[BZOJ1492][NOI2007][斜率优化][动态凸包][DP][分治]货币兑换cash

[题目]


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[算法]

dp+斜率优化 动态凸包维护 splay/分治

[分析]

由题意可得，每次买入或卖出时总是全部买入卖出。

易得转移方程 

f[i] = max(ans, f[j] * b[i] + f[j] * rate[j] * a[i]) / (rate[i] * a[i] + b[i])

其中a,b,rate含义如题目描述，f[i]表示第i天可以买到的最多的b券，ans则是截止i所能获得的最多资金

如果直接dp是n^2效率，化简式子可得

- (a[i] / b[i]) * rate[j] * f[j] + f[i] / b[i] = f[j]

满足斜率优化的格式，可以进行斜率优化。正常的做法是用splay维护动态上凸壳，但是由于太麻烦就没有写

这里参考陈丹琦 从​《​C​a​s​h​》​谈​一​类​分​治​算​法​的​应​用 ，用分治的方法解决

solve(left,right) 表示解决left ~right的f。solve(left, right)过程如下

1.solve(left, mid)

2.将left ~ mid中的元素压入栈，维护成一个上凸壳

3.更新一下mid+1~right 的 f * (rate[i] * a[i] + b[i]) (就是max那一坨) (注意这个时候mid+1 ~ right 已经按照斜率 -a[i] / b[i]排好序了，所以可以用单调栈)

4.solve(mid+1, right)

5.merge归并一下，保证f单调，可以用单调栈

这样问题就完美的解决了，代码的空间、时间和编程复杂度都非常令人满意

[注意]

这个地方提前将斜率排序比较有技巧。先将1~n排好序后（对编号排序），每次solve都根据mid将编号分到左右两组

[代码]

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;

#define MAXN 100100
#define x first
#define y second
#define EPS 1e-7

typedef pair<long double, long double> point;

point f[MAXN], temp[MAXN];
long double money[MAXN], k[MAXN];
long double a[MAXN], b[MAXN], rate[MAXN];
long double ans;
int pos[MAXN], stack[MAXN];
int n, sum;

long double max(long double a, long double b)
{
return a > b ? a : b;
}

bool cmp_pos(int a, int b)
{
return k[a] < k[b];
}

long double sell(point thePoint, int day)
{
return thePoint.x * a[day] + thePoint.y * b[day];
}

long double cross(point a, point b, point c)
{
return (b.x - a.x) * (c.y - b.y) - (b.y - a.y) * (c.x - b.x);
}

void solve(int left, int right)
{
if (left == right)
{
ans = max(ans, money[left]);
f[left].y = ans / (a[left] * rate[left] + b[left]);
f[left].x = f[left].y * rate[left];
return;
}
int mid = (left + right) >> 1;
int i = left - 1, j = mid;
memmove(stack + left, pos + left, (right - left + 1) * sizeof(int));
for (int k = left; k <= right; k++)
(stack[k] <= mid ? pos[++i] : pos[++j]) = stack[k];

solve(left, mid);
int top = 0;
for (int i = left; i <= mid; i++)
{
while (top > 1 && (cross(f[stack[top-1]], f[stack[top]], f[i]) >= 0))
top--;
stack[++top] = i;
}

for (int i = mid + 1; i <= right; i++)
{
while (top > 1 && sell(f[stack[top-1]], pos[i]) > sell(f[stack[top]], pos[i]))
top--;
money[pos[i]] = max(money[pos[i]], sell(f[stack[top]], pos[i]));
}
solve(mid + 1, right);

merge(f + left, f + mid + 1, f + mid + 1, f + right + 1, temp + left);
memmove(f + left, temp + left, sizeof(point)*(right - left + 1));
}

int main()
{
//freopen("input.txt","r",stdin);
double temp;
scanf("%d%lf", &n, &temp);
ans = temp;
for (int i = 1; i <= n; i++)
{
double tempa, tempb, tempc;
scanf("%lf%lf%lf", &tempa, &tempb, &tempc);
a[i] = tempa;
b[i] = tempb;
rate[i] = tempc;
k[i] = - a[i] / b[i];
pos[i] = i;
}
sort(pos + 1, pos + 1 + n, cmp_pos);
solve(1, n);
printf("%.3f\n", (double)ans);
}