spfa的SLF 和 LLL优化算法

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文章标签：

#SPFA #LLL优化 #SLF优化 #双向队列优化 #优化

SPFA判负环专栏收录该内容

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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: 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: 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

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.

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");
    }
}