题目大意
给出1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],变化以后的数字是B[i],那末花费是|A[i]?B[i]| 。求出1种方案，使得最大的花费最小。
思路
1眼就可以看出是2分，然后贪心甚么的随意yy1下就好了。
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;
}
波比源码 » BZOJ 3613 HEOI 2014 南园满地堆轻絮 二分+贪心
