POJ 3259-Wormholes（Bellman Ford或SPFA判负权值回路）

﹏ヽ暗。殇╰゛Y 2022-09-25 04:18 18阅读 0赞

Wormholes

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;44248</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;16280</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

N个地之间有M条路，W个虫洞。问能否从某块地出发，通过一些路径和虫洞，使得返回出发点的时候时间早于出发点。

F个测试用例。

对于M条路，给出M条从S到E的双向路径。

对于W各虫洞，给出W 条从S到E的单向负权值路径。

## 解题思路： ##

一个判负权值回路的裸题，判断一个带权值的有向网中是否存在负权值回路，存在就是能够早于出发点，反之不行。

## ①SPFA ##

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define MAXN 10010
    #define INF 0xfffffff
    using namespace std;
    
    struct ArcNode
    {
        int to;
        int weight;
        ArcNode *next;
    };
    queue<int> Q;//队列中的节点为顶点序号
    int n;//顶点个数
    int m,W;
    ArcNode * List[MAXN];//每个顶点的边链表表头指针
    int inq[MAXN];//每个顶点是否在队列中的标志
    int dist[MAXN],path[MAXN];
    int cnt[MAXN];
    bool SPFA(int src)
    {
        memset(cnt,0,sizeof(cnt));
        memset(inq,0,sizeof(inq));
        int i,u;//u为队列头顶点序号
        ArcNode * temp;
        for(i=1; i<=n; ++i)//初始化
        {
            dist[i]=INF;
            path[i]=src;
            inq[i]=0;
        }
        dist[src]=0;
        path[src]=src;
        ++inq[src];
        Q.push(src);
        ++cnt[src];
        while(!Q.empty())
        {
            u=Q.front();
            Q.pop();
            --inq[u];
            if(cnt[u]>n) return true;//存在负权回路
            temp=List[u];
            while(temp!=NULL)
            {
                int v=temp->to;
                if(dist[v]>dist[u]+temp->weight)
                {
                    dist[v]=dist[u]+temp->weight;
                    path[v]=u;
                    if(!inq[v])
                    {
                        Q.push(v);
                        ++inq[v];
                        ++cnt[v];
                    }
    
                }
                temp=temp->next;
            }
        }
        return false;
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/x/read.txt","r",stdin);
        freopen("G:/x/out.txt","w",stdout);
    #endif
        int i,t;
        cin>>t;
        while(t--)
        {
            int u,v,w;
            cin>>n>>m>>W;
            memset(List,0,sizeof(List));
            ArcNode *temp;
            while(!Q.empty()) Q.pop();
            while(m--)
            {
                cin>>u>>v>>w;
                temp=new ArcNode;
                temp->to=v;//构造邻接表
                temp->weight=w;
                temp->next=NULL;
                if(List[u]==NULL) List[u]=temp;
                else
                {
                    temp->next=List[u];
                    List[u]=temp;
                }
                //无向图，构建两条边
                temp=new ArcNode;
                temp->to=u;//构造邻接表
                temp->weight=w;
                temp->next=NULL;
                if(List[v]==NULL) List[v]=temp;
                else
                {
                    temp->next=List[v];
                    List[v]=temp;
                }
            }
            while(W--)//负权路
            {
                cin>>u>>v>>w;
                temp=new ArcNode;
                temp->to=v;//构造邻接表
                temp->weight=-w;
                temp->next=NULL;
                if(List[u]==NULL) List[u]=temp;
                else
                {
                    temp->next=List[u];
                    List[u]=temp;
                }
            }
            if(SPFA(1)) puts("YES");//求源点到其他顶点的最短路径
            else puts("NO");
            for(i=0; i<=n; ++i)
            {
                temp=List[i];
                while(temp!=NULL)
                {
                    List[i]=temp->next;
                    delete temp;
                    temp=List[i];
                }
            }
        }
        return 0;
    }
    /*
    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
    */

## ②[Bellman Ford][] 邻接表 ##

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <vector>
    #include <queue>
    #include <map>
    #include <algorithm>
    using namespace std;
    #define INF 0xfffffff
    #define MAXN 100010
    
    struct Edge
    {
        int u,v,w;
    };
    Edge edge[MAXN];//邻接表
    int n,m;//顶点数和边数
    int cnt=0;
    int dist[MAXN];//顶点s到其他顶点的最短路径、
    
    int Bellman(int s)//顶点s到其他顶点的最短路径
    {
        int i,j;
        for(i=0; i<n; ++i)//初始化
            dist[i]=INF;
        dist[s]=0;
        for(i=1; i<n; ++i)
            for(j=0; j<m; ++j)
            {
                Edge e=edge[j];
                if(dist[e.u]!=INF&&e.w+dist[e.u]<dist[e.v])//顶点k到j有直接路径而且途径k可以使得路径缩短
                    dist[e.v]=e.w+dist[e.u];
            }
        for(i=0; i<m; ++i)
            if(dist[edge[i].u]!=INF&&edge[i].w+dist[edge[i].u]<dist[edge[i].v])
                return 1;
        return 0;
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int t;
        cin>>t;
        while(t--)
        {
            cnt=0;
            int i,W;
            cin>>n>>m>>W;
            int u,v,va;
            for(i=0; i<m; ++i)
            {
                cin>>u>>v>>va;
                --u,--v;
                edge[cnt].u=u,edge[cnt].v=v,edge[cnt++].w=va;
                edge[cnt].u=v,edge[cnt].v=u,edge[cnt++].w=va;
            }
            for(i=0; i<W; ++i)
            {
                cin>>u>>v>>va;
                --u,--v;
                edge[cnt].u=u,edge[cnt].v=v,edge[cnt++].w=-va;
            }
            m=cnt;
            if(Bellman(0)) puts("YES");//顶点0到其他顶点的最短路
            else puts("NO");
        }
        return 0;
    }
    /**
    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
    **/

[USACO 2006 December Gold]: http://poj.org/searchproblem?field=source&key=USACO+2006+December+Gold
[Bellman Ford]: http://blog.csdn.net/mikasa3/article/details/52091082