poj2253Frogger——最短路变形

poj2253Frogge：http://poj.org/problem?id=2253

Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33131 Accepted: 10651

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping. 
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps. 
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence. 
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones. 

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone. 

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

通过做这个题，对Dijkstra和Floyd有了更深刻的理解，这算是最大的收获吧。

题意：读这个题就是累啊，读了n多遍。给你一些石头的坐标。第一、二个是两个青蛙的坐标，下面是n-2个石头坐标。第一只青蛙要到第二只青蛙那里。要求的是从第一只到第二只青蛙的所有路径中的最大的跳跃距离的最小值（不是指第一只到第二只青蛙的最短路或最长路）。什么意思呢？假如：存在路径：1 ——>(10)4——>5(20)——>2(15)。表示1到4长度为10……。在这条路径中可以得到最大跳跃距离是20；存在第二条路径：1——>3(14)——>4(9)——>2(25)。在这条路径中可以得到最大跳跃距离为25。那么最终的答案就是这两个的最小值20。

思路：可以用Dijkstra，也可以用Floyd。参考了大神的代码，自己写了一遍。

Floyd：http://blog.sina.com.cn/s/blog_5f5353cc0100gnxu.html

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <stack>#include <cmath>#include <algorithm>using namespace std;#define maxn 2000 + 10#define INF 10000001double grap[maxn][maxn];int n;void Folyd(){    for(int k = 1; k <= n; k++)        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                    grap[i][j] = min(grap[i][j],max(grap[i][k],grap[k][j]));}int s1[maxn],s2[maxn];double dis(int a, int b){    return sqrt((double)(s1[a]-s1[b])*(s1[a]-s1[b])+(double)(s2[a]-s2[b])*(s2[a]-s2[b]));}int main(){    int ca = 1;    while(~scanf("%d",&n) && n)    {        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                grap[i][j] = grap[j][i] = (i == j ? 0 : INF);        int i,j;        for(i=1; i<=n; i++)            scanf("%d %d", &s1[i], &s2[i]);        for(i=1; i<=n; i++)            for(j=i+1; j<=n; j++)            {                grap[i][j]=grap[j][i]=dis(i,j);            }        Folyd();        printf("Scenario #%d\nFrog Distance = %.3f\n\n",ca++,grap[2][1]);    }    return 0;}

Dijkstra：http://www.myexception.cn/program/1712988.html

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <stack>#include <cmath>#include <algorithm>using namespace std;#define maxn 2000 + 10#define INF 10000001int xx1,xx2,yy1,yy2;double d[maxn],grap[maxn][maxn];int vis[maxn],n;void Dij(int u0){    for(int i = 1; i <= n; i++)d[i] = INF;    memset(vis,0,sizeof(vis));    d[u0] = 0;    for(int i = 1; i <= n; i++)    {        int u = u0;        double minn = INF;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && d[j] < minn)            {                u = j;                minn = d[j];            }        }        vis[u] = 1;        for(int j = 1; j <= n; j++)            d[j] = min(d[j],max(grap[u][j],minn));    }}int s1[maxn],s2[maxn];double dis(int a, int b){    return sqrt((double)(s1[a]-s1[b])*(s1[a]-s1[b])+(double)(s2[a]-s2[b])*(s2[a]-s2[b]));}int main(){    int ca = 1;    while(~scanf("%d",&n) && n)    {        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                grap[i][j] = grap[j][i] = (i == j ? 0 : INF);        int i,j;        for(i=1; i<=n; i++)            scanf("%d %d", &s1[i], &s2[i]);        for(i=1; i<=n; i++)            for(j=i+1; j<=n; j++)            {                grap[i][j]=grap[j][i]=dis(i,j);            }        Dij(1);        printf("Scenario #%d\nFrog Distance = %.3f\n\n",ca++,d[2]);    }    return 0;}