【最小生成树】POJ 2421 Constructing Roads
来源:互联网 发布:网络创世纪 stones 编辑:程序博客网 时间:2024/05/27 00:25
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
本来是01矩阵输入,正常思维用prim,但是因为偏爱kruskal,于是就强行转成了kruskal:
#include<iostream>#include<algorithm>#include<cstdio>using namespace std;const int MAXN=100+10;int fa[MAXN];struct node{ int beg; int end; int w;};node a[MAXN*MAXN];void read(int &x){ x=0; char c=getchar(); while(c<'0'||c>'9')c=getchar(); while(c>='0'&&c<='9') { x=x*10+c-'0'; c=getchar(); }}int found(int x){ if(fa[x]!=x)fa[x]=found(fa[x]); return fa[x];}int comp(const node&i,const node&j){ return i.w<j.w;}int main(){ int n; read(n); int k=0; for(int i=1;i<=n;i++)fa[i]=i; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { k++; a[k].beg=i; a[k].end=j; read(a[k].w); } } int t; read(t); for(int i=1;i<=t;i++) { int x,y; read(x);read(y); fa[found(x)]=found(y); } sort(a+1,a+k+1,comp); int ans=0,s=0; for(int i=1;i<=k;i++) { if(found(a[i].beg)!=found(a[i].end)) { fa[found(a[i].beg)]=found(a[i].end); s++; ans+=a[i].w; } } cout<<ans; return 0;}
- POJ 2421 Constructing Roads(最小生成树)
- POJ 2421 Constructing Roads 最小生成树
- POJ 2421 Constructing Roads 最小生成树
- POJ 2421 Constructing Roads 最小生成树
- POJ--2421--Constructing Roads【最小生成树】
- POJ 2421 -Constructing Roads -最小生成树
- 【最小生成树】POJ 2421 Constructing Roads
- POJ 2421 Constructing Roads 最小生成树
- Constructing Roads 【poj-2421】【最小生成树】
- POJ-2421 Constructing Roads (最小生成树)
- poj 2421 Constructing Roads prim最小生成树 基础!!!
- POJ-2421(最小生成树模版)(Constructing Roads )
- Poj 2421 Constructing Roads(Prim 最小生成树)
- POJ 2421 Constructing Roads(简单最小生成树)
- poj 2421 Constructing Roads 并查集+最小生成树
- poj-2421 Constructing Roads(最小生成树 Kruskal算法)
- poj 2421 Constructing Roads(最小生成树 kruskal算法)
- poj 2421 Constructing Roads(最小生成树))
- 网络路径问题
- Java多线程-线程死锁
- mysql优化——mysqladmin命令
- 安全知识点
- LTspice introduction
- 【最小生成树】POJ 2421 Constructing Roads
- 贪心算法--阶乘之和
- [Leetcode] 103. Binary Tree Zigzag Level Order Traversal 解题报告
- jQuery开发过程中的一些技巧
- java.lang.OutOfMemoryError: PermGen space内存溢出解决方法
- Session实现防止会员下载资源被盗链
- JavaScript 的 join() 方法
- ubuntu软件安装 caffe相关软件安装 学习笔记
- NS2仿真分析无线网络的攻击防御(2)