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.
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
题意:一只青蛙在X1,Y1的石头上,他要到X2,Y2的石头上续人,池塘上除了1、2两块石头外还有n-2块石头坐标分别为X3,Y3;X4,Y4……Xn,Yn
定义青蛙的跳跃范围为从石头一到石头二路径上所跳跃的最远距离,求跳跃范围的最小值
题解:spfa的判断稍微换一下就行了~
但要注意poj上.3f与.3lf的区别…….3f是不能用的
代码如下:
#include<cmath> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; vector< pair<int,double> > g[1100]; double d[1100],x[1100],y[1100]; int vis[1100],n; void spfa(int u) { memset(vis,0,sizeof(vis)); for(int i=1; i<=n; i++) { d[i]=inf; } d[u]=0; queue<int> q; q.push(u); while(!q.empty()) { int x=q.front(); q.pop(); vis[x]=0; int sz=g[x].size(); for(int i=0; i<sz; i++) { int y=g[x][i].first; double w=g[x][i].second; if(max(d[x],w)<d[y]) { d[y]=max(d[x],w); if(!vis[y]) { q.push(y); vis[y]=1; } } } } } int main() { int ttt=0; while(scanf("%d",&n),n) { ttt++; for(int i=1;i<=n;i++) { g[i].clear(); } for(int i=1; i<=n; i++) { scanf("%lf%lf",&x[i],&y[i]); } for(int i=1; i<=n; i++) { for(int j=i+1; j<=n; j++) { double dx=x[i]-x[j],dy=y[i]-y[j]; double dist=sqrt(dx*dx+dy*dy); g[i].push_back(make_pair(j,dist)); g[j].push_back(make_pair(i,dist)); } } spfa(1); printf("Scenario #%d\n",ttt); printf("Frog Distance = %.3f\n",d[2]); printf("\n"); } }
