POJ 1258 AgriNet (最小生成树)

题目链接：https://vjudge.net/problem/POJ-1258

题目大意：有一个计算机网络的所有线路都坏了，网络中有n台计算机，现在你可以做两种操作，修理（O）和检测两台计算机是否连通（S），只有修理好的计算机才能连通。连通有个规则，两台计算机的距离不能超过给定的最大距离D（一开始会给你n台计算机的坐标）。检测的时候输出两台计算机是否能连通。

最小生成树裸题

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define zero(a) memset(a, 0, sizeof(a))
#define INF(a) fill(a, a+maxn, INF);
#define IOS ios::sync_with_stdio(false)
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<double, int> P2;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD =  1000000007LL;
const int INF = 0x3f3f3f3f;
const int _NAN = -0x3f3f3f3f;
const double EULC = 0.5772156649015328;
const int NIL = -1;
template<typename T> void read(T &x){
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxn = 1e2+10;
int p[maxn], vis[maxn];
vector<P> dis[maxn];
void prim(int n) {
    priority_queue<P> pq;
    zero(vis);
    int sum = 0;
    pq.push(make_pair(0, 0));
    while(!pq.empty()) {
        P t = pq.top();
        pq.pop();
        if (vis[t.second])
            continue;
        sum += -1*t.first;
        vis[t.second] = true;
        for (int i = 0; i<(int)dis[t.second].size(); ++i)
            if (!vis[dis[t.second][i].second])
                pq.push(make_pair(-1*dis[t.second][i].first, dis[t.second][i].second));
    }
    printf("%d\n", sum);
} 
int main(void) {
    int n;
    while(~scanf("%d", &n)) {
        for (int i = 0; i<n; ++i)
            for (int j = 0, d; j<n; ++j) {
                scanf("%d", &d);
                if (i != j)
                    dis[i].push_back(make_pair(d, j));
            }
        prim(n);
        for (int i = 0; i<maxn; ++i)
            dis[i].clear();
    }
    return 0;
}