HDU 4081 Qin Shi Huang's National Road System 最小生成树+倍增求LCA

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=4081

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5428    Accepted Submission(s): 1902

Problem Description

During the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in China ---- they were Qi, Chu, Yan, Han, Zhao, Wei and Qin. Ying Zheng was the king of the kingdom Qin. Through 9 years of wars, he finally conquered all six other kingdoms and became the first emperor of a unified China in 221 BC. That was Qin dynasty ---- the first imperial dynasty of China(not to be confused with the Qing Dynasty, the last dynasty of China). So Ying Zheng named himself "Qin Shi Huang" because "Shi Huang" means "the first emperor" in Chinese.

Qin Shi Huang undertook gigantic projects, including the first version of the Great Wall of China, the now famous city-sized mausoleum guarded by a life-sized Terracotta Army, and a massive national road system. There is a story about the road system:
There were n cities in China and Qin Shi Huang wanted them all be connected by n-1 roads, in order that he could go to every city from the capital city Xianyang.
Although Qin Shi Huang was a tyrant, he wanted the total length of all roads to be minimum,so that the road system may not cost too many people's life. A daoshi (some kind of monk) named Xu Fu told Qin Shi Huang that he could build a road by magic and that magic road would cost no money and no labor. But Xu Fu could only build ONE magic road for Qin Shi Huang. So Qin Shi Huang had to decide where to build the magic road. Qin Shi Huang wanted the total length of all none magic roads to be as small as possible, but Xu Fu wanted the magic road to benefit as many people as possible ---- So Qin Shi Huang decided that the value of A/B (the ratio of A to B) must be the maximum, which A is the total population of the two cites connected by the magic road, and B is the total length of none magic roads.
Would you help Qin Shi Huang?
A city can be considered as a point, and a road can be considered as a line segment connecting two points.

 

Input

The first line contains an integer t meaning that there are t test cases(t <= 10).
For each test case:
The first line is an integer n meaning that there are n cities(2 < n <= 1000).
Then n lines follow. Each line contains three integers X, Y and P ( 0 <= X, Y <= 1000, 0 < P < 100000). (X, Y) is the coordinate of a city and P is the population of that city.
It is guaranteed that each city has a distinct location.

 

Output

For each test case, print a line indicating the above mentioned maximum ratio A/B. The result should be rounded to 2 digits after decimal point.

 

Sample Input

2 4 1 1 20 1 2 30 200 2 80 200 1 100 3 1 1 20 1 2 30 2 2 40

 

Sample Output

65.00 70.00

 

Source

2011 Asia Beijing Regional Contest

 

Recommend

lcy

 题意：

给你平面上若干的点，每个点有个权值$a_i$，现在用这些点构成一棵树。存在一种操作，可以使得这棵树上的某条边的长度变为0，令这条边连接的两个点分别是$i,j$，这棵树的边权和为$B$，问你$(a_i+a_j)/B$的最小值。

题解：

看网上的做法都是求次小生成树，而我却不是这样做的。

首先使用Kruskal求出最小生成树，而后枚举两个点，将这两个点之间连接那条边，这样就构成了一个换，为了保持树的形态，必须删掉这两个点的唯一路径上最长的边。

找到这条边的方法是通过倍增来求解。

令$ancestor[u][i]$为节点$u$向上走$2^i$所能到达的节点，$maxEdge[u][i]$表示节点$u$向上走$2^i$所能碰到的最长的边，那么有以下转移：

$$ancestor[u][i]=ancestor[ancestor[u][i-1]][i-1]$$

$$maxEdge[u][i]=max(maxEdge[u][i-1],maxEdge[ancestor[u][i-1]][i-1])$$

预处理出这两个数组后，就能用类似LCA的思想求最长边了。

代码：

#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#define MAX_N 1234
#define MAX_D 25
using namespace std;

int father[MAX_N];
int n;

int popu[MAX_N];

struct edge {
    int to;
    double cost;

    edge(int t, double c) : to(t), cost(c) { }

    edge() { }
};

struct road {
    int from, to;
    double cost;

    road(int f, int t, double c) : from(f), to(t), cost(c) { }

    road() { }
};

bool cmp(road a,road b) {
    if (a.cost == b.cost)
        return popu[a.from] + popu[a.to] > popu[b.from] + popu[b.to];
    return a.cost < b.cost;
}

road ro[MAX_N*MAX_N];
int tot=0;
vector<edge> G[MAX_N];
double B;

int depth[MAX_N];
int ancestor[MAX_N][25];
double maxEdge[MAX_N][25];

struct Point {
    double x, y;

    Point(double xx, double yy) : x(xx), y(yy) { }

    Point() { }

    double dis(Point a) {
        return sqrt((x - a.x) * (x - a.x) + (y - a.y) * (y - a.y));
    }
};

Point po[MAX_N];

void init() {
    for (int i = 0; i <= n; i++)
        father[i] = i;
    for (int i = 0; i <= n; i++)G[i].clear();
    tot = 0;
    B = 0;
    memset(depth, 0, sizeof(depth));
    memset(ancestor, 0, sizeof(ancestor));
    memset(maxEdge, 0, sizeof(maxEdge));
}

int Find(int x){
    if(x==father[x])return x;
    return father[x]=Find(father[x]);
}

void unionSet(int x,int y) {
    int u = Find(x), v = Find(y);
    if (u == v)return;
    father[u] = v;
}

bool Same(int x,int y){
    return Find(x)==Find(y);
}

void Kruskal() {
    sort(ro + 1, ro + tot + 1, cmp);
    for (int i = 1; i <= tot; i++) {
        int u = ro[i].from, v = ro[i].to;
        if (Same(u, v))continue;
        unionSet(u, v);
        G[u].push_back(edge(v,ro[i].cost));
        G[v].push_back(edge(u,ro[i].cost));
        B += ro[i].cost;
    }
}

void dfs(int u,int p) {
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i].to;
        double c = G[u][i].cost;
        if (v == p)continue;
        depth[v] = depth[u] + 1;
        ancestor[v][0] = u;
        maxEdge[v][0] = c;
        dfs(v, u);
    }
}

void getAncestor() {
    for (int j = 1; j < MAX_D; j++) {
        for (int i = 1; i <= n; i++) {
            ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];
            maxEdge[i][j] = max(maxEdge[i][j - 1], maxEdge[ancestor[i][j - 1]][j - 1]);
        }
    }
}

double LCA(int u,int v) {
    double res = -1;
    if (depth[u] < depth[v])swap(u, v);
    for (int i = MAX_D - 1; i >= 0; i--) {
        if (depth[ancestor[u][i]] >= depth[v]) {
            res = max(res, maxEdge[u][i]);
            u = ancestor[u][i];
            if (depth[u] == depth[v])break;
        }
    }
    if (u == v)return res;
    for (int i = MAX_D - 1; i >= 0; i--) {
        if (ancestor[u][i] != ancestor[v][i]) {
            res = max(res, max(maxEdge[u][i], maxEdge[v][i]));
            u = ancestor[u][i];
            v = ancestor[v][i];
        }
    }
    return max(res, max(maxEdge[u][0], maxEdge[v][0]));
}

int T;

int main() {
    cin.sync_with_stdio(false);
    cin >> T;
    while (T--) {
        cin >> n;
        init();
        for (int i = 1; i <= n; i++)
            cin >> po[i].x >> po[i].y >> popu[i];
        for (int i = 1; i <= n; i++)
            for (int j = i + 1; j <= n; j++)
                ro[++tot] = road(i, j, po[i].dis(po[j]));
        Kruskal();
        depth[1] = 1;
        dfs(1, 0);
        getAncestor();
        double ans = -1;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                ans = max(ans, (popu[i] + popu[j]) / (B - LCA(i, j)));
        cout << setprecision(2) << fixed << ans << endl;
    }
    return 0;
}

PS：第一次写手写倍增，居然一发就AC了