[POJ2728] Desert King

2019-11-10 21:13:13

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题意

n个点有三维坐标，没两点之间有一条边，花费为两点之间的高度差，获利为两点之间水平直线距离。最小化花费和 / 获利和

题解

01分数规划问题，最优比率生成树模型，最大化或最小化 ∑value/∑cost $/sum value / /sum cost$ ，令value=花费,cost=距离 $value=花费,cost = 距离$ 。最小化其比值，讲解在写在这一篇中两种算法，二分法和Dinkelbach迭代。二分精度设为至少1E-6才过_(:зゝ∠)_ 另外，二分上界大致设在 40 ~ 100W之间，精度1E-5可过

但是讲道理上界应该老老实实设为 1000W 对不对？因为比赛不是刷题

否则这组数据

input:20 0 01 0 100000000output:10000000

肯定WA嘛

代码

1:二分法1822MS

/// by ztx/// blog.csdn.net/hzoi_ztx//#include <bits/stdc++.h>// minimize sumcost / sumdistance// minimize sumvalue / sumcost// value:cost// cost:distance#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)typedef long long ll ;typedef double lf ;int CH , NEG ;template <typename TP>inline void read(TP& ret) { ret = NEG = 0 ; while (CH=getchar() , CH<'!') ; if (CH == '-') NEG = true , CH = getchar() ; while (ret = ret*10+CH-'0' , 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];lf v[maxn][maxn], c[maxn][maxn];bool vis[maxn];lf w[maxn];inline lf PRim(lf M) { int i, j, k; lf minf, minw; memset(vis,0,sizeof vis); Rep (i,2,n) w[i] = v[1][i]-M*c[1][i]; vis[1] = true, minf = 0; rep (i,1,n) { minw = infi; Rep (j,1,n) if (!vis[j] && w[j]<minw) minw = w[j], k = j; minf += minw, vis[k] = true; Rep (j,1,n) if (!vis[j]) MI(w[j],v[k][j]-M*c[k][j]); } return minf;}int main() { int i, j; lf L, M, R; lf maxv, maxc, minv, minc; while (scanf("%d", &n)!=EOF && n) { Rep (i,1,n) scanf("%d%d%d", &x[i], &y[i], &h[i]); maxv = maxc = -infi, minv = minc = infi; rep (i,1,n) Rep (j,i+1,n) { c[i][j] = c[j][i] = sqrt(sqr((lf)x[i]-x[j])+sqr((lf)y[i]-y[j])); v[i][j] = v[j][i] = abs((lf)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 (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 0;}

2:Dinkelbach算法235MS

/// by ztx/// blog.csdn.net/hzoi_ztx//#include <bits/stdc++.h>// minimize sumcost / sumdistance// minimize sumvalue / sumcost// value:cost// cost:distance/* "///"表示改动过的地方 */#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)typedef long long ll ;typedef double lf ;int CH , NEG ;template <typename TP>inline void read(TP& ret) { ret = NEG = 0 ; while (CH=getchar() , CH<'!') ; if (CH == '-') NEG = true , CH = getchar() ; while (ret = ret*10+CH-'0' , 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];lf v[maxn][maxn], c[maxn][maxn];bool vis[maxn];lf w[maxn];lf rv[maxn];///inline lf prim(lf M) { int i, j, k; lf minf, minw;lf sumc = 0, sumv = 0;/// memset(vis,0,sizeof vis); Rep (i,2,n) w[i] = v[1][i]-M*c[1][i],rv[i] = v[1][i];/// vis[1] = true, minf = 0; rep (i,1,n) { minw = infi; Rep (j,1,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,1,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; lf L, M, R; lf maxv, maxc, minv, minc; while (scanf("%d", &n)!=EOF && n) { Rep (i,1,n) scanf("%d%d%d", &x[i], &y[i], &h[i]); maxv = maxc = -infi, minv = minc = infi; rep (i,1,n) Rep (j,i+1,n) { c[i][j] = c[j][i] = sqrt(sqr((lf)x[i]-x[j])+sqr((lf)y[i]-y[j])); v[i][j] = v[j][i] = abs((lf)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 0;}

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