HDU 1863 畅通工程

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include<set>
#include<stack>
#define bug puts("bugbugbug");
using namespace std;
typedef long long ll;
//适合稠密图 复杂度n*n n为点数
const int INF=1000000010;
const int N=105;//N 为点的数量
int vis
,dis
;
vector<int> V
;//邻接表存边
int g

={0};//邻接矩阵存边
int n,m;
int prim() {
    memset(vis, 0,sizeof(vis));
    for (int i = 1; i <= n; i++) dis[i] = i==1? 0 : INF;//初始化
    int ans = 0;
    int i;
    for (i = 1; i <= n; i++) {
        int k = -1, minn = INF;
        for (int j = 1; j <= n; j++)
            if (!vis[j] && dis[j] < minn) {
            minn = dis[j];
            k = j;
        }
        if (k == -1) break;
        vis[k] = 1, ans += dis[k];
        for (int j = 0; j < V[k].size(); j++) {
            int u = V[k][j];
            if (!vis[u] && g[k][u] < dis[u]) dis[u] = g[k][u];
        }
    }
    if(i!=n+1)return -1;
    return ans;
}
int main()
{
    while(~scanf("%d%d",&m,&n)&&m)
    {
        for(int i=0;i<=n;i++)
        V[i].clear();
        memset(g,0,sizeof(g));
        for(int i=0;i<m;i++){
                int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
        V[a].push_back(b);
         V[b].push_back(a);
         g[a][b]=g[b][a]=c;
        }
             int ans=prim();
             if(ans==-1) puts("?");
             else printf("%d\n",ans);
    }
}

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include<set>
#include<stack>
#define bug puts("bugbugbug");
using namespace std;
typedef long long ll;
const int maxm=10005;
int f[105];
int n,m;
struct node
{
    int x, y, v;
    bool operator<(const node &b)const{
        if(this->v!=b.v)
    return (this->v<b.v);
    if(this->x!=b.x)
    return (this->x<b.x);
    return (this->y<b.y);
    }
} e[maxm];
int find(int x){ return f[x]= f[x]==x?x:find(f[x]); }
int Kruscal()
{
    sort(e, e + m);
    int cnt = n, ans = 0;
    for(int i = 0; i < n; i++) f[i] = i;
    for (int i = 0; i < m; i++)
    {
        int x = find(e[i].x);
        int y = find(e[i].y);
        if (x != y)
        {
            f[x] = y;
            cnt--;
            ans += e[i].v;
            if (cnt == 1) break;
        }
    }
    if(cnt!=1)return -1;
    return ans;
}
int main()
{
    while(~scanf("%d%d",&m,&n)&&m)
    {
        for(int i=0;i<m;i++)
            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v);
             int ans=Kruscal();
             if(ans==-1) puts("?");
             else printf("%d\n",ans);
    }
}