POJ
来源:互联网 发布:暗黑破坏神2修改器mac 编辑:程序博客网 时间:2024/06/05 22:29
Description
Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Input
The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.
Output
For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
Sample Input
40 4 9 214 0 8 179 8 0 1621 17 16 0
Sample Output
28
题解:
最小生成树的板子 以前都是用克鲁斯卡尔算法写的 这次 用prime
AC代码
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL long long#define CLR(a,b) memset(a,(b),sizeof(a))const int INF = 0x3f3f3f3f;const int N = 1e2+10;int dis[N];int G[N][N];bool vis[N];int n, m;//void init()//{// for(int i = 1; i <= n; i++) {// for(int j = 1; j <= n; j++) {// if(i == j) G[i][j] = 0;// else G[i][j] = INF;// }// }//}int prim(int s){ int ans = 0; CLR(vis,false); for(int i = 1; i <= n; i++) { dis[i] = INF; } dis[s] = 0; for(int i = 1; i <= n; i++) { int minx = INF; int next; for(int j = 1; j <= n; j++) { if(!vis[j] && dis[j]<minx) { minx = dis[j]; next = j; } } // if(minx == INF) break; ans += minx; vis[next] = true; for(int j = 1; j <= n; j++) { if(!vis[j] && dis[j]>G[next][j]) { dis[j] = G[next][j]; } } } return ans;}int main(){ while(~scanf("%d",&n)) { for(int i = 1;i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d",&G[i][j]); } } printf("%d\n",prim(1)); }return 0;}
- POJ
- poj
- POJ
- POJ
- poj
- poj
- POJ
- POJ
- poj
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- 搞定!Firefox收藏夹默认在当前tab打开
- Android常用框架
- WPS 2016安装密钥
- Baidu Map 周边雷达
- Chrome 控制台高级用法
- POJ
- CentOS Mongodb安装使用
- 详细Oracle 安装文档
- Mongodb DB.prototype._authOrThrow@src/mongo/shell/db.js:1441:20
- 【memset函数】
- utf-8 和 utf8的区别
- 数据结构:将二叉搜索树转换成一个排序的双向链表
- WebStorm集成Mocha + Chai进行js单元测试
- leetcode[Find Mode in Binary Search Tree]//待整理多种解法