POJ 3255 Wormholes（最短路最负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20639		Accepted: 7342

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

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

Source

USACO 2006 December Gold

 

 

 

 

这题一方面是找负环。。其实环需要判断起点和负环是不是连通的。因为要求是回到起点。。有可能有负环，但是回不去起点。。。

暂时还没有好的办法判断负环和起点相连，或者起点是负环的一部分。。。路过的大牛知道的话指点下。

 

但是本题数据都是连通的，直接SPFA或者BF做

 

Bellman_Ford

/*
POJ 3259
就是判断负环回路。
用SPFA和BellmanFord都可以

但是一直想不通的是题目要求是要回到出发点，
有负环可能回不去啊。。。。数据弱爆了。。。Orz

*/
// G++ 732K 94ms
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=550;
const int MAXE=5050;
const int INF=0x3f3f3f3f;
int dist[MAXN];

int edge[MAXE][3];
int tol;

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++)
    {
        bool flag=false;
        for(int j=0;j<tol;j++)
        {
            if(dist[edge[j][1]]>dist[edge[j][0]]+edge[j][2])
            {
                dist[edge[j][1]]=dist[edge[j][0]]+edge[j][2];
                flag=true;
            }
        }
        if(!flag)return false;//没有负环
    }

    for(int j=0;j<tol;j++)
       if(dist[edge[j][1]]>dist[edge[j][0]]+edge[j][2])
          return true;
    return false;
}
int main()
{
    int n,M,W;
    int a,b,c;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&M,&W);
        tol=0;
        while(M--)
        {
            scanf("%d%d%d",&a,&b,&c);
            edge[tol][0]=a;
            edge[tol][1]=b;
            edge[tol][2]=c;
            tol++;
            edge[tol][0]=b;
            edge[tol][1]=a;
            edge[tol][2]=c;
            tol++;
        }
        while(W--)
        {
            scanf("%d%d%d",&a,&b,&c);
            edge[tol][0]=a;
            edge[tol][1]=b;
            edge[tol][2]=-c;
            tol++;
        }
        if(bellman(1,n))printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

SPFA:

/*
POJ 3259

SPFA判断负环
*/

//G++ 756K 157ms
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=550;
const int MAXE=5050;//这个5000会RE
const int INF=0x3f3f3f3f;
int head[MAXN];
int dist[MAXN];
int cnt[MAXN];
int que[MAXN];
bool vis[MAXN];

struct edge
{
    int to;
    int next;
    int v;
}edge[MAXE];
int tol;

void add(int a,int b,int c)
{
    edge[tol].to=b;
    edge[tol].v=c;
    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++)
    {
        if(v==start)
        {
            dist[v]=0;
            que[rear++]=v;
            vis[v]=true;
            cnt[v]=1;
        }
        else
        {
            dist[v]=INF;
            vis[v]=false;
            cnt[v]=0;
        }
    }
    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 true;
                }
            }
        }
    }
    return false;
}
int main()
{
   // freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    int n,M,W;
    int a,b,c;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&M,&W);
        tol=0;
        memset(head,-1,sizeof(head));//这个不能忘记！！！
        while(M--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        while(W--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        if(SPFA(1,n))printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}