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 50010using 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;}