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Language: Default Desert King
Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel As King David‘s prime scientist and programmer, you are asked to find out the best solution to build the channels. Input There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the Output For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point. Sample Input 4 0 0 0 0 1 1 1 1 2 1 0 3 0 Sample Output 1.000 Source |
题意：将n个村庄连在一起，告诉每个村庄的三维坐标，村庄之间的距离为水平方向上的距离，花费为垂直方向上的高度差，求把村庄连接起来的最小的花费与长度之比为多少。
思路：经典的01分数规划问题，参考这位大神的讲解应该就能明白了：http://www.cnblogs.com/Fatedayt/archive/2012/03/05/2380888.html
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;
const int N = 1005;
double x[maxn],y[maxn],z[maxn];
double dist[maxn],mp[maxn][maxn],len[maxn][maxn],cost[maxn][maxn];
bool vis[maxn];
int pre[maxn];
int n;
double Dis(int i,int j)
{
return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
double prim(double r)
{
int i,j,now;
double mi,c=0,l=0;
for (i=0;i<n;i++)
{
dist[i]=INF;
for (j=0;j<n;j++)
{
mp[i][j]=cost[i][j]-r*len[i][j];
}
}
for (i=0;i<n;i++)
{
dist[i]=mp[i][0];
pre[i]=0;
vis[i]=false;
}
dist[0]=0;
vis[0]=true;
for (i=1;i<n;i++)
{
mi=INF;now=-1;
for (j=0;j<n;j++)
{
if (!vis[j]&&mi>dist[j])
{
mi=dist[j];
now=j;
}
}
if (now==-1) break;
vis[now]=true;
c+=cost[pre[now]][now];
l+=len[pre[now]][now];
for (j=0;j<n;j++)
{
if (!vis[j]&&dist[j]>mp[now][j])
{
dist[j]=mp[now][j];
pre[j]=now;
}
}
}
return c/l;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j;
while (sf(n))
{
if (n==0) break;
for (i=0;i<n;i++)
scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
for (i=0;i<n;i++)
{
for (j=0;j<n;j++)
{
len[i][j]=Dis(i,j);
cost[i][j]=fabs(z[i]-z[j]);
}
}
double r=0,rate; //r迭代初值为0
while (1)
{
rate=r;
r=prim(r);
if (fabs(r-rate)<eps) break;
}
printf("%.3f\n",r);
}
return 0;
}
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