Poj 2253 Frogger

Multiple methods can be understood as the shortest path deformation.

It can also be used as the Minimum Spanning Tree.

It is easy to understand the meaning of the question.

The maximum Hop Distance between frog a and frog B.

Two shortest paths are written. Spfa and Dijkstra.

By the way, you should be familiar with the minimal spanning tree and write a Kruskal.

Spfa:

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<queue>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6using namespace std;int n,m;struct node{    double x,y;}l[201];struct lx{    int v;    double d;};vector<lx> g[201];void SPFA(int start){    double dis[201];    bool vis[201];    for(int i=1;i<=n;i++)        dis[i]=INF,vis[i]=0;    queue<int>q;    dis[start]=0,vis[start]=1;    q.push(start);    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int j=0;j<g[u].size();j++)        {            int v=g[u][j].v;            double d=g[u][j].d;            if(dis[v]>max(dis[u],d))            {                dis[v]=max(dis[u],d);                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    printf("Scenario #%d\nFrog Distance = %.3f\n\n",m++,dis[2]);}int main(){    m=1;    while(scanf("%d",&n),n)    {        for(int i=1;i<=n;i++)        {            scanf("%lf%lf",&l[i].x,&l[i].y);            g[i].clear();        }        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)            {                double d=sqrt(pow(l[i].x-l[j].x,2)+pow(l[i].y-l[j].y,2));                lx now;                now.d=d;                now.v=j;g[i].push_back(now);                now.v=i;g[j].push_back(now);            }        }        SPFA(1);    }}

Dijkstra:

#include <cstdio>#include <cmath>#include <algorithm>#define inf 0x7ffffff#define MAXV 210using namespace std;struct lx{    double x,y;} point[MAXV];double d[MAXV];int n;void dijkstra(){    int i,j,vis[MAXV],v=1;    double min;    for(i=1; i<=n; i++)    {        d[i]=inf;        vis[i]=0;    }    d[1]=0;    for(i=1; i<=n; i++)    {        min=inf;        for(j=1; j<=n; j++)            if(!vis[j] && d[j]<min)            {                min=d[j];                v=j;            }        vis[v]=1;        if(v==2) break;        for(j=1; j<=n; j++)        {            double len=sqrt(pow(point[v].x-point[j].x,2)+pow(point[v].y-point[j].y,2));            if(!vis[j] && d[j]>max(d[v],len))            {                d[j]=max(d[v],len);            }        }    }}int main(){    int i,cnt=1;    while(scanf("%d",&n) ,n)    {        for(i=1; i<=n; i++) scanf("%lf%lf",&point[i].x,&point[i].y);        dijkstra();        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",cnt++,d[2]);    }    return 0;}

Kruskal:

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<queue>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6using namespace std;int n,m;int fa[201];struct lx{    double x,y;}point[201];struct edge{    int u,v;    double len;}l[20101];int father(int x){    if(x!=fa[x])        fa[x]=father(fa[x]);    return fa[x];}bool cmp(edge a,edge b){    return a.len<b.len;}int main(){    int m=1;    while(scanf("%d",&n),n)    {        for(int i=1;i<=n;i++)            scanf("%lf%lf",&point[i].x,&point[i].y);        for(int i=1;i<=n;i++)            fa[i]=i;        int edgecot=0;        for(int i=1;i<=n;i++)            for(int j=i+1;j<=n;j++)        {            l[edgecot].u=i;            l[edgecot].v=j;            l[edgecot++].len=sqrt(pow(point[i].x-point[j].x,2)+pow(point[i].y-point[j].y,2));        }        sort(l,l+edgecot,cmp);        double ans=0;        for(int i=0;i<edgecot;i++)        {            int x=l[i].u;            int y=l[i].v;            double len=l[i].len;            int fx=father(x);            int fy=father(y);            if(fx==fy)continue;            fa[fy]=fx;            if(father(1)==father(2))            {                ans=len;break;            }        }        printf("Scenario #%d\nFrog Distance = %.3f\n\n",m++,ans);    }}