spfa的SLF 和 LLL优化算法
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。
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total 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 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
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 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
#include
#include
#include
#include
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;iq;
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
#include
#include
#include
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;iq;
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] 最后，我们用SLF加上LLL优化，跑完程序，是47ms。
#include
#include
#include
#include
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;iq;
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]