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  • POJ 2728 Desert King (最优比率树)

    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
    ", 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
    ", a);
            break;
          }
          a = b;
        }
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7745597.html
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