POJ 1258 Agri-Net 最小生成树 Prim 算法
2015-10-22 19:18
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Prim 算法描述
1).输入:一个加权连通图,其中顶点集合为V,边集合为E;
2).初始化:Vnew = {x},其中x为集合V中的任一节点(起始点),Enew =
{},为空;
3).重复下列操作,直到Vnew = V:
a.在集合E中选取权值最小的边<u, v>,其中u为集合Vnew中的元素,而v不在Vnew集合当中,并且v∈V(如果存在有多条满足前述条件即具有相同权值的边,则可任意选取其中之一);
b.将v加入集合Vnew中,将<u, v>边加入集合Enew中;
4).输出:使用集合Vnew和Enew来描述所得到的最小生成树。
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//感谢bin巨的模板
ACcode:
1).输入:一个加权连通图,其中顶点集合为V,边集合为E;
2).初始化:Vnew = {x},其中x为集合V中的任一节点(起始点),Enew =
{},为空;
3).重复下列操作,直到Vnew = V:
a.在集合E中选取权值最小的边<u, v>,其中u为集合Vnew中的元素,而v不在Vnew集合当中,并且v∈V(如果存在有多条满足前述条件即具有相同权值的边,则可任意选取其中之一);
b.将v加入集合Vnew中,将<u, v>边加入集合Enew中;
4).输出:使用集合Vnew和Enew来描述所得到的最小生成树。
Agri-Net
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 4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0 Sample Output 28 Source USACO 102 |
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裸最小生成树
//感谢bin巨的模板
ACcode:
#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 105
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI acos(-1.0)
#define E exp(1)
using namespace std;
bool vis[maxn];
int lowc[maxn];
int a[maxn][maxn];
int Prim(int cost[][maxn],int n){///点是1~n
int ans=0;
MT(vis,false);
vis[0]=true;
FOR(i,1,n-1)lowc[i]=cost[0][i];
FOR(i,1,n-1){
int minc=INF;
int p=-1;
FOR(j,0,n-1)
if(!vis[j]&&minc>lowc[j]){
minc=lowc[j];
p=j;
}
if(minc==INF)return -1;///原图不连接
ans+=minc;
vis[p]=true;
FOR(j,0,n-1)
if(!vis[j]&&lowc[j]>cost[p][j])
lowc[j]=cost[p][j];
}
return ans;
}
int main(){
int n;
while(rd(n)!=EOF){
FOR(i,0,n-1)
FOR(j,0,n-1)
rd(a[i][j]);
printf("%d\n",Prim(a,n));
}
return 0;
}
/*
4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0*/
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