[NOIP2011] 聪明的质监员

嘟嘟嘟

看到绝对值最小这种逼近的问题，应该能想到二分。

容易发现，随着W的变大，符合条件的矿石越来越少，即满足单调关系。因此我们二分W，然后在W确定时求出x = ΣY_i，如果x >= S，说明W取小了，向右二分；否则向左二分。每一次都更新答案。

求x用前缀和就解决了。

时间复杂度：二分O(logn)，求值O(n)，总共O(nlogn)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e6 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) {last = ch; ch = getchar();}
  while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m;
ll S;
struct Node
{
  ll w, v;
}t[maxn];
struct Node2
{
  int L, R;
}q[maxn];

int num[maxn];
ll sum[maxn];
ll calc(ll x)
{
  for(int i = 1; i <= n; ++i)
    {
      if(t[i].w >= x)
    {
      num[i] = num[i - 1] + 1;
      sum[i] = sum[i - 1] + t[i].v;
    }
      else
    {
      num[i] = num[i - 1];
      sum[i] = sum[i - 1];
    }
    }
  ll ret = 0;
  for(int i = 1; i <= m; ++i)
    ret += (num[q[i].R] - num[q[i].L - 1]) * (sum[q[i].R] - sum[q[i].L - 1]);
  return ret;
}

ll ans = (ll)INF * (ll)INF;

int main()
{
  n = read(); m = read(); S = read();
  ll l = INF, r = -INF;
  for(int i = 1; i <= n; ++i)
    {
      t[i].w = read(), t[i].v = read();
      l = min(l, t[i].w);
      r = max(r, t[i].w);
    }
  for(int i = 1; i <= m; ++i) q[i].L = read(), q[i].R = read();
  while(l < r)
    {
      ll mid = (l + r + 1) >> 1;
      ll x = calc(mid);
      if(x <= S) r = mid - 1, ans = min(ans, S - x);
      else l = mid, ans = min(ans, x - S);
    }
  write(ans), enter;
  return 0;
}