[POJ2728] Desert King

題意

n個點有三維座標，沒兩點之間有一條邊，花費爲兩點之間的高度差，獲利爲兩點之間水平直線距離。最小化 花費和 / 獲利和

題解

01分數規劃問題，最優比率生成樹模型，最大化或最小化 ∑value/∑cost ，令value=花費,cost=距離 。最小化其比值，講解在寫在這一篇中
兩種算法，二分法和Dinkelbach迭代。
二分精度設爲至少1E-6才過_(:зゝ∠)_
另外，二分上界大致設在 40 ~ 100W之間，精度1E-5可過

但是講道理上界應該老老實實設爲 1000W 對不對？因爲比賽不是刷題

否則這組數據

input:
2
0 0 0
1 0 10000000
0
output:
10000000

肯定WA嘛

代碼

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
/*     "///"表示改動過的地方     */

#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;
}