HDU4081 Qin Shi Huang's National Road System 【次小生成树】

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3395    Accepted Submission(s): 1189

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

241 1 201 2 30200 2 80200 1 10031 1 201 2 302 2 40

 

Sample Output

65.0070.00

 

Source

2011 Asia Beijing Regional Contest 

题意：给定n个点的点权及相互间的边权，求一棵树，其中一条边的边权变为0，树的比率值为该0值边所连的两点的点权和/剩下的树边和，求这个值最大是多少。

题解：这题要用到次小生成树的思想，即找到最小生成树，然后添加一条边构成环，再删掉环中属于最小树的最大边，用这种方法遍历所有边以找到最终ans。

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>#define maxn 1002#define maxm (maxn * maxn) >> 1using std::sort;struct Node{int u, v;double dist;} E[maxm]; //存储边struct Node2{int x, y, peo;} V[maxn]; //存储顶点int head[maxn], end[maxn]; //存储集合首尾struct Node3{ int u, next;} G[maxn]; //链式前向星存集合归属信息int pre[maxn]; //并查集double max[maxn][maxn]; //存储最小树上的两点间最长的一条单元路double maxv(double a, double b){return a > b ? a : b;}double calDist(int i, int j){double x = V[i].x - V[j].x;double y = V[i].y - V[j].y;return sqrt(x * x + y * y);}bool cmp(Node a, Node b){return a.dist < b.dist;}int ufind(int k){int a = k, b;while(pre[k] != -1) k = pre[k];while(a != k){b = pre[a];pre[a] = k;a = b;}return k;}double Kruskal(int n, int m){memset(pre, -1, sizeof(pre));int count = n, i, j, k, u, v;double len = 0, dist;for(i = 0; i < n; ++i){ //初始化每个点的集合只有其本身G[i].u = i; G[i].next = -1;head[i] = end[i] = i;}for(i = 0; i < m; ++i){u = E[i].u; v = E[i].v;dist = E[i].dist;u = ufind(u); v = ufind(v);if(u != v){for(j = head[u]; j != -1; j = G[j].next)for(k = head[v]; k != -1; k = G[k].next)max[G[j].u][G[k].u] = max[G[k].u][G[j].u] = dist;//合并集合G[end[v]].next = head[u]; head[u] = head[v];pre[v] = u; --count; len += dist;if(1 == count) break; //最小树生成}}return len;}int main(){//freopen("stdin.txt", "r", stdin);int t, n, i, j, id;double minLen, ans;scanf("%d", &t);while(t--){scanf("%d", &n);for(i = 0; i < n; ++i)scanf("%d%d%d", &V[i].x, &V[i].y, &V[i].peo);for(i = id = 0; i < n; ++i)for(j = i + 1; j < n; ++j){E[id].u = i; E[id].v = j;E[id++].dist = calDist(i, j);}sort(E, E + id, cmp);minLen = Kruskal(n, id);ans = 0;for(i = 0; i < n; ++i) //枚举所有魔法边for(j = i + 1; j < n; ++j){ans = maxv(ans, (V[i].peo + V[j].peo) / (minLen - max[i][j]));}printf("%.2lf\n", ans);}return 0;}

附之前写的一个用prim的超时代码：

#include <stdio.h>#include <string.h>#include <math.h>#include <limits.h>#define maxn 1002double map[maxn][maxn], max[maxn][maxn];bool vis[maxn];struct Node{    int x, y, peo;} node[maxn];double calDist(int i, int j){    int x = node[i].x - node[j].x;    int y = node[i].y - node[j].y;    return sqrt((double)(x * x + y * y));}double prim(int n){    memset(vis, 0, sizeof(vis));    int count = 0, i, j, v;        double tmp, len = 0;    vis[0] = 1;    while(count < n - 1){        for(i = 0, tmp = INT_MAX; i < n; ++i){            if(!vis[i]) continue;            for(j = 0; j < n; ++j){                if(vis[j]) continue;                if(map[i][j] < tmp){                    tmp = map[i][j]; v = j;                }            }        }        for(i = 0; i < n; ++i)        if(vis[i]) max[i][v] = max[v][i] = tmp;        ++count; vis[v] = 1; len += tmp;    }    return len;}double getAns(int n, double minLen){int i, j, peo;double ans = 0, tmp;for(i = 0; i < n; ++i)for(j = i + 1; j < n; ++j){tmp = (node[i].peo + node[j].peo) / (minLen - max[i][j]);if(tmp > ans) ans = tmp;}return ans;}int main(){    int t, n, i, j, peo;    scanf("%d", &t);    while(t--){        scanf("%d", &n);        for(i = 0; i < n; ++i)            scanf("%d%d%d", &node[i].x, &node[i].y, &node[i].peo);        for(i = 0; i < n; ++i)        for(j = i + 1; j < n; ++j)            map[i][j] = map[j][i] = calDist(i, j);                printf("%.2lf\n", getAns(n, prim(n)));    }    return 0;}