POJ3259(ford判环)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39078		Accepted: 14369

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: N, M, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. 
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

题意:每个农场有N各区域,连接所有区域的是M个双向路径和W个单向时空隧道,从S->E若为路径则花费T秒,若为时空隧道则倒退T秒。问是否可以从某点出发，转一圈回来，回到出发时刻之前。
思路:因为时空隧道实现倒退,所以将其权值设为负值,利用ford判断是否存在负环。

#include"cstdio"
#include"cstring"
using namespace std;
const int MAXN=10005;
const int INF=0x3fffffff;
struct Edge{
    int from,to,cost;
}es[MAXN];
int N,M,W;
int E;
int d[MAXN];
bool ford(int s)
{
    for(int i=1;i<=N;i++)    d[i]=INF;
    d[s]=0;
    
    int n=N;
    while(n--)
    {
        bool update=false;
        for(int i=0;i<E;i++)
        {
            Edge e=es[i];
            if(d[e.from]!=INF&&d[e.to]>d[e.from]+e.cost)
            {
                d[e.to]=d[e.from]+e.cost;
                update=true;
            }
        }
        if(!update)    break;
        
    }
    
    if(n==-1)    return true;
    else return false;
}
int main()
{
    int F;
    scanf("%d",&F);
    while(F--)
    {
        E=0;
        scanf("%d%d%d",&N,&M,&W);
        for(int i=0;i<M;i++)
        {
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            es[E].from=u,es[E].to=v,es[E++].cost=c;
            es[E].from=v,es[E].to=u,es[E++].cost=c;
        }
        for(int i=0;i<W;i++)
        {
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            es[E].from=u,es[E].to=v,es[E++].cost=-c;//倒退c秒 
        }
        
        if(ford(1))    printf("YES
");
        else printf("NO
");
    }    
    
    return 0;
}

 spfa+前向星可解决重边问题

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN=505;
const int INF=0x3f3f3f3f;
struct Edge{
    int v,w,next;
}es[6006];
int head[MAXN],tot;
void addedge(int u,int v,int w)
{
    es[tot].v=v;
    es[tot].w=w;
    es[tot].next=head[u];
    head[u]=tot++;
}
int d[MAXN],vis[MAXN],cnt[MAXN];
int n,m,k;
bool spfa(int s)
{
    for(int i=1;i<=n;i++)
    {
        d[i]=INF;
        vis[i]=0;
        cnt[i]=0;
    }
    d[s]=0;
    queue<int> que;
    que.push(s);
    vis[s]=1;
    cnt[s]++;
    while(!que.empty())
    {
        int u=que.front();que.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=es[i].next)
        {
            Edge e=es[i];
            if(d[e.v]>d[u]+e.w)
            {
                d[e.v]=d[u]+e.w;
                if(!vis[e.v])
                {
                    vis[e.v]=1;
                    que.push(e.v);
                    cnt[e.v]++;
                    if(cnt[e.v]>=n)    return true;
                }
            }            
        }
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(head,-1,sizeof(head));
        tot=0;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        for(int i=0;i<k;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,-w);
        }
        if(spfa(1))    printf("YES
");
        else    printf("NO
");    
    }
    return 0;
}