Qin Shi Huang's National Road System（最小生成树 + LCA）

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3750    Accepted Submission(s): 1305

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：

       给出 T 组样例，每个样例都有 N 个点，后给出 N 个点的坐标和值。要求从中选出 n - 1 条路，选中其中一条为魔法路，这时候的 A = 魔法路所连接两个点的总和值， B = n - 1 条路的总和长度值 - 魔法路的长度值。求 A / B 最大。

 

       思路：

       最小生成树 + LCA。因为要求 A / B 最大，相当于求 B 最小，即求最小生成树。求完之后并没有结束，因为此刻还未能确定答案。之后开始枚举任意两个点，使这两个点连边（相当于在枚举魔法边），如果这两个点连接的边并不是生成树上的边，添加之后则必定会生成环，这时候要删除这个环上权值最大的那条边以保证 B 最小（B = min_tree - max_val）；如果已经是生成树上的边，则直接删除这条边即可（B = min_tree - G [ i ] [ j ] ）。找环则通过 LCA 寻找路径来找，边往上边比较最大值即可。注意数据类型！！！！因为保存边的时候一不小心用了 int 所以导致无限 WA。

 

        AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int VMAX = 1010;
const int EMAX = 1000010;

typedef struct {
    int root;
    double len;
} fath;

typedef struct {
    double x, y, num;
} node;

typedef struct {
    double len;
    int a, b;
} Map;

int G_ind, n;
double Min_sum;
node no[VMAX];
Map G[EMAX];
double G1[VMAX][VMAX];

int root[VMAX];

int ind;
int fir[VMAX], next[EMAX], v[EMAX];
double w[EMAX];

int cnt;
bool vis[VMAX];
int vs[VMAX * 2], dep[VMAX * 2], id[VMAX];
fath fa[VMAX];
int dp[VMAX * 2][25];

bool cmp (Map a, Map b) { return a.len < b.len; }

void init () {
    G_ind = ind = Min_sum = cnt = 0;
    for (int i = 1; i <= n; ++i) {
        fa[i].root = root[i] = i;
    }
    memset(fir, -1, sizeof(fir));
    memset(G1, 0, sizeof(G1));
    memset(vis, 0, sizeof(vis));
}

void add_edge (int f, int t, double val) {
    v[ind] = t;
    w[ind] = val;
    next[ind] = fir[f];
    fir[f] = ind;
    ++ind;
}

int Find (int x) {
    return root[x] == x ? x : root[x] = Find(root[x]);
}

void Min_tree () {
    sort(G, G + G_ind, cmp);

    int ans = 0;
    for (int i = 0; i < G_ind; ++i) {
        if (ans == n - 1) break;

        int a = Find(G[i].a);
        int b = Find(G[i].b);
        if (a != b) {
            ++ans;
            Min_sum += G[i].len;
            root[a] = b;
            add_edge(G[i].a, G[i].b, G[i].len);
            add_edge(G[i].b, G[i].a, G[i].len);
            G1[G[i].a][G[i].b] = G[i].len;
            G1[G[i].b][G[i].a] = G[i].len;
        } else continue;
    }

}

void dfs (int x, int d) {
    id[x] = cnt;
    dep[cnt] = d;
    vs[cnt++] = x;
    vis[x] = 1;

    for (int e = fir[x]; e != -1; e = next[e]) {
        int V = v[e];
        if (!vis[V]) {
            fa[V].root = x;
            fa[V].len = w[e];
            dfs (V, d + 1);
            vs[cnt] = x;
            dep[cnt++] = d;
        }
    }
}

void RMQ_init () {
    for (int i = 0; i < cnt; ++i) dp[i][0] = i;

    for (int j = 1; (1 << j) <= cnt; ++j) {
        for (int i = 0; i + (1 << j) < cnt; ++i) {
            int a = dp[i][j - 1];
            int b = dp[i + (1 << (j - 1))][j - 1];

            if (dep[a] < dep[b]) dp[i][j] = a;
            else dp[i][j] = b;
        }
    }
}

int RMQ (int L, int R) {
    int len = 0;
    while ((1 << (1 + len)) <= R - L + 1) ++len;

    int a = dp[L][len];
    int b = dp[R - (1 << len) + 1][len];

    if (dep[a] < dep[b]) return a;
    return b;
}

int LCA (int a, int b) {
    int L = min(id[a], id[b]);
    int R = max(id[a], id[b]);

    int node = RMQ(L, R);
    return vs[node];
}

double Find_max_road (int a, int b) {
    int lca = LCA(a, b);

    double Max = 0;
    while (a != lca) {
        Max = max(Max, fa[a].len);
        a = fa[a].root;
    }

    while (b != lca) {
        Max = max(Max, fa[b].len);
        b = fa[b].root;
    }

    return Max;
}

int main() {

    int t;
    scanf("%d", &t);

    while (t--) {
        scanf("%d", &n);

        init();

        for (int i = 1; i <= n; ++i) {
            scanf("%lf%lf%lf", &no[i].x, &no[i].y, &no[i].num);
        }

        for (int i = 1; i <= n - 1; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                double x = (double)no[i].x - (double)no[j].x;
                double y = (double)no[i].y - (double)no[j].y;
                double len = sqrt(x * x + y * y);
                G[G_ind].a = i;
                G[G_ind].b = j;
                G[G_ind++].len = len;
            }
        }

        Min_tree();
        dfs (1, 1);
        RMQ_init();

        double ans = 0;
        for (int i = 1; i <= n - 1; ++i) {
            for (int j = i + 1; j <= n; ++j) {

                double Max;

                if (G1[i][j]) Max = G1[i][j];
                else Max = Find_max_road(i , j);

                double ans1 = Min_sum - Max;
                ans1 = (no[i].num + no[j].num) / ans1;

                ans = max(ans, ans1);
            }
        }

        printf("%.2lf\n", ans);

    }

    return 0;
}