Desert King（01分数规划问题）（最优斜率生成树）

Desert King

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions:33847		Accepted: 9208

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 ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.

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 channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.

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 between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can't share a lifter. Channels can intersect safely and no three villages are on the same line.

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 position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.

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

题意

有带权图G, 对于图中每条边e[i], 都有benifit[i](收入)和cost[i](花费), 我们要求的是一棵生成树T, 它使得 ∑(benifit[i]) / ∑(cost[i]), i∈T 最大(或最小).

这显然是一个具有现实意义的问题.

题解

解法之一 0-1分数规划

设x[i]等于1或0, 表示边e[i]是否属于生成树.

则我们所求的比率 r = ∑(benifit[i] * x[i]) / ∑(cost[i] * x[i]), 0≤i<m .

为了使 r 最大, 设计一个子问题---> 让 z = ∑(benifit[i] * x[i]) - l * ∑(cost[i] * x[i]) = ∑(d[i] * x[i]) 最大 (d[i] = benifit[i] - l * cost[i]) , 并记为z(l). 我们可以兴高采烈地把z(l)看做以d为边权的最大生成树的总权值.

然后明确两个性质:

　1.  z单调递减

　　证明: 因为cost为正数, 所以z随l的减小而增大.

　2.  z( max(r) ) = 0

　　证明: 若z( max(r) ) < 0, ∑(benifit[i] * x[i]) - max(r) * ∑(cost[i] * x[i]) < 0, 可化为 max(r) < max(r). 矛盾;

　　        若z( max(r) ) >= 0, 根据性质1, 当z = 0 时r最大.

到了这个地步, 七窍全已打通, 喜欢二分的上二分, 喜欢Dinkelbach的就Dinkelbach.

复杂度

时间 O( O(MST) * log max(r) )

空间 O( O(MST) )

C++代码

二分法

/*
*@Author:   Agnel-Cynthia
*@Language: C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=+;
const int M=+;
const int MOD=1e9+;
const double PI = acos(-1.0);
const double EXP = 1E-;
const int INF = 0x3f3f3f3f;
//int t,n,m,k,p,l,r,u,v;
const int maxn = ;
//ll a[maxn],b[maxn];

struct node
{
    int x , y ,z ;
}edge[maxn];

int n ;

double mp[maxn][maxn];

double dis(double x1 ,double y1,double x2,double y2){
    return sqrt(1.0*(x1-x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2));
}

void creat(){
    for(int i = ;i <= n ;i ++){
        for(int j = ;j <= n ; j++){
            mp[i][j] = dis(edge[i].x,edge[i].y,edge[j].x,edge[j].y);
        }
    }
}

double d[maxn];
bool vis[maxn];

double prime(double mid){
    memset(vis,,sizeof vis);
    for(int i = ;i <= n ; i++){
        d[i] = abs(edge[].z - edge[i].z) - mp[][i] * mid;
    }
    vis[] = true;
    double ans = ;
    for(int i = ;i < n ; i++){
        int v = -;double MIN = INF;
        for(int j = ;j <= n ; j++){
            if(MIN >= d[j] && !vis[j]){
                v = j;
                MIN = d[j];
            }
        }
        if(v == -)
            break;
        vis[v] = true;
        ans += MIN;
        for(int j = ;j <= n ; j++){
            if(!vis[j] && (fabs(edge[v].z - edge[j].z) - mp[v][j] * mid) < d[j])
                d[j] = (fabs(edge[v].z - edge[j].z) - mp[v][j] * mid);
        }
    }
    return ans ;
}

int main()
{
#ifdef DEBUG
    freopen("input.in", "r", stdin);
    //freopen("output.out", "w", stdout);
#endif
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    while(cin >> n && n){
        for(int i = ;i <= n ; i++){
            cin >> edge[i].x >> edge[i].y >> edge[i].z;
        }
        double  l = , r = 40.0;
        double mid = ;
        creat();
        while(fabs(r - l) > EXP){
            mid = (l + r) / ;
            if(prime(mid) >= )
                l = mid;
            else
                r = mid;
        }
        printf("%.3lf\n",mid );
    }
#ifdef DEBUG
    printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return ;
}

二分法与Dinkelbach算法

Dinkelbach

/*
*@Author:   Agnel-Cynthia
*@Language: C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=+;
const int M=+;
const int MOD=1e9+;
const double PI = acos(-1.0);
const double EXP = 1E-;
const int INF = 0x3f3f3f3f;
//int t,n,m,k,p,l,r,u,v;
//ll a[maxn],b[maxn];

#define Rep(i,l,r) for(i=(l);i<=(r);i++)
#define rep(i,l,r) for(i=(l);i< (r);i++)
#define Rev(i,r,l) for(i=(r);i>=(l);i--)
#define rev(i,r,l) for(i=(r);i> (l);i--)
#define Each(i,v)  for(i=v.begin();i!=v.end();i++)
#define r(x)   read(x)

int CH , NEG ;
template <typename TP>inline void read(TP& ret) {
ret = NEG =  ; while (CH=getchar() , CH<'!') ;
if (CH == '-') NEG = true , CH = getchar() ;
while (ret = ret*+CH-'' , CH=getchar() , CH>'!') ;
if (NEG) ret = -ret ;
}
#define  maxn  1010LL
#define  infi  100000000LL
#define  eps   1E-8F
#define  sqr(x) ((x)*(x))

template <typename TP>inline bool MA(TP&a,const TP&b) { return a < b ? a = b, true : false; }
template <typename TP>inline bool MI(TP&a,const TP&b) { return a > b ? a = b, true : false; }

int n;
int x[maxn], y[maxn], h[maxn];
double v[maxn][maxn], c[maxn][maxn];

bool vis[maxn];
double w[maxn];
double rv[maxn];///
inline double prim(double M) {
    int i, j, k;
    double minf, minw;
double sumc = , sumv = ;///
memset(vis,,sizeof vis);
Rep (i,,n) w[i] = v[][i]-M*c[][i],
rv[i] = v[][i];///
vis[] = true, minf = ;
rep (i,,n) {
    minw = infi;
    Rep (j,,n) if (!vis[j] && w[j]<minw)
    minw = w[j], k = j;
sumv += rv[k], sumc += rv[k]-w[k];///
minf += minw, vis[k] = true;
Rep (j,,n) if (!vis[j])
if (MI(w[j],v[k][j]-M*c[k][j]))
rv[j] = v[k][j];///
}
return sumv*M/sumc;///
return minf;
}

int main() {
    int i, j;
    double L, M, R;
    double maxv, maxc, minv, minc;
    while (scanf("%d", &n)!=EOF && n) {
        Rep (i,,n)
        scanf("%d%d%d", &x[i], &y[i], &h[i]);
        maxv = maxc = -infi, minv = minc = infi;
        rep (i,,n) Rep (j,i+,n) {
            c[i][j] = c[j][i] = sqrt(sqr((double)x[i]-x[j])+sqr((double)y[i]-y[j]));
            v[i][j] = v[j][i] = abs((double)h[i]-h[j]);
            MA(maxv,v[i][j]), MI(minv,v[i][j]);
            MA(maxc,c[i][j]), MI(minc,c[i][j]);
        }
        L = minv/maxc, R = maxv/minc;
while (true) {///
    R = prim(L);///
    if (fabs(L-R) < eps) break;///
    L = R;///
}///
        /*while (R-L > 1E-6) { // L:minf>0  R:minf<=0
            M = (L+R)/2.0;
            if (prim(M) > eps) L = M;
            else R = M;
        }*/
printf("%.3f\n", R);
}
//END: getchar(), getchar();
return ;
}