Optimal proportion spanning tree (0/1 score Planning)

First, what is the problem to be solved: a fully Weighted Graph with each edge having its own cost value cost [I] and benefit value benifit [I]? if X [I] is used to represent an edge or not, a spanning tree is obtained. Requirements: r = (Σ cost [I] * X [I])/(Σ benifit [I] * X [I]) minimum.

Typical question: poj2728-desert king

How to solve the problem: here we use the 0-1 score planning idea. The above formula can be transformed to Z (R) = Sigma cost [I] * X [I]-R * Sigma benifit [I] * X [I]. While Z (R) = 0 is what we want.

Here is a very important conclusion: Z (R) is a monotonic decreasing function, so it is linear. So "we can happily regard Z (r) as the total weight of the Minimum Spanning Tree with cost [I]-R * benifit [I] As the edge weight ". Obviously, the request is Z (max (R) = 0. Here, Max (R) is determined by Z = 0.

There are two algorithms:

1. dinkelbanch algorithm, that is, iteration method. Take an initial value of R, which is generally 0. After a spanning tree is obtained, r = (Σ benifit [I] * X [I]) /(Σ cost [I] * X [I]), after iteration of ...... R tends to the maximum value, while Z tends to 0.

2. the binary method. If Z (R) = 0 is what we want, and it is a monotonic function, then binary R is enough. It is said that binary is much slower than iteration.

 

 

Template dinkelbach poj 2728:

#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <cstring>#include <algorithm>#include <string>#include <set>#include <ctime>#include <queue>#include <map>#include <sstream>#define CL(arr, val)    memset(arr, val, sizeof(arr))#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))#define L(x)    (x) << 1#define R(x)    (x) << 1 | 1#define MID(l, r)   (l + r) >> 1#define Min(x, y)   x < y ? x : y#define Max(x, y)   x < y ? y : x#define E(x)    (1 << (x))const double eps = 1e-6;typedef long long LL;using namespace std;const int N = 1<<10;const int inf = ~0u>>2;struct node {    int x, y, z;} point[N];double cost[N][N];double benifit[N][N];double mi[N];int pre[N];bool vis[N];double c, d;int n;double prim(double ans) {    int i, j, f;    double mx;    c = d = 0;    for(i = 1; i < n; ++i) {        mi[i] = cost[0][i] - ans*benifit[0][i];        pre[i] = 0;        vis[i] = false;    }    mi[0] = 0;    vis[0] = true;    for(i = 0; i < n - 1; ++i) {        mx = inf; f = 0;        for(j = 0; j < n; ++j) {            if(!vis[j] && mx > mi[j]) {                mx = mi[j];                f = j;            }        }        c += cost[pre[f]][f];        d += benifit[pre[f]][f];        vis[f] = true;        for(j = 1; j < n; ++j) {            double val = cost[f][j] - ans*benifit[f][j];            if(!vis[j] && mi[j] > val) {                pre[j] = f;                mi[j] = val;            }        }    }    return c/d;}int iabs(int x) {    return x < 0 ? -x : x;}void init() {    int i, j;    for(i = 0; i < n; ++i) {        for(j = 0; j < n; ++j) {            cost[i][j] = iabs(point[i].z - point[j].z);            benifit[i][j] = sqrt(1.*(point[i].x - point[j].x)*(point[i].x - point[j].x) +                                 1.*(point[i].y - point[j].y)*(point[i].y - point[j].y));        }    }}void solve() {    double ans = 0, tmp;    while(1) {        tmp = prim(ans);        if(fabs(ans - tmp) < eps)   break;        else    ans = tmp;    }    printf("%.3f\n", ans);}int main() {    freopen("data.in", "r", stdin);    int i;    while(scanf("%d", &n), n) {        for(i = 0; i < n; ++i) {            scanf("%d%d%d", &point[i].x, &point[i].y, &point[i].z);        }        init();        solve();    }    return 0;}