标签：sam des 1.4 最小生成树 names follow reac eve rop
题目链接
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
2
0 0
3 4
3
17 4
19 4
18 5
0
Sample Output
Scenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414
Source
可用二分法与BFS法解决，但有更好的算法。先求出最小生成树，起点和终点在树上的唯一路径就是我们所要找的路径。需要注意的是这道题只用求最短的那条边，所以如果用kruskal算法不用将最小生成树求出来，只要起点和终点连通即可。
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
struct aa{
int x;
int y;
aa(int x1,int y1):x(x1),y(y1){}
};
struct bb{
int a;
int b;
int c;
bb(int a1,int b1,int c1):a(a1),b(b1),c(c1){}
bool operator < (const bb &rhs)const{
return c > rhs.c;
}
};
vector<aa>s;
priority_queue<bb>z;
int p[251];
int find_z(int x){
return p[x]==x?x:p[x]=find_z(p[x]);
}
int main(){
int n;
int yy=0;
while(~scanf("%d",&n)){
if(n==0)break;
yy++;
s.clear();
while(!z.empty())z.pop();
for(int i=0,q,w;i<n;i++){
scanf("%d%d",&q,&w);
s.push_back(aa(q,w));
}
for(int i=0,o;i<n;i++){
for(int j=0;j<n;j++){
o=(s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
z.push(bb(i,j,o));
}
}
for(int i=0;i<n;i++)p[i]=i;
while(!z.empty()){
bb zz=z.top(); z.pop();
if( find_z(zz.a) != find_z(zz.b) ){
p[find_z(zz.a)]=find_z(zz.b);
}
if(find_z(0)==find_z(1)){
printf("Scenario #%d\n",yy);
double sss=(double)zz.c;
printf("Frog Distance = %.3lf\n\n",sqrt(sss));
break;
}
}
}
return 0;
}
poj-2253(最小瓶颈路问题)
标签：sam des 1.4 最小生成树 names follow reac eve rop
原文地址：https://www.cnblogs.com/wz-archer/p/10568974.html
