        if(dist[edge[j][1]] > dist[edge[j][0]]+edge[j][2])
          return false;
    return true;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    int ML,MD;
    int a,b,c;
    while(scanf("%d%d%d",&n,&ML,&MD)!=EOF)
    {
        tol=0;
        while(ML--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a>b)swap(a,b);
            edge[tol][0]=a;
            edge[tol][1]=b;
            edge[tol][2]=c;
            tol++;
        }
        while(MD--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a<b)swap(a,b);
            edge[tol][0]=a;
            edge[tol][1]=b;
            edge[tol][2]=-c;
            tol++;
        }
        if(!bellman(1,n))printf("-1\n");
        else if(dist[n]==INF)printf("-2\n");
        else printf("%d\n",dist[n]);
    }
    return 0;
}