POJ - 2421 Constructing Roads(最小生成树 prim算法)

本文介绍了一道关于构造道路的算法题目,通过Prim算法求解最小生成树问题。题目要求在已知村庄间距离的情况下,新建道路使得所有村庄连接且总长度最短。

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Constructing Roads
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 25971 Accepted: 11367

Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected. 

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j. 

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output

179

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 110
#define INF 0x3f3f3f3f

using namespace std;

int map[N][N];
int vis[N];
int dis[N];
int ans,n;

void prim()
{
    int i,j,min,pos;
    for(i=1; i<=n; i++)
    {
        dis[i] = map[1][i];
        vis[i] = 0;
    }
    vis[1] = 1;
    for(i=1; i<=n-1; i++)
    {
        min = INF;
        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] < min)
            {
                min = dis[j];
                pos = j;
            }
        }
        ans += min;
        vis[pos] = 1;
        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] > map[pos][j])
            {
                dis[j] = map[pos][j];
            }
        }
    }
}

int main()
{
    int m,i,j,a,b;
    while(~scanf("%d",&n))
    {
        ans = 0;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d %d",&a,&b);
            map[a][b] = map[b][a] = 0;
        }
        prim();
        printf("%d\n",ans);
    }
    return 0;
}

反思:

题目大意:多组输入,给出图的点数n和图的邻接矩阵,接下来有m组数据,每组包含a和b两个整数,说明a和b两个村庄连通。最小生成树模板题,因为题目给出了邻接矩阵,所以连图的初始化都省了♪(゚▽^*)ノ⌒☆。已经连通的两个村庄权值置为0,然后用prim即可。

内容概要:本文档为《400_IB Specification Vol 2-Release-2.0-Final-2025-07-31.pdf》,主要描述了InfiniBand架构2.0版本的物理层规范。文档详细规定了链路初始化、配置与训练流程,包括但不限于传输序列(TS1、TS2、TS3)、链路去偏斜、波特率、前向纠错(FEC)支持、链路速度协商及扩展速度选项等。此外,还介绍了链路状态机的不同状态(如禁用、轮询、配置等),以及各状态下应遵循的规则命令。针对不同数据速率(从SDR到XDR)的链路格式化规则也有详细说明,确保数据包格式控制符号在多条物理通道上的一致性正确性。文档还涵盖了链路性能监控错误检测机制。 适用人群:适用于从事网络硬件设计、开发及维护的技术人员,尤其是那些需要深入了解InfiniBand物理层细节的专业人士。 使用场景及目标:① 设计实现支持多种数据速率编码方式的InfiniBand设备;② 开发链路初始化训练算法,确保链路两端设备能够正确配置并优化通信质量;③ 实现链路性能监控错误检测,提高系统的可靠性稳定性。 其他说明:本文档属于InfiniBand贸易协会所有,为专有信息,仅供内部参考技术交流使用。文档内容详尽,对于理解实施InfiniBand接口具有重要指导意义。读者应结合相关背景资料进行学习,以确保正确理解应用规范中的各项技术要
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