HDU 5697 刷题计划

Problem Description

n道题目，第 

i(1≤i≤n) 道题目能提升 

ai 点智力值，代码量为 

bi KB，无聊值为 

ci ，求至少提升

m点智力值的情况下，所做题目代码量之和

∗无聊值之和最小为多少。

样例解释：

选择第1、3、4道题

Input

多组测试数据

第一行两个整数

n,m(0<n≤400)

接下来

n行每行三个数，

ai,bi,ci(0<bi,ci≤1000,0<ai≤800)。

0<m≤∑ai≤800

Output

m点智力值的情况下，所做题目代码量之和

∗无聊值之和的最小值。

Sample Input

4 6
3 7 2
2 2 4
1 3 3
2 5 1      

Sample Output

90

最小乘积生成树，差不多理解了这个算法的思想了，不过实现起来还是有点困难，数学太渣了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<bitset>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef __int64 LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 8e2 + 10;
LL ans, dp[maxn];
int x[maxn], y[maxn], z[maxn];
int n, m, s;

struct pii
{
  int x, y;
  pii(int x = 0, int y = 0) :x(x), y(y) {}
  void read() { scanf("%d%d", &x, &y); }
}f[maxn];

pii operator +(pii a, pii b)
{
  return pii(a.x + b.x, a.y + b.y);
}

pii get(int X, int Y)
{
  dp[0] = 0;  f[0] = pii(0, 0);
  for (int i = 1; i <= s; i++) dp[i] = 1LL << 62;
  for (int i = 1; i <= n; i++)
  {
    LL v = (LL)y[i] * X + (LL)z[i] * Y;
    for (int j = s; j >= x[i]; j--)
    {
      if (dp[j] <= dp[j - x[i]] + v) continue;
      dp[j] = dp[j - x[i]] + v;
      f[j] = f[j - x[i]] + pii(y[i], z[i]);
    }
  }
  int k = m;
  for (int i = m; i <= s; i++) if (dp[i] < dp[k]) k = i;
  ans = min(ans, (LL)f[k].x*f[k].y);
  return f[k];
}

void solve(pii x, pii y)
{
  pii t = get(x.y - y.y, y.x - x.x);
  if (1LL * (x.x - y.x)*(t.y - y.y) <= 1LL * (x.y - y.y)*(t.x - y.x)) return;
  solve(x, t), solve(t, y);
}

int main()
{
  while (scanf("%d%d", &n, &m) != EOF)
  {
    ans = 1LL << 62;  s = 0;
    for (int i = 1; i <= n; i++)
    {
      scanf("%d%d%d", &x[i], &y[i], &z[i]);
      s += x[i];
    }
    solve(get(1, 0), get(0, 1));
    printf("%I64d\n", ans);
  }
  return 0;
}