POJ--3259--spfa spfa写法

poj-3259

题目

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 8

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.

大意：就是有一个人有n个农场，农场与农场间共有M条路，长度为val，有的有虫洞，从虫洞过去后会回到val时间前，问你是否看到他自己

最短路，判负环，不用多说spfa
SPFA 判负环
bfs：判断入队次数是否超过顶点数
dfs： 判断是否重复递归

邻接矩阵
BFS 和SLF优化

SLF 优化：就是把队列改成双端队列，再入队前先判断，如果队首大于就放到队尾，小于的话放到队前。。。。。

#include<iostream>#include<cstring>#include<cstdio>#include<deque>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int map[502][502];int n,m,b;void spfa(){   int a[502]={0};    int vis[520]={0};    int dis[520];    memset(dis,INF,sizeof(dis));    vis[1]=1;    dis[1]=0;    deque<int>q;    q.push_front(1);    while(!q.empty())    {        int now=q.front();        q.pop_front();        vis[now]=0;        for(int i=1;i<=n;i++)        {               if(dis[i]>dis[now]+map[now][i])            {                dis[i]=dis[now]+map[now][i];                if(!vis[i])                {   a[i]++;                    if(a[i]>=n)                     {                        printf("YES\n");return;                    }                    vis[i]=1;                    if(!q.empty())                    {                        if(dis[i]>dis[q.front()])                            {                                q.push_back(i);                             }                            else                            q.push_front(i);                    }                    else                    q.push_front(i);                }            }        }    }    printf("NO\n"); }int main(){       std::ios::sync_with_stdio(false);    int t;    cin>>t;    while(t--)    {           cin>>n>>m>>b;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(i==j) map[i][j]=map[j][i]=0;                else map[i][j]=map[j][i]=INF;            }        }        for(int i=1;i<=m;i++)        {            int x,y,v;            cin>>x>>y>>v;            if(map[x][y]>v)            map[x][y]=map[y][x]=v;        }        for(int i=1;i<=b;i++)        {            int x,y,v;            cin>>x>>y>>v;            if(map[x][y]>-v)            map[x][y]=-v;        }        spfa();    }    return 0;}

DFS （超时，不知道是不是这道题的问题）

 //poj  3259#include<iostream>#include<cstring>#include<cstdio>#include<deque>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int map[502][502];int v[502]={0};int n,m,b;int flag=0;int d[502]={0};void spfa(int i){       if(flag) return;            v[i]=1;        for(int j=1;j<=n;j++)        {            if(d[j]>d[i]+map[i][j])            {                d[j]=d[i]+map[i][j];                if(v[j])                {                    flag=1;return;                }                if(flag) return;                    spfa(j);                }         }          v[i]=0;}int main(){       std::ios::sync_with_stdio(false);    int t;    cin>>t;    while(t--)    {           cin>>n>>m>>b;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(i==j) map[i][j]=map[j][i]=0;                else map[i][j]=map[j][i]=INF;            }        }        for(int i=1;i<=m;i++)        {            int x,y,v;            cin>>x>>y>>v;            if(map[x][y]>v)            map[x][y]=map[y][x]=v;        }        for(int i=1;i<=b;i++)        {            int x,y,v;            cin>>x>>y>>v;            if(map[x][y]>-v)            map[x][y]=-v;        }        memset(v,0,sizeof(v));        memset(d,INF,sizeof(d));        d[1]=0;flag=0;        spfa(1);        if(flag)        cout<<"YES"<<endl;        else        {            cout<<"NO"<<endl;        }    }    return 0;}

邻接表
BFS

#include<cstring>#include<cstdio>#include<queue>#define INF 0x3f3f3f3fint m,n,w;using namespace std;int head[522];int u;  int dis[520];struct sp{    int v,va,next;}q[5200];bool spfa(){    memset(dis,INF,sizeof(dis));    bool vis[520]={0};    int cur;    int cnt[520]={0};    vis[1]=true;    dis[1]=0;    cnt[1]=1;    queue<int>p;    p.push(1);    while(!p.empty())    {         cur=p.front();        p.pop();        vis[cur]=false;        for(int i=head[cur];i!=-1;i=q[i].next)        {            int id=q[i].v;            if(dis[cur]+q[i].va<dis[id])            {                dis[id]=dis[cur]+q[i].va;                if(!vis[id])                {                    cnt[id]++;                    vis[id]=true;                    p.push(id);                    if(cnt[id]>n)                    {                        return true;                    }                }            }        }    }    return false;}void add(int x,int y,int z){    q[u].v=y;    q[u].va=z;    q[u].next=head[x];    head[x]=u++;}int main(){    int t;    scanf("%d",&t);    while(t--)    {   u=0;        memset(head,-1,sizeof(head));        scanf("%d%d%d",&m,&n,&w);            for(int j=0;j<n;j++)            {                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 a,b,va;                scanf("%d%d%d",&a,&b,&va);                 add(a,b,-va);                   }               if(spfa())        {   printf("YES\n");        }        else            printf("NO\n");    }    return 0;}

dfs

#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int n,m,w,o;#define INF 0x3f3f3f3fstruct pp{    int v,va,next;}ss[5200];int head[520];int vis[520];int dis[520];void add(int u,int v,int val){    ss[o].v=v;    ss[o].va=val;    ss[o].next=head[u];    head[u]=o++;}int flag=0;void spfa(int x){   if(flag) return;    vis[x]=1;    for(int i=head[x];i!=-1;i=ss[i].next)    {   int v=ss[i].v;int val=ss[i].va;        if(dis[v]>val+dis[x])        {            dis[v]=val+dis[x];            if(vis[v])            {                flag=1;                return;            }            spfa(v);            if(flag) return;        }    }    vis[x]=0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {   o=0;        memset(head,-1,sizeof(head));        scanf("%d%d%d",&n,&m,&w);        for(int i=0;i<m;i++)        {            int u,v,val;            scanf("%d%d%d",&u,&v,&val);            add(u,v,val);            add(v,u,val);               }        for(int i=0;i<w;i++)        {            int u,v,val;            scanf("%d%d%d",&u,&v,&val);            add(u,v,-val);        }           memset(dis,INF,sizeof(dis));            dis[1]=0;        flag=0;        memset(vis,0,sizeof(vis));        spfa(1);        if(flag)        printf("YES\n");        else        printf("NO\n");    }    return 0;}

vector
和邻接表类似，只不过换用存法了

BFS

#include<cstring>#include<iostream>#include<cstdio>#include<queue>#include<vector>#define INF 0x3f3f3f3fint m,n,w;using namespace std;struct pp{    int v,va;}ss;vector<pp>qq[5200];void spfa(){    int dis[520];    memset(dis,INF,sizeof(dis)) ;    int vis[520]={0};    dis[1]=0;    vis[1]=1;    int cnt[520]={0};    queue<int>q1;    q1.push(1);    cnt[1]=1;    while(!q1.empty())    {        int now=q1.front();        q1.pop();        vis[now]=0;        int len=qq[now].size();        for(int i=0;i<len;i++)        {   int v=qq[now][i].v;int va=qq[now][i].va;            if(dis[v]>va+dis[now])            {                   dis[v]=va+dis[now];                if(!vis[v])                {                    cnt[v]++;                    q1.push(v);                    vis[v]=1;                    if(cnt[v]>=m)                    {                        printf("YES\n");                        return;                    }                }            }        }    }        printf("NO\n");}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&m,&n,&w);        for(int i=0;i<m+10;i++)        qq[i].clear();        for(int i=0;i<n;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            ss.v=y;ss.va=z;            qq[x].push_back(ss);            ss.v=x;            qq[y].push_back(ss);        }        for(int j=0;j<w;j++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            ss.v=y;ss.va=-z;            qq[x].push_back(ss);        }        spfa();     }     return 0; } 

DFS

#include<cstring>#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<vector>using namespace std;int n,m,w,o;#define INF 0x3f3f3f3fstruct pp{    int v,val;}ss;int flag=1;int dis[520],vis[520];vector<pp>q[520];void spfa( int x){   if(flag) return;    vis[x]=1;    int len=q[x].size();    for(int i=0;i<len;i++)    {           int v=q[x][i].v;        int val=q[x][i].val;        if(dis[v]>dis[x]+val)        {            dis[v]=dis[x]+val;            if(vis[v])            {                flag=1;                return;            }            spfa(v);            if(flag) return;        }    }    vis[x]=0;}int main(){     //  ios::sync_with_stdio(false);       int t;    //   cin>>t;     scanf("%d",&t);       while(t--)       {            cin>>n>>m>>w;            for(int i=1;i<=n;i++)            q[i].clear();            for(int i=0;i<m;i++)            {   int u,v;            //  cin>>u>>v>>ss.val;                scanf("%d%d%d",&u,&v,&ss.val);                ss.v=u;                q[v].push_back(ss);                ss.v=v;                q[u].push_back(ss);            }            for(int i=0;i<w;i++)            {   int u,v,val;            //  cin>>u>>v>>val;                scanf("%d%d%d",&u,&v,&val);                ss.v=v;ss.val=-val;                q[u].push_back(ss);            }            memset(vis,0,sizeof(vis));            memset(dis,INF,sizeof(dis));            dis[1]=0;            flag=0;            spfa(1);            if(flag) printf("YES\n");            else printf("NO\n");       }    return 0;}

vector真是个好东西！
还有这道题cin输出取消同步后比printf多300ms。

学习。。。。。