一. spfa的SLF优化，就是双向队列优化，在spfa压入队列时，判断要压入队首还是队尾。这个优化可以优化15%~20%。
二. spfa的LLL优化，就是记录现在队列中元素所代表值的平均值，和要压入元素的值相比较，如果大于平均值，直接压入对列尾部，LLL优化+SLF优化可以优化大概50%。
我们以一道判负环的题为例子，来讲解一下。
Wormholes Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 53294 Accepted: 19856
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 8Sample Output
NO YESHint
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 首先 ：我们用的是普通spfa。跑完程序是141ms #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; #define inf 0x3f3f3f3f struct s1{ int to; int val; int next; }a[10000]; int head[10000]; int dis[10000]; int book[10000]; int cnt; void add(int u,int v,int w) { a[++cnt].to=v; a[cnt].val=w; a[cnt].next=head[u]; head[u]=cnt; } int main() { int F; scanf("%d",&F); while(F--) { int n,m,w; cnt=0; scanf("%d%d%d",&n,&m,&w); memset(book,0,sizeof(book)); memset(head,-1,sizeof(head)); memset(a,0,sizeof(a)); for(int i=0;i<m;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,z); add(y,x,z); } for(int i=0;i<w;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,-z); } queue<int>q; memset(dis,inf,sizeof(dis)); dis[1]=0; book[1]=1; q.push(1); int cn[10000]; int flag=0; memset(cn,0,sizeof(cn)); cn[1]=1; while(!q.empty()) { int p=q.front(); q.pop(); book[p]=0; for(int i=head[p];i!=-1;i=a[i].next) { if(dis[a[i].to]>dis[p]+a[i].val) { dis[a[i].to]=dis[p]+a[i].val; if(book[a[i].to]==0) { book[a[i].to]=1; cn[a[i].to]++; if(cn[a[i].to]>=n) { flag=1; break; } q.push(a[i].to); } } } if(flag==1) break; } if(flag==1) printf("YES\n"); else printf("NO\n"); } } 然后 我们加上 SLF优化 ，跑完程序是94ms。 #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; #define inf 0x3f3f3f3f struct s1{ int to; int val; int next; }a[10000]; int head[10000]; int dis[10000]; int book[10000]; int cnt; void add(int u,int v,int w) { a[++cnt].to=v; a[cnt].val=w; a[cnt].next=head[u]; head[u]=cnt; } int main() { int F; scanf("%d",&F); while(F--) { int n,m,w; cnt=0; scanf("%d%d%d",&n,&m,&w); memset(book,0,sizeof(book)); memset(head,-1,sizeof(head)); memset(a,0,sizeof(a)); for(int i=0;i<m;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,z); add(y,x,z); } for(int i=0;i<w;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,-z); } deque<int>q; memset(dis,inf,sizeof(dis)); dis[1]=0; book[1]=1; q.push_back(1); int cn[10000]; int flag=0; memset(cn,0,sizeof(cn)); cn[1]=1; while(!q.empty()) { int p=q.front(); q.pop_front(); book[p]=0; for(int i=head[p];i!=-1;i=a[i].next) { if(dis[a[i].to]>dis[p]+a[i].val) { dis[a[i].to]=dis[p]+a[i].val; if(book[a[i].to]==0) { book[a[i].to]=1; cn[a[i].to]++; if(cn[a[i].to]>=n) { flag=1; break; } if(!q.empty()&&dis[a[i].to]<dis[q.front()]) //判断是要压入队尾还是队首 （SLF优化） { q.push_front(a[i].to); } else { q.push_back(a[i].to); } } } } if(flag==1) break; } if(flag==1) printf("YES\n"); else printf("NO\n"); } } 最后，我们用SLF加上LLL优化，跑完程序，是47ms。 #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; #define inf 0x3f3f3f3f struct s1{ int to; int val; int next; }a[10000]; int head[10000]; int dis[10000]; int book[10000]; int cnt; void add(int u,int v,int w) { a[++cnt].to=v; a[cnt].val=w; a[cnt].next=head[u]; head[u]=cnt; } int main() { int F; scanf("%d",&F); while(F--) { int n,m,w; cnt=0; scanf("%d%d%d",&n,&m,&w); memset(book,0,sizeof(book)); memset(head,-1,sizeof(head)); memset(a,0,sizeof(a)); for(int i=0;i<m;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,z); add(y,x,z); } for(int i=0;i<w;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,-z); } deque<int>q; memset(dis,inf,sizeof(dis)); dis[1]=0; book[1]=1; q.push_back(1); int cn[10000]; int flag=0; memset(cn,0,sizeof(cn)); cn[1]=1; long long sum=0; int len=1; while(!q.empty()) { int p=q.front(); q.pop_front(); if(dis[p]*len>sum) //LLL优化 { q.push_back(p); continue; } sum-=dis[p]; book[p]=0; len--; for(int i=head[p];i!=-1;i=a[i].next) { if(dis[a[i].to]>dis[p]+a[i].val) { dis[a[i].to]=dis[p]+a[i].val; if(book[a[i].to]==0) { book[a[i].to]=1; cn[a[i].to]++; if(cn[a[i].to]>=n) { flag=1; break; } if(!q.empty()&&dis[a[i].to]<dis[q.front()]) //SLF优化 { q.push_front(a[i].to); } else { q.push_back(a[i].to); } sum += dis[a[i].to]; len++; } } } if(flag==1) break; } if(flag==1) printf("YES\n"); else printf("NO\n"); } }