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+7using 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\n",totaldis);    }    return 0;}