poj 1287 Networking (最小生成树)
来源:互联网 发布:华为办公网络 编辑:程序博客网 时间:2024/06/07 02:39
http://poj.org/problem?id=1287
Networking
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8028 Accepted: 4417
Description
You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.
Input
The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.
Output
For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.
Sample Input
1 02 31 2 372 1 171 2 683 71 2 192 3 113 1 71 3 52 3 893 1 911 2 325 71 2 52 3 72 4 84 5 113 5 101 5 64 2 120
Sample Output
0171626
思路:最基本的prim,kruskal,但要注意的是有重边,所以用prim写的时候需要处理一下,kruskal的话就不需要了
kruskal
#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;#define N 2100#define INF 0x3f3f3f3f#define met(a, b) memset (a, b, sizeof(a))struct node{ int x, y, d;}stu[N];bool cmp (node a, node b){ return a.d < b.d;}int n, m, f[60];int Find (int x){ if (x!=f[x]) f[x] = Find (f[x]); return f[x];}void kruskal (){ int sum = 0; for (int i=0; i<m; i++) { int xx = Find (stu[i].x); int yy = Find (stu[i].y); if (xx != yy) { f[xx] = yy; sum += stu[i].d; } } printf ("%d\n", sum);}int main (){ while (scanf ("%d", &n), n) { met (stu, 0); scanf ("%d", &m); for (int i=0; i<=n; i++) f[i] = i; for (int i=0; i<m; i++) scanf ("%d%d%d", &stu[i].x, &stu[i].y, &stu[i].d); sort (stu, stu+m, cmp); kruskal (); } return 0;}
prim
#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;#define N 200#define INF 0x3f3f3f3f#define met(a, b) memset (a, b, sizeof(a))int g[N][N], dist[N], vis[N], n, m;void prim (int sa){ for (int i=1; i<=n; i++) dist[i] = g[sa][i]; met (vis, 0); dist[sa] = 0; vis[sa] = 1; int ans = 0; for (int i=1; i<=n; i++) { int minx = INF, Index = 0; for (int j=1; j<=n; j++) if (!vis[j] && dist[j] < minx) minx = dist[Index = j]; vis[Index] = 1; ans += dist[Index]; for (int j=1; j<=n; j++) if (!vis[j] && dist[j] > g[Index][j]) dist[j] = g[Index][j]; } printf ("%d\n", ans);}int main (){ while (scanf ("%d", &n), n) { int x, y, d; scanf ("%d", &m); for (int i=1; i<=n; i++) { dist[i] = INF; for (int j=1; j<=n; j++) g[i][j] = INF; } while (m--) { scanf ("%d %d %d", &x, &y, &d); g[y][x] = g[x][y] = min (g[x][y], min (g[y][x], d)); } prim (1); } return 0;}
1 0
- POJ 1287 最小生成树 Networking
- poj 1287 Networking 最小生成树
- POJ 1287 Networking(最小生成树)
- poj 1287 Networking(图论:最小生成树)
- poj-1287 Networking 最小生成树
- POJ 1287 Networking(最小生成树)
- POJ 1287 Networking(最小生成树)
- poj 1287 Networking(最小生成树)
- poj 1287 Networking ->最小生成树
- Poj 1287 Networking【最小生成树】
- POJ - 1287 Networking(最小生成树)
- poj 1287 Networking(最小生成树)
- POJ 1287 Networking(最小生成树)
- poj 1287 Networking (最小生成树)
- POJ 1287 Networking(最小生成树)
- 【POJ】-1287-Networking(最小生成树)
- poj-1287-Networking【最小生成树】
- POJ 1287 Networking (最小生成树)
- SQLITE数据库修复
- xFire 实现webservice
- Android开发编程Lint工具应用布局Xml文件
- SDUT 2410 Mine Number DFS+回溯 (扫雷)
- PHP获取多个checkbox的值
- poj 1287 Networking (最小生成树)
- 百度echarts 的使用
- Building Android Kernel for the Nexus 5 — AOSP(6.0.1)
- 淘宝notify-消息中间件(2)
- 最短路径算法
- Pixhawk 四旋翼状态估计
- hdu1711 Number Sequence 求模式串在主串中的位置
- servlet 生命周期 和 作原理详解
- UE4 引擎 package 可执行文件过程中碰到fatal error的常见解决方法