08-图7 公路村村通

思路：该题为最小生成树，利用Prime算法每次找到离最小生成树最近的没有被访问过的点加入最小生成树，直至所有点都加入了最小生成树。cost数组记录了点到最小生成树的距离，最开始是其他点到起点的距离，每次选择一个加入一个点后，检查没有加入的点的到新加入的点距离是否比之前到最小生成树的距离短，若小则更新。#include<iostream>
using namespace std;
#define MaxNum 1001
#define INF 999999
int N, M,sumCost=0;
int map[MaxNum][MaxNum],cost[MaxNum];
int findmin() {//找到与最小生成树最近的点
int mincost = INF;
int k, j;
for (k = 0, j = 1; j <= N; j++) {
if (cost[j] && cost[j] < mincost) {
mincost = cost[j];
k = j;
}
}
return k;
}
int Prime() {
for (int i = 1; i <= N; i++) {
cost[i] = map[1][i];
}
cost[1] = 0;
for (int i = 1; i < N; i++) {//还需要找到N-1个点
int newnode = findmin();
if (newnode) {
sumCost += cost[newnode];
cost[newnode] = 0;
for (int j = 2; j <= N; j++) {//更新剩余所有结点到最小生成树的距离
if (cost[j] && map[newnode][j] < cost[j]) {
cost[j] = map[newnode][j];
}
}

}else{
return -1;
}
}
return sumCost;
}
void init() {
for (int i = 1; i <= N; i++) {
cost[i] = 0;
for (int j = 1; j <= N; j++) {
if (i == j) {
map[i][j] = 0;
}
else {
map[i][j] = INF;
}
}
}
for (int k = 0; k < M; k++) {
int i, j, cost;
cin >> i >> j >> cost;
map[i][j] = map[j][i] = cost;
}

}
int main()
{
cin >> N >> M;
init();
cout<< Prime()<<endl;
return 0;
}