2011 regional Beijing Site A —— Qin Shi Huang's National Road System

题目：DescriptionDuring the Warring States Period of ancient China(476 BC to 221 BC), there were sevenkingdoms in China ---- they were Qi, Chu, Yan, Han, Zhao, Wei and Qin. Ying Zheng wasthe king of the kingdom Qin. Through 9 years of wars, he finally conquered all six otherkingdoms 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 lastdynasty 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, thenow famous city-sized mausoleum guarded by alife-sized Terracotta Army, and a massive national roadsystem. There is a story about the road system:There were n cities in China and Qin Shi Huangwanted them all be connected by n-1 roads, in order thathe could go to every city from the capital city Xianyang.Although Qin Shi Huang was a tyrant, he wanted thetotal length of all roads to be minimum,so that the roadsystem may not cost too many people's life. A daoshi(some kind of monk) named Xu Fu told Qin Shi Huangthat he could build a road by magic and that magic roadwould cost no money and no labor. But Xu Fu couldonly build ONE magic road for Qin Shi Huang. So QinShi Huang had to decide where to build the magic road. Qin Shi Huang wanted the totallength of all none magic roads to be as small as possible, but Xu Fu wanted the magic road tobenefit as many people as possible ---- So Qin Shi Huang decided that the value of A/B (theratio of A to B) must be the maximum, which A is the total population of the two citesconnected 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 segmentconnecting two points.InputThe 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).2Then 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.OutputFor each test case, print a line indicating the above mentioned maximum ratio A/B. Theresult should be rounded to 2 digits after decimal point.Sample Input2411 201 2 30200 2 80200 1 100311 201 2 302 2 40Sample Output65.0070.00

就是给你N个城市，完全图，求一个比率，两个的城市人口/其他城市相连的代价 ，求这个比率最大。要建立N-1条路。

所以就是相当于，求一棵最小生成树，然后替换掉其中一条边枚举求最大比率。

用到替换边肯定是替换从u->v 路上最大的边，所以先BFS或者DFS求出最小生成树里任意两点最大的边，然后替换掉求比率即可。

用kruskal稍不留神就会TLE，我过不去 = =。。用Prim过的。中间需要记录哪一条是生成树的边。

#include <set>#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <string>#include <algorithm>#define MID(x,y) ( ( x + y ) >> 1 )#define L(x) ( x << 1 )#define R(x) ( x << 1 | 1 )#define FOR(i,s,t) for(int i=(s); i<(t); i++)#define BUG puts("here!!!")#define STOP system("pause")#define file_r(x) freopen(x, "r", stdin)#define file_w(x) freopen(x, "w", stdout)using namespace std;const int MAX = 1010;struct point{int x, y;};struct TNODE{int v;double len;TNODE *next;};TNODE *t[MAX], tnode[MAX*2];double medge[MAX][MAX];point p[MAX];int val[MAX];int cou;double map[MAX][MAX];pair<int,int> pii[MAX];double disp2p(point &a, point &b){return sqrt((a.x - b.x)*(a.x - b.x)*1.0 + (a.y - b.y)*(a.y - b.y));}void Add(int u, int v, double l){tnode[cou].v = v;tnode[cou].len = l;tnode[cou].next = t[u];t[u] = &tnode[cou++];}double Prim(int n){    int used[MAX], i, j, now;double sum = 0, dis[MAX], min;          memset(used, 0, sizeof(used));    fill(dis, dis+MAX, 1e20);        now = 0; dis[now] = 0; used[now] = 1;    memset(t, 0, sizeof(t));    for(i=1; i<n; i++)      {          for(j=0; j<n; j++)              if( !used[j] && dis[j] > map[now][j] ){                  dis[j] = map[now][j];pii[j] = make_pair(now, j);//记录最小边的两个点 }          min = 1e20;        for(j=0; j<n; j++)              if( !used[j] && dis[j] < min )min = dis[now = j];         sum += min;        used[now] = 1;                int u = pii[now].first;// 构造邻接表的生成树，为下面搜索做准备         int v = pii[now].second;        Add(u, v, min);Add(v, u, min);}      return sum; }bool used[MAX];int q[MAX];void BFS(int x)// 查找一个点在树上到达另一个点的最大边，存入medge数组 {memset(used, 0, sizeof(used));int top = 0, end = 1;q[0] = x;used[x] = true;while( top < end ){int now = q[top++];TNODE *head = t[now];while( head ){int v = head->v;if( !used[v] ){used[v] = true;q[end++] = v;double len = head->len;medge[x][v] = max(len, medge[x][now]);}head = head->next;}}}double solve( int n ){FOR(i, 0, n){map[i][i] = 0;FOR(k, i+1, n){double l = disp2p(p[i], p[k]);map[i][k] = map[k][i] = l;}}double sum = Prim( n );FOR(i, 0, n)FOR(k, 0, n)medge[i][k] = 0;FOR(i, 0, n)BFS(i);double ratio = 0;FOR(i, 0, n)FOR(k, i+1, n){int sumval = val[i] + val[k];ratio = max(ratio, sumval/(sum - medge[i][k]));}return ratio;}int main(){int ncases, n;scanf("%d", &ncases);while( ncases-- ){scanf("%d", &n);cou = 0;FOR(i, 0, n)scanf("%d%d%d", &p[i].x, &p[i].y, &val[i]);double ans = solve( n );printf("%.2lf\n", ans);}return 0;}