HDU.1233 还是畅通工程（Prim）

HDU.1233 还是畅通工程（Prim）

题意分析

1. 首先给出n，代表村庄的个数
2. 然后出n*(n-1)/2个信息，每个信息包括村庄的起点，终点，距离，
3. 要求求出最小生成树的权值之和。
4. 注意村庄的编号从1开始即可
5. 直接跑prim

代码总览

#include <bits/stdc++.h>
#define nmax 105
#define inf 1e8+7
using namespace std;
int mp[nmax][nmax];
int n;
int totaldis =0;
void prim()
{
    int lowcost[nmax];
    int path[nmax];
    for(int j = 1;j<=n;++j){
        lowcost[j] = mp[1][j];
        path[j] = 1;
    }
    path[1] = 0;
    int nowmin,nowminid;
    for(int i =2 ;i<=n;++i){
        nowmin = inf;
        nowminid = 0;
        for(int j = 2;j<=n;++j){
            if(nowmin > lowcost[j] && lowcost[j] != 0){
                nowmin = lowcost[j];
                nowminid = j;
            }
        }
        totaldis += nowmin;
        lowcost[nowminid] = 0;
        for(int j = 2;j<=n;++j){
            if(mp[nowminid][j] < lowcost[j]){
                lowcost[j] = mp[nowminid][j];
                path[j] = nowminid;
            }
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&n) != EOF && n){
        int sta,end,dis;
        totaldis = 0;
        for(int i = 1; i<=n;++i)
            for(int j = 1;j<=n;++j)
                mp[i][j] = inf;
        for(int i = 1; i<=n*(n-1)/2 ;++i){
            scanf("%d %d %d",&sta,&end,&dis);
            mp[sta][end] = mp[end][sta] = dis;
        }
        prim();
        printf("%d
",totaldis);
    }
    return 0;
}