Hihocoder Hiho The 26th week minimum spanning tree one · (Prim algorithm)

Topic 1: Minimum spanning tree one · Prim algorithm time limit:10000msSingle Point time limit:1000msMemory Limit:256MBDescribe

Recently, little hi is very fond of playing a game simulation of the city opened a new mod, in this mod, the player can have more than one city!

But the problem also follows--Little hi now has n cities in hand, and it is known that the cost of building roads between any two cities in these n cities, little hi would like to know that the minimum cost would allow any two cities to reach each other through the roads they built (assuming a, B, c three cities, Only the road between AB and BC can be built, and the two roads will be connected by AC.

Tip: I don't know why prim algorithm and Dijstra algorithm are like σ (っ°д°;) our store. Input

Each test point (input file) has and has only one set of test data.

In a set of test data:

The 1th behavior is 1 integer n, which indicates the number of cities owned by small hi.

The next n rows, for a n*n matrix A, describe the cost of building a road between any two cities, with the number of J in line I being AIJ, which represents the cost of building roads between the city of Block I and Block J.

For 100% of data, satisfies n<=10^3, for any I, satisfies aii=0, for any I, J satisfies Aij=aji, 0<aij<10^4.

Output

For each set of test data, Output 1 integer ans, indicating that in order for any two cities to reach each other at least the required construction costs through the roads they build.

Sample input

Sample output

4178

#include <stdio.h> #include <string.h> #include <iostream> #include <string> #include < algorithm> #define INF 999999999#define n 1001using namespace std;int map[n][n];bool vis[n];int dis[n];int N, sum;void p    Rim () {sum=0;//minimum spanning tree weights and initialization 0 int mm;    int I, J;    Memset (Vis, false, sizeof (VIS));    for (i=0; i<n; i++) {dis[i]=map[0][i];    } vis[0]=true;    int POS;        for (i=0; i<n-1; i++) {mm=inf;                 for (j=0; j<n; J + +) {if (Vis[j]==false && dis[j]<mm) {mm=dis[j];            Pos=j;        }} sum+=mm;        Vis[pos]=true;                for (j=0; j<n; J + +) {if (Vis[j]==false && Dis[j]>map[j][pos]) {            Dis[j]=map[j][pos]; }}} printf ("%d\n", sum);}    int main () {int i, J;    scanf ("%d", &n);          For (i=0, i<n; i++) {for (j=0; j<n; J + +) {  scanf ("%d", &map[i][j]);    }} prim (); return 0;}

Hihocoder Hiho The 26th week minimum spanning tree one · (Prim algorithm)