#include
#include
#include
#include
#define INF 1000000000
using namespace std;
struct p
{
double x,y;
}spot[110];
double cost[110][110];
double mincost[110];
bool used[110];
int n;
double prim()
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2022-08-05 15:48:21
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prim#include#include#include#include#include#include#include#ix;
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2023-07-27 18:41:44
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题目大意：给你N个点的坐标，求能使这N个点相连的所有边的最小距离是多少。思路：先求出每个点和其他点的距离，存到图中，用Prim模板来做。
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2014-12-25 16:18:10
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http://acm.hdu.edu.cn/showproblem.php?pid=1162Problem DescriptionEddy begins to like painting pictures rec...
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2019-06-17 14:53:00
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http://acm.hdu.edu.cn/showproblem.php?pid=1162Problem DescriptionEddy begins to like painting pictures rec...
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2019-06-17 14:53:00
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Eddy's pictureTime Limit: 2000/1000 MS (Jav
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2022-08-08 18:42:51
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Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested
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2013-08-17 23:13:00
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#include#include#include#define N 200#define inf 999999999double map[N][N];int visit[N],next[N];int main() {int n,m,i,j,k,cnt;double a[N],b[N],dis,min...
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2013-11-10 15:54:00
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最小生成树//prim 求最小生成树#include #include #include #define MAXN 105#define UPPERDIS 999999double lowcost[MAXN],vist[MAXN];double cost[MAXN][MAXN];int n;double prim(int v0){ int i, j, minone;
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2023-03-03 12:45:08
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http://acm.hdu.edu.cn/showproblem.php?pid=1162 1 #include 2 #include 3 #include 4 #include 5 #include 6 #define maxn 200 7 using namespace std; 8...
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2014-04-13 20:26:00
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Prim算法#include #include #include int n;double map[210][210],a[210];int b[210];void met(){ int i,t=1; a[t]=0; while(b[t]==0){ b[t]=1; for(i=1;imap[t][i]) a[i]=map[t][i]; double min=200000000; for(i=1;ia[i])min=a[i],t=i; }}int main(){ int i,j,k...
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2014-03-10 10:42:00
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Eddy's pictureTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s):iptionEddy begins to
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2022-12-02 00:24:57
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Eddy's picture Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8797 Accepted Submission(s): 4476
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2021-07-21 15:56:18
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Eddy's pictureTime Limit: 2000/1000 MS (Java/Others) Memory
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2023-02-20 15:55:18
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Eddy's picture
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7530 Accepted Submission(s): 3824
Problem Description
Eddy begins to
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2023-04-24 02:37:49
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Problem B Time Limit : 2000/1000ms (Java/Other) Memory Limit :65536/32768K (Java/Other)Total Submission(s) : 3tionEddy begins to like paintingpictures rec...
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2023-02-07 11:21:52
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题目大意：输入一个整数n表示表示有n个点。在接下来的n行中，每行有两个整数x ， y 。分别表示一个点的横坐标以及纵坐标。求距离最小的连线解题思路:1)二维----->>一维for(i = 1 ; i <= n ; ++i){ scanf("%lf%lf",&point[i].x,&point[i].y); point[i].id = i; }2)求
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2023-04-11 14:58:36
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Eddy's pictureTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9101 Accepted Submission(s): 4607Problem DescriptionEddy begins to li
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2022-08-11 15:52:23
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思路1，暴力法，超时。
1 import sys
2 class Solution:
3 def maxDistance(self, grid: 'List[List[int]]') -> int:
4 row = len(grid)
5 column = len(grid[0])
6 waters = []
7
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2019-08-18 13:02:00
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我去写评测机的人好阴险啊竟然不让使用List....W(￣_￣)W 然后Java强大的APi根本挡不住的好伐..... 用C写一大坨的java一行调用现成轮子搞定~ AC代码
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2021-07-27 12:00:50
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