POJ

传送门：POJ3111 

题意：给定n种物品，每种物品有不同的vi和wi，让你从中选出k种来使得∑v/∑w最大。

思路：设∑v/∑w = x，我们的目标就是最大化x，因此我们可二分x，然后判断是否存在一个集合使得∑v/∑w >= x，关键是如何进行判断，显然直接枚举所有的集合是不现实的，可以考虑将不等式变形成∑(v - x * w) >= 0 ，这样我们就可以针对vi - x * wi 排序，然后选前k个判断他们的和是否大于零就好了。（可以参考挑战程序设计P144）

还可以证明∑(v - x * w)是收敛且有界的，因此可以用牛顿迭代。

证明请戳：点击打开链接

二分代码：

#include<stdio.h>#include<iostream>#include<vector>#include<algorithm>#define ll long long#define pb push_back#define fi first#define se second#define pi acos(-1)#define inf 0x3f3f3f3f#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define rep(i,x,n) for(int i=x;i<n;i++)#define per(i,n,x) for(int i=n;i>=x;i--)using namespace std;typedef pair<int,int>P;const int MAXN=100010;int gcd(int a,int b){return b?gcd(b,a%b):a;}int n, k;int v[MAXN], w[MAXN];vector<int>ans;vector<pair<double, int> > p;bool check(double mid){p.clear();for(int i = 0; i < n; i++)p.pb(make_pair(v[i] - mid * w[i], i));sort(p.begin(), p.end());double sum = 0;for(int i = n - k; i < n; i++)sum += p[i].fi;return sum > 0;}int main(){cin >> n >> k;for(int i = 0; i < n; i++)scanf("%d %d", v + i, w + i);double l = 0, r = 1e7, mid;for(int i = 0; i < 50; i++){mid = (l + r) / 2;if(check(mid))l = mid;elser = mid;}for(int i = n - k; i < n; i++)printf("%d%c", p[i].se + 1, " \n"[i == n - 1]); return 0;}

牛顿迭代代码：

#include<stdio.h>#include<iostream>#include<vector>#include<algorithm>#include<math.h>#define eps 1e-8#define ll long long#define pb push_back#define fi first#define se second#define pi acos(-1)#define inf 0x3f3f3f3f#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define rep(i,x,n) for(int i=x;i<n;i++)#define per(i,n,x) for(int i=n;i>=x;i--)using namespace std;typedef pair<int,int>P;const int MAXN=100010;int gcd(int a,int b){return b?gcd(b,a%b):a;}int n, k;int v[MAXN], w[MAXN];pair<double, int> p[MAXN];double get_x(){double s1 = 0, s2 = 0;sort(p, p + n);for(int i = n - k; i < n; i++){s1 += v[p[i].se];s2 += w[p[i].se];}return s1 / s2;}int main(){cin >> n >> k;double x1 = 0, x2 = 0;for(int i = 0; i < n; i++){scanf("%d %d", v + i, w + i);}do{x1 = x2;for(int i = 0; i < n; i++){p[i] = make_pair(v[i] - x1 * w[i], i);}x2 = get_x();}while(fabs(x1 - x2) > eps);for(int i = n - k; i < n; i++)printf("%d ", p[i].se + 1); return 0;}

事实证明本题用牛顿迭代能比二分块8 - 10 倍。

传送门：POJ2976

题意：题意和3111差不多，只不过这次是去掉k个使剩下的最大。

思路：思路和上面一样，不过这题数据量小，二分就够够的了。

需要特别注意的就是poj的g++只能交%f过，c++ %lf或者%f都行，气得我想骂娘。

代码：

#include<iostream>#include<stdio.h>#include<algorithm>#define ll long long#define pb push_back#define fi first#define se second#define pi acos(-1)#define inf 0x3f3f3f3f#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define rep(i,x,n) for(int i=x;i<n;i++)#define per(i,n,x) for(int i=n;i>=x;i--)using namespace std;typedef pair<int,int>P;const int MAXN=100010;int gcd(int a,int b){return b?gcd(b,a%b):a;}struct node{int a, b;}p[MAXN];double tmp[MAXN];int n, k;bool check(double mid){double sum = 0;for(int i = 0; i < n; i++)tmp[i] = p[i].a - mid * p[i].b;sort(tmp, tmp + n);for(int i = k; i < n; i++)sum += tmp[i];return sum > 0;}int main(){while(cin >> n >> k, n + k){ll a = 0, b = 0;for(int i = 0; i < n; i++)scanf("%d", &p[i].a);for(int i = 0; i < n; i++)scanf("%d", &p[i].b);double l = 0, r = 1, mid;for(int i = 0; i <= 15; i++){mid = (l + r) / 2;if(check(mid))l = mid;elser = mid;}printf("%.0lf\n", mid * 100);} return 0;}