poj 3259 Bellman-ford + SPFA

Wormholes

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 43809 Accepted: 16085

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..NM (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.

题意:
虫洞问题,现在有n个点,m条边,代表现在可以走的通路,比如从a到b和从b到a需要花费c时间
现在在地上出现了w个虫洞,虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)
现在问你从某个点开始走,能回到从前

Bellman-ford

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define INF 0x3f3f3f3f

int top;
struct node
{
    int u,v,t;
}path[6010];
int n,m,w;

void add(int u,int v,int t)
{
    path[top].u=u;
    path[top].v=v;
    path[top++].t=t;
}

int bellman_ford(int n)
{
    int weight[520];
    for(int i=0;i<=n;i++)
        weight[i]=INF;
    weight[1]=0;

    int u,v,t;
    for(int i=0;i<n-1;i++){
        for(int j=0;j<top;j++){
            u=path[j].u;
            v=path[j].v;
            t=path[j].t;

            if(weight[u]+t<weight[v])
                weight[v]=weight[u]+t;
        }
    }

    for(int j=0;j<top;j++){
        u=path[j].u;
        v=path[j].v;
        t=path[j].t;
        if(weight[u]+t<weight[v])
            return 0;
    }
    return 1;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&w);

        top=0;
        int u,v,t;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&t);
            add(u,v,t);
            add(v,u,t);
        }
        for(int i=0;i<w;i++){
            scanf("%d%d%d",&u,&v,&t);
            add(u,v,-t);
        }

        if(!bellman_ford(n))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

SPFA


建立一个队列,初始时队列里只有起始点,再建立一个表格记录起始点到所有点的最短路径(该表格的初始值要赋为极大值,该点到他本身的路径赋为0)。然后执行松弛操作,用队列里有的点作为起始点去刷新到所有点的最短路,如果刷新成功且被刷新点不在队列中则把该点加入到队列最后。重复执行直到队列为空。

判断有无负环:
  如果某个点进入队列的次数超过N次则存在负环(SPFA无法处理带负环的图)

 

《poj 3259 Bellman-ford + SPFA》

 

 

 

首先建立起始点a到其余各点的
最短路径表格

                                  《poj 3259 Bellman-ford + SPFA》

首先源点a入队,当队列非空时:
 1、队首元素(a)出队,对以a为起始点的所有边的终点依次进行松弛操作(此处有b,c,d三个点),此时路径表格状态为:

                                  《poj 3259 Bellman-ford + SPFA》

在松弛时三个点的最短路径估值变小了,而这些点队列中都没有出现,这些点
需要入队,此时,队列中新入队了三个结点b,c,d

队首元素b点出队,对以b为起始点的所有边的终点依次进行松弛操作(此处只有e点),此时路径表格状态为:

                                 《poj 3259 Bellman-ford + SPFA》

在最短路径表中,e的最短路径估值也变小了,e在队列中不存在,因此e也要
入队,此时队列中的元素为c,d,e

队首元素c点出队,对以c为起始点的所有边的终点依次进行松弛操作(此处有e,f两个点),此时路径表格状态为:

                                 《poj 3259 Bellman-ford + SPFA》

在最短路径表中,e,f的最短路径估值变小了,e在队列中存在,f不存在。因此
e不用入队了,f要入队,此时队列中的元素为d,e,f

 队首元素d点出队,对以d为起始点的所有边的终点依次进行松弛操作(此处只有g这个点),此时路径表格状态为:

 

 

                               《poj 3259 Bellman-ford + SPFA》

在最短路径表中,g的最短路径估值没有变小(松弛不成功),没有新结点入队,队列中元素为f,g

队首元素f点出队,对以f为起始点的所有边的终点依次进行松弛操作(此处有d,e,g三个点),此时路径表格状态为:

                               《poj 3259 Bellman-ford + SPFA》

在最短路径表中,e,g的最短路径估值又变小,队列中无e点,e入队,队列中存在g这个点,g不用入队,此时队列中元素为g,e

队首元素g点出队,对以g为起始点的所有边的终点依次进行松弛操作(此处只有b点),此时路径表格状态为:

                           《poj 3259 Bellman-ford + SPFA》

在最短路径表中,b的最短路径估值又变小,队列中无b点,b入队,此时队列中元素为e,b
队首元素e点出队,对以e为起始点的所有边的终点依次进行松弛操作(此处只有g这个点),此时路径表格状态为:

 

                          《poj 3259 Bellman-ford + SPFA》

在最短路径表中,g的最短路径估值没变化(松弛不成功),此时队列中元素为b

队首元素b点出队,对以b为起始点的所有边的终点依次进行松弛操作(此处只有e这个点),此时路径表格状态为:

                         《poj 3259 Bellman-ford + SPFA》

在最短路径表中,e的最短路径估值没变化(松弛不成功),此时队列为空了

最终a到g的最短路径为14

#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define INF 0x3f3f3f3f
#define MAXN 10005

int map[MAXN][MAXN];
int dis[MAXN];
int num[MAXN],vis[MAXN];
int n,m,w;

bool SPFA()
{
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++)
        dis[i]=INF;

    queue<int>q;
    dis[0]=0;     vis[0]=1;
    q.push(0);    num[0]++;

    while(q.size())
    {
        int p=q.front();
        q.pop();
        vis[p]=0;

        for(int i=1;i<=n;i++){
            if(dis[p]+map[p][i]<dis[i]){
                dis[i]=dis[p]+map[p][i];

                if(vis[i]==0){
                    vis[i]=1;
                    num[i]++;
                    q.push(i);
                    if(num[i]>n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    int T;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=i;j++)
                map[i][j]=map[j][i]=INF;

        int a,b,c;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            if(c<map[a][b])
                map[a][b]=map[b][a]=c;
        }
        for(int i=1;i<=w;i++){
            scanf("%d%d%d",&a,&b,&c);
            map[a][b]=-c;
        }

        for(int i=1;i<=n;i++)
            map[0][i]=0;
        if(SPFA())
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define INF 0x3f3f3f3f
#define MAXN 6000

int dis[MAXN],head[MAXN],num[MAXN],vis[MAXN];
int n,m,w;
int top;

struct node
{
    int v,w,next;
}path[MAXN];

void add(int u,int v,int w)
{
    path[top].v=v;
    path[top].w=w;
    path[top].next=head[u];
    head[u]=top++;
}

bool SPFA()
{
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++)
        dis[i]=INF;

    queue<int>q;
    vis[1]=1;    dis[1]=0;
    q.push(1);   num[1]++;

    while(q.size())
    {
        int p=q.front();
        q.pop();
        vis[p]=0;

        for(int i=head[p];i!=-1;i=path[i].next){
            int t=path[i].v;

            if(dis[t]>dis[p]+path[i].w){
                dis[t]=dis[p]+path[i].w;
                if(vis[t]==0){
                    vis[t]=1;
                    q.push(t);
                    num[t]++;
                    if(num[t]>n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    int T;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        top=0;
        scanf("%d%d%d",&n,&m,&w);
        memset(head,-1,sizeof(head));

        int a,b,c;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        for(int i=1;i<=w;i++){
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }

        if(SPFA())
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

    原文作者:Bellman - ford算法
    原文地址: https://blog.csdn.net/Summer__show_/article/details/52092737
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞