POJ 2728 Desert King (最小比率生成树，二分/迭代）

 
题意：沙漠里的王国需要修建水渠，连接国都与村庄····。说白了求一棵树，每个点有三个坐标（x,y,z)。边的benifit为两点之间的距离，cost为两点的高度差。现在要求一棵树使得 cost / benift 最小。

题解：很显然任意两点之间都有边，所以是一个很稠密的图。用Prime。二分的话2800ms+, 迭代300ms+。

#include <cmath>
#include <iostream>
using namespace std;

#define MAX 1009
#define INF 99999999999
#define eps 1.0e-6  // 精度除以（0.001/2 )

struct NODE
{
int x, y, z;
} node[MAX];

double e[MAX][MAX];
double dis[MAX];
bool vis[MAX];

void build_map ( int n, double l )
{
for ( int i = 1; i <= n; i++ )
for ( int j = i + 1; j <= n; j++ )
{
double len = sqrt(0.0 + (node[i].x-node[j].x)

* (node[i].x-node[j].x) + (node[i].y-node[j].y) * (node[i].y-node

[j].y));
double cost = fabs(0.0+node[i].z - node[j].z);
e[i][j] = e[j][i] = cost - l * len;
}
}

double prime ( int n )
{
int i, j, k;
double minc, ans = 0.0;
for ( i = 1; i <= n; i++ )
{
dis[i] = e[1][i];
vis[i] = false;
}
dis[1] = 0; vis[1] = true;

for ( i = 2; i <= n; i++ )
{
minc = INF; k = -1;
for ( j = 1; j <= n; j++ )
{
if ( !vis[j] && minc > dis[j] )
{
minc = dis[j];
k = j;
}
}
if ( minc == INF ) break;
ans += minc;
vis[k] = true;
for ( j = 1; j <= n; j++ )
if ( ! vis[j] && dis[j] > e[k][j] )
dis[j] = e[k][j];
}
return ans;
}

int main()
{
int n;
while ( scanf("%d",&n) && n )
{
for ( int i = 1; i <= n; i++ )
scanf("%d%d%d",&node[i].x, &node[i].y, &node

[i].z);

double mid, temp;
double high = 100, low = 0.0;
while ( 1 )
{
mid = ( high + low ) / 2;
build_map ( n, mid ) ;
temp = prime ( n );
if ( fabs(temp) <= eps ) break;
if ( temp < 0.0 )
high = mid - eps;
else
low = mid + eps;
}
printf("%.3lf\n",mid);
}
return 0;
}

 

迭代法：

#include <cmath>
#include <iostream>
using namespace std;

#define MAX 1001
#define INF 99999999999
#define eps 1.0e-6

struct NODE
{
int x, y, z;
} node[MAX];

double edge[MAX][MAX];
double len[MAX][MAX];
double cost[MAX][MAX];
double dis[MAX];
bool vis[MAX];
int pre[MAX];

void build_map ( int n, double l )
{
for ( int i = 1; i <= n; i++ )
for ( int j = i + 1; j <= n; j++ )
{
len[i][j] = len[j][i] = sqrt(0.0 + (node[i].x-node[j].x) * (node[i].x-node[j].x) + (node[i].y-node[j].y) * (node[i].y-node[j].y));
cost[i][j] = cost[j][i] = fabs ( 0.0 + node[i].z - node[j].z);
edge[i][j] = edge[j][i] = cost[i][j] - l * len[i][j];
}
}

double prime ( int n )  // 注意，与普通prime有区别
{
int i, j, k;
double minc, b = 0, c = 0;
for ( i = 1; i <= n; i++ )
{
dis[i] = edge[1][i];
pre[i] = 1;     // 所有点的前驱初始化为1
vis[i] = false;
}

dis[1] = 0;
vis[1] = true;
for ( i = 2; i <= n; i++ )
{
minc = INF; k = -1;
for ( j = 1; j <= n; j++ )
{
if ( ! vis[j] && minc > dis[j] )
{
minc = dis[j];
k = j;
}
}
if ( minc == INF ) break;
b += len[pre[k]][k];     //*********
c += cost[pre[k]][k];    //*********
vis[k] = true;
for ( j = 1; j <= n; j++ )
if ( ! vis[j] && dis[j] > edge[k][j] )
{
dis[j] = edge[k][j];
pre[j] = k;   //修改前驱
}
}
return c / b;
}

int main()
{
int n;
while ( scanf("%d",&n) && n )
{
for ( int i = 1; i <= n; i++ )
scanf("%d%d%d",&node[i].x, &node[i].y, &node[i].z);

double a = 0, b;
while ( 1 )
{
build_map ( n, a );
b = prime ( n );
if ( fabs(b-a) <= eps ) break;
a = b;
}
printf("%.3lf\n",b);
}
return 0;
}