POJ 2728 Desert King (最优比率树)
Posted dwtfukgv
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 2728 Desert King (最优比率树)相关的知识,希望对你有一定的参考价值。
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; }
以上是关于POJ 2728 Desert King (最优比率树)的主要内容,如果未能解决你的问题,请参考以下文章
POJ 2728 Desert King(最优比率生成树 01分数规划)
(最优比率生成树)POJ 2728 - Desert King