poj 2253 dijkstra

题意：一只青蛙要通过在石头间的跳跃到达一个地方 ， 求跳跃宽度的最大最小值？

关键在于 ， 思考出宽度和距离的相同之处

我们只需要把距离变成每天路径上的最大宽度就行 ， 然后在用dijkstra求出每个点的最小宽度

代码：
#include
#include
#include
#include
using namespace std;

#define INF 0xfffffff
#define maxn 310
//#define max(x,y) (x)>(y)?(x):(y)

double grap[maxn][maxn];
int pre[maxn] ;
double dist[maxn];
int xy[maxn][2];
int n;

void init()
{
    int i ,j;
    for(i = 1; i<= n; i++)
       for(j = 1; j<= n; j++)
          grap[i][j] =INF*1.0 , grap[i][i] = 0.0;
}

inline double max(double x , double y)
{
    if(x> y)   returnx;
    returny;
}

void dijkstra(int u)
{
    memset(pre ,0 , sizeof(pre));
    int i ,j;
    for(i = 1; i<= n; i++)
       dist[i] =grap[u][i];
    dist[u] =0.0; //初始化对源点

    pre[u] =1;
    int x =u;
    for(j = 1; j< n; j++)
    {
       double maxs= INF*1.0;

       for(i = 1; i<= n; i++)
          if(!pre[i]&& maxs >dist[i])
             x = i , maxs= dist[i];

       pre[x] =1;

       for(i = 1; i<= n; i++)
          if(!pre[i]&& dist[i] >max(dist[x],grap[x][i]))
             dist[i] =max(dist[x],grap[x][i]);
    }
}

int main()
{
    int gh =1;
   while(scanf("%d" , &n)&& n)
    {
      init();
       int i , x ,j , y ;
       doublez;
       for(i = 1; i<= n; i++)
       {
          scanf("%d%d" , &xy[i][0] , &xy[i][1]);
       }
       for(i = 1; i<= n; i++)
          for(j = i+1;j <= n; j++)
          {
             x =xy[i][0]-xy[j][0] , y = xy[i][1] - xy[j][1];
             z =sqrt(x*x*1.0+y*y*1.0);
             grap[i][j] =grap[j][i] = z;
         //   cout<<z<<endl;
          }

      dijkstra(1);

      printf("Scenario #%d\n" , gh++);
       printf("FrogDistance = %.3lf\n" , dist[2]);

      cout<<endl;

    }
   cout<<endl;
    return0;
}

这个题目同时能用bfs 或则 bfs+二分来做
下面是bfs+二分的代码：
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <queue>
using namespace std;

#define INF 0xfffffff
#define maxn 310
//#define max(x,y) (x)>(y)?(x):(y)

double grap[maxn][maxn];
int pre[maxn] ;
double dist[maxn];
int xy[maxn][2];
int n;

void init()
{
    int i ,j;
    for(i = 1; i<= n; i++)
       for(j = 1; j<= n; j++)
          grap[i][j] =-1.0;
    for(i = 1; i<= n; i++)
       dist[i] =INF*1.0;
   
}

inline double max(double x , double y)
{
    if(x> y)   returnx;
    returny;
}

int bfs(double maxz)
{
    intpre[maxn];
    memset(pre ,0 , sizeof(pre));
   
    int i;
   
    dist[1] =0.0;
   queue<int>q;
   q.push(1);
    pre[1] =1;
   while(!q.empty())
    {
       int u =q.front() ;  q.pop();
   //    pre[u] =0;
       if(dist[2]< maxz)  return 1;
       for(i = 1; i<= n; i++)
       {
          if(i != u&& grap[u][i] > 0&& max(dist[u],grap[u][i])< maxz)
          {
             dist[i] =max(dist[u] , grap[u][i]);
            if(!pre[i])  q.push(i) , pre[i] = 1;
          }
       }
    }
    return0;
}

int main()
{
    int gh =1;
   while(scanf("%d" , &n)&& n)
    {
      init();
       int i , x ,j , y ;
       double z ,maxz = 0.0;
       for(i = 1; i<= n; i++)
       {
          scanf("%d%d" , &xy[i][0] , &xy[i][1]);
       }
       for(i = 1; i<= n; i++)
          for(j = i+1;j <= n; j++)
          {
             x =xy[i][0]-xy[j][0] , y = xy[i][1] - xy[j][1];
             z =sqrt(x*x*1.0+y*y*1.0);
             grap[i][j] =grap[j][i] = z;
             if(i == 1&& j == 2)  maxz =z;
             //maxz =max(maxz , z);
         //   cout<<z<<endl;
          }
       double l =0.0 , r = maxz;
      while(r-l>0.0001)
       {
          double mind= (l+r)/2;
         //cout<<mind<<endl;
         if(bfs(mind))  r = mind;
          else l =mind;
      //   cout<<dist[2]<<endl;

       }

      printf("Scenario #%d\n" , gh++);
       printf("FrogDistance = %.3lf\n" , l);

      cout<<endl;

    }
   cout<<endl;
    return0;
}