poj 1944 Qin Shi Huang's National Road System(次小生成树）

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UVA - 1494点我点我:-)

题目的意思是：n个城市，需要修建一些道路使得任意两个城市联通，还可以修一条魔法道路，不花钱，设魔法路连接的城市的人口之和为A，所有道路总长为B，求A/B的最大值

也用到了D[i][j]数组。。。不难不难。。。

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>#include<vector>using namespace std;#define MAXN (1000+5)struct node{    int u, v, dis;        bool operator <(const node &x)const{        return dis < x.dis;    }};int X[MAXN], Y[MAXN], peo[MAXN], f[MAXN], D[MAXN][MAXN];vector<node> G[MAXN];node edges[MAXN*MAXN];bool vis[MAXN];void init(){    for(int i = 0; i < MAXN; i++) G[i].clear(), f[i] = i;    memset(D, 0, sizeof(D));}int calc(int x1, int y1, int x2, int y2){    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);}int find(int x){    return x == f[x]? x: (f[x] = find(f[x]));}void Bspfa(int s){memset(vis, 0, sizeof(vis));queue<int> q;q.push(s); D[s][s] = 0; vis[s] = true;while(!q.empty()){int now = q.front(); q.pop();for(int i = 0; i < G[now].size(); i++){node next = G[now][i];int v = G[now][i].v, dis = G[now][i].dis;if(vis[v]) continue;D[s][v] = max(D[s][now], dis);q.push(v);vis[v] = true;}}}int main(){    freopen("test.in", "r", stdin);    freopen("test.out", "w", stdout);        int T;    scanf("%d", &T);        while(T--){        init();                int n, m = 0;        scanf("%d", &n);        for(int i = 1; i <= n; i++){            scanf("%d%d%d", &X[i], &Y[i], &peo[i]);                        for(int j = 1; j < i; j++)                 edges[++m] = ((node){i, j, calc(X[i], Y[i], X[j], Y[j])});         }        sort(edges+1, edges+m+1);                 double sum = 0;        int tot = 0;;        for(int i = 1; i <= m; i++){            node now = edges[i];            int u = now. u, v = now.v, dis = now.dis;            int x = find(u), y = find(v);            if(x == y) continue;                        f[x] = y;            G[x].push_back((node){x, y, dis});            G[y].push_back((node){y, x, dis});            sum += sqrt((double)dis); tot++;                        if(tot >= n-1) break;         }                for(int i = 1; i <= n; i++) Bspfa(i);                double ans = 0;        for(int i = 1; i <= n; i++)          for(int j = i+1; j <= n; j++)             ans = max(ans, (peo[i]+peo[j])*1.0 / (sum-sqrt(D[i][j])));                  printf("%.2lf\n", ans);    }        return 0;}

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