# 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.

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;

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

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

SPFA

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）

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：

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：

``````#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 n,m,w;
int top;

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

{
path[top].v=v;
path[top].w=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;
vis[1]=1;    dis[1]=0;
q.push(1);   num[1]++;

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

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);

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