POJ 2728 Desert King (最优比率树)

2023-07-27 21:11:34

题意：有n个村庄，村庄在不同坐标和海拔，现在要对所有村庄供水，只要两个村庄之间有一条路即可，建造水管距离为坐标之间的欧几里德距离，费用为海拔之差，现在要求方案使得费用与距离的比值最小，很显然，这个题目是要求一棵最优比率生成树。

析：也就是求 r = sigma(x[i] * d) / sigma(x[i] * dist)这个值最小，变形一下就可以得到 d * r - dist <= 0，当r 最小时，取到等号，也就是求最大生成树，然后进行判断，有两种方法，一种是二分，这个题时间长一点，另一种是迭代，这个比较快。

代码如下：

二分：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-4;

const int maxn = 1000 + 10;

const int maxm = 1e5 + 10;

const int mod = 50007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Point{

  int x, y, d;

};

Point a[maxn];

double dist[maxn][maxn];

double dis[maxn];

bool vis[maxn];

bool judge(double mid){

  ms(vis, 0);

  for(int i = 1; i <= n; ++i)  dis[i] = -inf;

  dis[1] = 0;

  double ans = 0;

  for(int i = 1; i <= n; ++i){

    int mark = -1;

    for(int j = 1; j <= n; ++j)  if(!vis[j]){

      if(mark == -1)  mark = j;

      else if(dis[j] > dis[mark])  mark = j;

    }

    if(mark == -1)  break;

    vis[mark] = 1;

    ans += dis[mark];

    for(int j = 1; j <= n; ++j)  if(!vis[j]){

      dis[j] = max(dis[j], dist[j][mark] * mid - abs(a[mark].d - a[j].d));

    }

  }

  return ans >= 0.0;

}

int main(){

  while(scanf("%d", &n) == 1 && n){

    for(int i = 1; i <= n; ++i){

      scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].d);

      for(int j = 1; j < i; ++j)

        dist[i][j] = dist[j][i] = sqrt(sqr(a[i].x*1.-a[j].x) + sqr(a[i].y*1.-a[j].y));

    }

    double l = 0.0, r = 1e6;

    while(r - l > eps){

      double m = (l + r) / 2.0;

      if(judge(m))  r = m;

      else l = m;

    }

    printf("%.3f\n", l);

  }

  return 0;

}

　迭代：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-4;

const int maxn = 1000 + 10;

const int maxm = 1e5 + 10;

const int mod = 50007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Point{

  int x, y, d;

};

Point a[maxn];

double dist[maxn][maxn];

double dis[maxn];

double A[maxn], B[maxn];

bool vis[maxn];

double judge(double mid){

  ms(vis, 0);

  for(int i = 1; i <= n; ++i)  dis[i] = -inf;

  dis[1] = 0;  A[1] = B[1] = 0;

  double ans = 0, aa = 0, bb = 0;

  for(int i = 1; i <= n; ++i){

    int mark = -1;

    for(int j = 1; j <= n; ++j)  if(!vis[j]){

      if(mark == -1)  mark = j;

      else if(dis[j] > dis[mark])  mark = j;

    }

    if(mark == -1)  break;

    vis[mark] = 1;

    ans += dis[mark];

    aa += A[mark];

    bb += B[mark];

    for(int j = 1; j <= n; ++j)  if(!vis[j]){

      if(dis[j] < dist[j][mark] * mid - abs(a[mark].d - a[j].d)){

        dis[j] = dist[j][mark] * mid - abs(a[mark].d - a[j].d);

        A[j] = dist[j][mark];

        B[j] = abs(a[mark].d - a[j].d);

      }

    }

  }

  return bb / aa;

}

int main(){

  while(scanf("%d", &n) == 1 && n){

    for(int i = 1; i <= n; ++i){

      scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].d);

      for(int j = 1; j < i; ++j)

        dist[i][j] = dist[j][i] = sqrt(sqr(a[i].x*1.-a[j].x) + sqr(a[i].y*1.-a[j].y));

    }

    double a = 0.0;

    while(1){

      double b = judge(a);

      if(fabs(b - a) < eps){

        printf("%.3f\n", a);

        break;

      }

      a = b;

    }

  }

  return 0;

}

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