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
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<=(n); i++)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxm = 1e6 + ;
const double PI = acos(-1.0);
const double eps = 1e-;
const int dx[] = {-,,,,,,-,-};
const int dy[] = {,,,-,,-,,-};
int dir[][] = {{,},{,-},{-,},{,}};
const int mon[] = {, , , , , , , , , , , , };
const int monn[] = {, , , , , , , , , , , , };
const int mod = ;
#define inf 0x3f3f3f3f
#define ll long long
const int maxn = ;int u,v,w;
int n,m,ans,k,sum,cnt;
int a[][];
struct node
{
int u,v,w;
}e[maxn];
int fa[maxn];
int Find(int x)
{
if(fa[x]!=x)
fa[x]=Find(fa[x]);
return fa[x];
}
void join(int x,int y)
{
int xx = Find(x);
int yy = Find(y);
fa[xx]=yy;
}
bool cmp(node a,node b)
{
return a.w < b.w;
}
void kruskal()
{
cnt=;
rep(i,,m)
{
int x=e[i].u;
int y=e[i].v;
if(Find(x)!=Find(y))
{
join(x,y);
cnt++;
sum += e[i].w;
}
if(cnt == n-) break;
}
printf("%d\n",sum);
}
int main()
{
while(~scanf("%d",&n))
{
sum=,cnt=,m=,ms(e,);
rep(i,,n)
fa[i]=i;
rep(i,,n)
{
rep(j,,n)
{
scanf("%d",&a[i][j]);
if(j<i) //对称的无向图，建一半即可
{
e[++m].u = i;
e[m].v = j;
e[m].w = a[i][j]; //注意是m条边
}
}
}
sort(e+, e+m+, cmp);
kruskal();
}
}
/*
【题意】
给你n*n矩阵表示i(行)和j(列)之间的权值，求该图的MST。【类型】
最小生成树模板题【分析】【时间复杂度&&优化】【trick】
*/
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.
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.
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 3
1 2 37
2 1 17
1 2 683 7
1 2 19
2 3 11
3 1 7
1 3 5
2 3 89
3 1 91
1 2 325 7
1 2 5
2 3 7
2 4 8
4 5 11
3 5 10
1 5 6
4 2 120
Sample Output
0
17
16
26
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<=(n); i++)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxm = 1e6 + ;
const double PI = acos(-1.0);
const double eps = 1e-;
const int dx[] = {-,,,,,,-,-};
const int dy[] = {,,,-,,-,,-};
int dir[][] = {{,},{,-},{-,},{,}};
const int mon[] = {, , , , , , , , , , , , };
const int monn[] = {, , , , , , , , , , , , };
const int mod = ;
#define inf 0x3f3f3f3f
#define ll long long
const int maxn = ;int u,v,w;
int n,m,ans,k,sum,cnt;
int a[][];
struct node
{
int u,v,w;
}e[maxn];int fa[maxn];int Find(int x)
{
if(fa[x]!=x)
fa[x]=Find(fa[x]);
return fa[x];
}
void join(int x,int y)
{
int xx = Find(x);
int yy = Find(y);
fa[xx]=yy;
}
bool cmp(node a,node b)
{
return a.w < b.w;
}
void kruskal()
{
cnt=;
rep(i,,m)
{
int x=e[i].u;
int y=e[i].v;
if(Find(x)!=Find(y))
{
join(x,y);
cnt++;
sum += e[i].w;
}
if(cnt == n-) break;
}
printf("%d\n",sum);
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
if(n==) break;
sum=,cnt=;
rep(i,,n)
fa[i]=i;
for(int i=;i<=m;i++)
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
sort(e+, e+m+, cmp);
kruskal();
}
}
/*
【题意】
给你u v w表示u和v之间的权值w，求该图的MST。【类型】
最小生成树模板题【分析】【时间复杂度&&优化】【trick】
*/
