[模板]poj3259（判断是否存在负环）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 51623 Accepted: 19175

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NO
YES
bellman-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。
DFS
spfa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<<"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;}