BZOJ 3613 HEOI 2014 南园满地堆轻絮 二分+贪心

题目大意

给出一个数字序列，要求将这个数字序列变成单调不降的序列。若原来的数字是A[i],变化之后的数字是B[i],那么花费是$|A[i] - B[i]|$ 。求出一种方案，使得最大的花费最小。

思路

一眼就能看出是二分，然后贪心什么的随便yy一下就行了。

CODE

#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 50010
using namespace std;
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)

struct Cow{
    int x,y,c;
    int st,ed;

    bool operator <(const Cow &a)const {
        return y > a.y;
    }
    void Read() {
        scanf("%d%d%d", &x, &y, &c);
        x *= -1;
        st = c * (x - 1);
        ed = st + (c << 1);
    }
}src[MAX];

int cows;
pair<int,int *> xx[MAX << 1];
int cnt,t;

int tree[MAX << 4];

inline void PushDown(int pos)
{
    if(tree[pos]) {
        tree[LEFT] = tree[pos];
        tree[RIGHT] = tree[pos];
        tree[pos] = 0;
    }
}

void Modify(int l, int r, int x, int y, int c, int pos)
{
    if(l == x && y == r) {
        tree[pos] = c;
        return ;
    }
    PushDown(pos);
    int mid = (l + r) >> 1;
    if(y <= mid) Modify(l, mid, x, y, c, LEFT);
    else if(x > mid) Modify(mid + 1, r, x, y, c, RIGHT);
    else {
        Modify(l, mid, x, mid, c, LEFT);
        Modify(mid + 1, r, mid + 1, y, c, RIGHT);
    }
}

inline int Ask(int l, int r, int x, int pos)
{
    if(l == r)  return tree[pos];
    PushDown(pos);
    int mid = (l + r) >> 1;
    if(x <= mid) return Ask(l, mid, x, LEFT);
    return Ask(mid + 1, r, x, RIGHT);
}

bool v[MAX];

int main()
{
    cin >> cows;
    for(int i = 1; i <= cows; ++i)
        src[i].Read();
    sort(src + 1, src + cows + 1);
    for(int i = 1; i <= cows; ++i) {
        xx[++cnt] = make_pair(src[i].st, &src[i].st);
        xx[++cnt] = make_pair(src[i].ed, &src[i].ed);
    }
    sort(xx + 1, xx + cnt + 1);
    for(int i = 1; i <= cnt; ++i) {
        if(i == 1 || xx[i].first != xx[i - 1].first)
            ++t;
        *xx[i].second = t;
    }
    for(int i = 1; i <= cows; ++i)
        Modify(1, cnt, src[i].st, src[i].ed , i, 1);
    for(int i = 1; i <= cnt; ++i)
        v[Ask(1, cnt, i, 1)] = true;
    int ans = 0;
    for(int i = 1; i <= cows; ++i)
        ans += v[i];
    cout << ans << endl;
    return 0;
}