Networking
Time Limit:1000MSMemory Limit:10000K
Total Submissions:6589Accepted:3590
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.
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 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
Source
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<algorithm>#include<cstdlib>#include<set>#include<queue>#include<stack>#include<vector>#include<map>#define N 10010#define Mod 10000007#define lson l,mid,idx<<1#define rson mid+1,r,idx<<1|1#define lc idx<<1#define rc idx<<1|1const double EPS = 1e-11;const double PI = acos ( -1.0 );//const double E = 2.718281828;typedef long long ll;const int INF = 1000010;using namespace std;int n,m;int mp[N][N];int par[N],Rank[N];struct edge {int u,v,cost;};bool cmp(const edge& e1,const edge& e2) {return e1.cost<e2.cost;}edge es[N];int E;void init(int n) {for(int i=0; i<=n; i++) {par[i]=i;Rank[i]=0;}}int finds(int x) {if(par[x]==x)return x;return par[x]=finds(par[x]);}void unite(int x,int y) {x=finds(x);y=finds(y);if(x==y)return;if(Rank[x]<Rank[y])par[x]=y;else {par[y]=x;if(Rank[x]==Rank[y])Rank[x]++;}}bool same(int x,int y) {return finds(x)==finds(y);}int kruskal() {sort(es,es+E,cmp);init(E);int res=0;for(int i=0; i<E; i++) {edge e=es[i];if(!same(e.u,e.v)) {res+=e.cost;unite(e.u,e.v);}}return res;}int main() {while(cin>>n&&n) {E=0;cin>>m;for(int i=0; i<=n; i++)for(int j=0; j<=n; j++) {//if(i==j)mp[i][j]=0;//elsemp[i][j]=INF;}int x,y,cost;for(int i=1; i<=m; i++) {scanf("%d%d%d",&x,&y,&cost);if(mp[x][y]>cost)mp[x][y]=cost;}for(int i=1; i<=n; i++)for(int j=1; j<=n; j++) {if(i==j||mp[i][j]==INF)continue;es[E].u=i;es[E].v=j;es[E++].cost=mp[i][j];}int ans=kruskal();cout<<ans<<endl;}return 0;}
，勇气执着的背负起那厚重的行囊，奔向远方。
