[模板]poj3259（判断是否存在负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51623		Accepted: 19175

DescriptionWhile 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 farmscomprises 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 10,000seconds.InputLine 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 connectedby 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.OutputLines 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

YESbellman-ford判负环，如果每个边都松弛了n-1还能松弛，则有负环，效率低判断给定的有向图中是否存在负环。      利用 spfa 算法判断负环有两种方法：      1） spfa 的 dfs 形式，判断条件是存在一点在一条路径上出现多次。      2） spfa 的 bfs 形式，判断条件是存在一点入队次数大于总顶点数。DFSspfa的SLF优化：SLF：Small LabelFirst 策略，设要加入的节点是j，队首元素为i，若dist(j)<dist(i)，则将j插入队首，否则插入队尾 //注意一开始的图是双向的//bellman-ford判负环//时间复杂度分析：log(n*m)//通过本次wa，我明白了cin的执行顺序是从右到左的//此代码效率莫名的高，可能是操作简单吧#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int mn=502,mm=5300;int n,m,ww,tot;int u[mm],v[mm],w[mm];int dis[mn];bool bellman_ford(){for(int i=1;i<=n;i++) dis[i]=499999999;dis[1]=0;for(int i=1;i<n;i++){int f=1;for(int i=0;i<tot;i++)if(dis[v[i]]>dis[u[i]]+w[i]){dis[v[i]]=dis[u[i]]+w[i];f=0;}if(f) return false;}for(int i=0;i<tot;i++)if(dis[v[i]]>dis[u[i]]+w[i]) return true;return false;}int main(){ios::sync_with_stdio(0);int T;cin>>T;while(T--){cin>>n>>m>>ww;tot=0;while(m--){cin>>u[tot]>>v[tot]>>w[tot];tot++;v[tot]=u[tot-1],u[tot]=v[tot-1],w[tot]=w[tot-1];tot++;}while(ww--){cin>>u[tot]>>v[tot]>>w[tot];w[tot]=-w[tot];tot++;}if(bellman_ford()) cout<<"YES\n";else cout<<"NO\n";}return 0;}

//spfa判负环dfs，不推荐（效率低）
#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int mn=502,mm=5300;
struct node{
int to,w,next;
}edge[mm];
int n,m,ww,tot;
int dis[mn],vis[mn],head[mn];
void add(int u,int v,int w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
bool spfa(int u)
{
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dis[v]>dis[u]+edge[i].w)
{
if(vis[v]) return true;
vis[v]=1;
dis[v]=dis[u]+edge[i].w;
if(spfa(v)) return true;
vis[v]=0;
}
}
return false;
}
int main()
{
int u,v,w;
int T;
ios::sync_with_stdio(0);
cin>>T;
while(T--)
{
tot=0;
cin>>n>>m>>ww;
memset(head,-1,sizeof(head));
for(int i=0;i<m;i++)
{
cin>>u>>v>>w;
add(u,v,w);
add(v,u,w);
}
for(int i=0;i<ww;i++)
{
cin>>u>>v>>w;
add(u,v,-w);
}
for(int i=1;i<=n;i++) dis[i]=49999999;
dis[1]=0;
memset(vis,0,sizeof(vis));
vis[1]=1;
if(spfa(1)) cout<<"YES\n";
else cout<<"NO\n";
}
return 0;
}

//spfa判负环bfs，时间复杂度ke，e为边数，k为入队列次数
#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int mn=502,mm=5300;
struct node{
int to,w,next;
}edge[mm];
int n,m,ww,tot;
int dis[mn],inq[mn],head[mn],cnt[mn];
void add(int u,int v,int w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
bool spfa()
{
for(int i=1;i<=n;i++) dis[i]=49999999;
memset(cnt,0,sizeof(cnt));
memset(inq,0,sizeof(inq));
dis[1]=0,inq[1]=cnt[1]=1;;
queue<int> q;
q.push(1);
while(q.size())
{
int x=q.front();
q.pop();
inq[x]=0;
for(int i=head[x];i!=-1;i=edge[i].next)
{
int u=edge[i].to;
if(dis[u]>dis[x]+edge[i].w)
{
dis[u]=dis[x]+edge[i].w;
if(!inq[u])
{
if(++cnt[u]>=n) return true;
inq[u]=1;
q.push(u);
}
}
}
}
return false;
}
int main()
{
int u,v,w;
int T;
ios::sync_with_stdio(0);
cin>>T;
while(T--)
{
tot=0;
cin>>n>>m>>ww;
memset(head,-1,sizeof(head));
for(int i=0;i<m;i++)
{
cin>>u>>v>>w;
add(u,v,w);
add(v,u,w);
}
for(int i=0;i<ww;i++)
{
cin>>u>>v>>w;
add(u,v,-w);
}
if(spfa()) cout<<"YES\n";
else cout<&l
4000
t;"NO\n";
}
return 0;
}

//spfa判负环bfs+slf优化
#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int mn=502,mm=5300;
struct node{
int to,w,next;
}edge[mm];
int n,m,ww,tot;
int dis[mn],inq[mn],head[mn],cnt[mn];
void add(int u,int v,int w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
bool spfa()
{
for(int i=1;i<=n;i++) dis[i]=49999999;
memset(cnt,0,sizeof(cnt));
memset(inq,0,sizeof(inq));
dis[1]=0,inq[1]=cnt[1]=1;;
deque<int> q;
q.push_back(1);
while(q.size())
{
int x=q.front();
q.pop_front();
inq[x]=0;
for(int i=head[x];i!=-1;i=edge[i].next)
{
int u=edge[i].to;
if(dis[u]>dis[x]+edge[i].w)
{
dis[u]=dis[x]+edge[i].w;
if(!inq[u])
{
if(++cnt[u]>=n) return true;
inq[u]=1;
if(q.size()&&dis[q.front()]>dis[u])
q.push_front(u);
else q.push_back(u);
}
}
}
}
return false;
}
int main()
{
int u,v,w;
int T;
ios::sync_with_stdio(0);
cin>>T;
while(T--)
{
tot=0;
cin>>n>>m>>ww;
memset(head,-1,sizeof(head));
for(int i=0;i<m;i++)
{
cin>>u>>v>>w;
add(u,v,w);
add(v,u,w);
}
for(int i=0;i<ww;i++)
{
cin>>u>>v>>w;
add(u,v,-w);
}
if(spfa()) cout<<"YES\n";
else cout<<"NO\n";
}
return 0;
}