大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
解决报告
意甲冠军：
乞讨0至1所有最大的道路值的最小数量。
思维：
floyd。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define inf 0x3f3f3f3f
using namespace std;
int n,m,q;
double mmap[210][210];
struct node {
double x,y;
} p[210];
double dis(node p1,node p2) {
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
void floyd() {
for(int k=0; k<n; k++)
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
mmap[i][j]=min(mmap[i][j],max(mmap[i][k],mmap[k][j]));
}
int main() {
int i,j,u,v,w,k=1;
while(~scanf("%d",&n)) {
if(!n)break;
for(i=0; i<n; i++) {
for(j=0; j<n; j++)
mmap[i][j]=(double)inf;
mmap[i][i]=0;
}
for(i=0; i<n; i++) {
scanf("%lf%lf",&p[i].x,&p[i].y);
}
for(i=0; i<n; i++) {
for(j=0; j<n; j++) {
mmap[i][j]=dis(p[i],p[j]);
}
}
floyd();
printf("Scenario #%d\n",k++);
printf("Frog Distance = %.3lf\n",mmap[0][1]);
printf("\n");
}
return 0;
}
Frogger
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 25958 | Accepted: 8431 |
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.
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
版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 举报，一经查实，本站将立刻删除。
发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/116705.html原文链接：https://javaforall.cn
【正版授权，激活自己账号】： Jetbrains全家桶Ide使用，1年售后保障，每天仅需1毛
【官方授权 正版激活】： 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...
