poj2253

题意：青蛙要从点1出发到点2去，每条边之间距离为权值，中间可能经过很多点，有不同路径，问每条1到2可行路中某一条路径的最大值的最小情况是多少。类似最短路问题，更改一下几个变量定义，用dijkstra,floyed都可以做，也可以等价求最小生成树里最大边的问题。

dijkstra

这里dist[i]定义是，第i个点到起点路径中最大边的边，由于求的是最小的情况，松弛还是用每次确定的最短边进行松弛!

[code]#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
double dist[210];
bool visit[210];
const double inf=99999999;
struct Point
{
    int x,y;
}point[210];

double Distance(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int main()
{
  int cnt=0;
  while(~scanf("%d",&n)&&n)
  {
      cnt++;
      for(int i=1;i<=n;i++)
        scanf("%d %d",&point[i].x,&point[i].y);
      memset(visit,false,sizeof(visit));
      for(int i=1;i<=n;i++)
        dist[i]=inf;
      dist[1]=0;
      for(int i=1;i<=n;i++)
      {
          int minn=inf,index;
          for(int j=1;j<=n;j++)
          {
              if(!visit[j]&&dist[j]<minn)
              {
                  minn=dist[j];
                  index=j;
              }
          }
          visit[index]=1;
          for(int j=1;j<=n;j++)
          {
              if(!visit[j]&&dist[j]>max(dist[index],Distance(point[j],point[index])))
                dist[j]=max(dist[index],Distance(point[j],point[index]));
          }
      }
      printf("Scenario #%d\n",cnt);
      printf("Frog Distance = %.3f\n\n",dist[2]);
  }
  return 0;
}

floyed

这里M[i][j]定义是i j两点间最大边的松弛后最小值， M[i][j]=min(M[i][j],max(M[i][k],M[k][j]));。时间复杂度n^3，由于这题只有200个数据，可以用！

[code]#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
double M[210][210];
const double inf=99999999;
struct Point
{
    int x,y;
}point[210];

double Distance(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void floyd()
{
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
           M[i][j]=min(M[i][j],max(M[i][k],M[k][j]));
}

int main()
{
  int cnt=0;
  while(~scanf("%d",&n)&&n)
  {
      cnt++;
      for(int i=1;i<=n;i++)
        scanf("%d %d",&point[i].x,&point[i].y);
      for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        M[i][j]=(double)Distance(point[i],point[j]);
      floyd();
      printf("Scenario #%d\n",cnt);
      printf("Frog Distance = %.3f\n\n",M[1][2]);
  }
  return 0;
}

最小生成树

构建一个最小生成树，输出最大值！bingo

[code]#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
double dist[210],ans;
bool visit[210];
const double inf=99999999;
struct Point
{
    int x,y;
}point[210];

double Distance(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void prime()
{
   ans=0;
   memset(visit,0,sizeof(visit));
   for(int i=1;i<=n;i++)
        dist[i]=Distance(point[1],point[i]);
   visit[1]=1;
   for(int i=1;i<=n-1;i++)
   {
       double minn=99999999.0;int index;
       for(int j=1;j<=n;j++)
       {
           if(!visit[j]&&dist[j]<minn)
           {
               index=j;
               minn=dist[j];
           }
       }
       visit[index]=1;
       ans=max(ans,minn);
       if(index==2)break;
       for(int j=1;j<=n;j++)
       {
         if(!visit[j]&&dist[j]>Distance(point[index],point[j]))
                dist[j]=Distance(point[index],point[j]);
       }
   }
}

int main()
{
  int cnt=0;
  while(~scanf("%d",&n)&&n)
  {
      cnt++;
      for(int i=1;i<=n;i++)
        scanf("%d %d",&point[i].x,&point[i].y);
      prime();
      printf("Scenario #%d\n",cnt);
      printf("Frog Distance = %.3f\n\n",ans);
  }
  return 0;
}