http://poj.org/problem?id=2253
Frogger
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 37886 | Accepted: 12196 |
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:
给定两个青蛙的坐标,A青蛙拜访B青蛙,要求
拜访路程选择的路线上可以跳的最大距离,小于其它任何路线上可以跳的最大距离。
如图:
思路:
按照前天做过一类型的题目,找出最短路径下的最大权值边。后来一想,第二组测试数据 ans=2 昂,就卡壳了。
更重要的是你要知道这是最短路径!
其求的就是所选择的路线上可以跳的最大距离,小于其它任何路线上可以跳的最大距离
参考CODE:
提交选择C++
#include<stdio.h>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MYDD=1103;
const int INF=0x3f3f3f3f;
struct Q {
int x,y;
} node[MYDD];
double Dis(Q a,Q b) {/*两点距离*/
double X=(a.x-b.x)*1.0;
double Y=(a.y-b.y)*1.0;
return sqrt(X*X+Y*Y);
}
double dis[MYDD];
bool vis[MYDD];
void dijk(int n) {
memset(vis,0,sizeof(vis));
for(int j=0; j<=n; j++)
dis[j]=INF;
dis[1]=0;
for(int j=1; j<=n; j++) {
int Tp=INF,choose=-1;
for(int k=1; k<=n; k++) {
if(!vis[k]&&dis[k]<Tp) {
Tp=dis[k];
choose=k;
}
}
vis[choose]=1;
if(choose==2) break;/*找到第二只青蛙*/
for(int k=1; k<=n; k++) {
if(!vis[k]&&dis[k]>max(dis[choose],Dis(node[choose],node[k])))
dis[k]=max(dis[choose],Dis(node[k],node[choose]));
}/*[Error] expected ')' before 'dis' 英语渣渣的 me */
}
}
int main() {
int n,kc=1;
while(scanf("%d",&n)&&n) {
for(int j=1; j<=n; j++) {
scanf("%d%d",&node[j].x,&node[j].y);
}
dijk(n);
printf("Scenario #%d\n",kc++);
printf("Frog Distance = %.3lf\n\n",dis[2]);
}
return 0;
}