1112: [POI2008]砖块Klo

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1112: [POI2008]砖块Klo

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 1664 Solved: 583
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Description

N柱砖,希望有连续K柱的高度是一样的. 你可以选择以下两个动作 1:从某柱砖的顶端拿一块砖出来,丢掉不要了. 2:从仓库中拿出一块砖,放到另一柱.仓库无限大. 现在希望用最小次数的动作完成任务.

Input

第一行给出N,K. (1 ≤ k ≤ n ≤ 100000), 下面N行,每行代表这柱砖的高度.0 ≤ hi ≤ 1000000

Output

最小的动作次数

Sample Input

5 3
3
9
2
3
1

Sample Output

HINT

原题还要求输出结束状态时,每柱砖的高度.本题略去.

Source

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提问：给定一维坐标系上的k个点，找一个点，使得它到其它点的距离和最小找中位数啊。。。。维护一个treap就行

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E5 + 10;typedef long long LL;int n,k,cnt,Root,a[maxn],ch[maxn][2],key[maxn],va[maxn],siz[maxn];LL Ans = 1E18,sum[maxn];LL Sum(int x,int tot){if (!tot) return 0;if (siz[x] <= tot) return sum[x];int Left = siz[ch[x][0]] + 1;if (Left > tot) return Sum(ch[x][0],tot);else return sum[ch[x][0]] + 1LL*va[x] + Sum(ch[x][1],tot - Left);}int Rank(int x,int tot){int Left = siz[ch[x][0]] + 1;if (Left == tot) return va[x];else if (Left > tot) return Rank(ch[x][0],tot);else return Rank(ch[x][1],tot - Left);}void Solve(LL Now,int mid1,int mid2){LL z = mid1,sum = Sum(Root,mid1);LL tot = Now*z - sum;sum = Sum(Root,k) - Sum(Root,mid1);tot += (sum - Now*z);Ans = min(Ans,tot);}void maintain(int x){sum[x] = va[x];siz[x] = 1;for (int i = 0; i < 2; i++)if (ch[x][i]) {siz[x] += siz[ch[x][i]];sum[x] += sum[ch[x][i]];}}void rotate(int &x,int d){int y = ch[x][d];ch[x][d] = ch[y][d^1];maintain(x);ch[y][d^1] = x;x = y; maintain(x);}int cmp(int x,int w){if (va[x] == w) return -1;if (va[x] < w) return 1;return 0;}int New(int w){int ret = ++cnt;ch[ret][0] = ch[ret][1] = 0;key[ret] = rand();va[ret] = sum[ret] = w;siz[ret] = 1;return ret;}void Insert(int &x,int w){if (!x) {x = New(w);return;}int d = cmp(x,w);d = max(d,0);Insert(ch[x][d],w);maintain(x);if (key[ch[x][d]] > key[x]) rotate(x,d);}void Remove(int &x,int w){int d = cmp(x,w);if (d == -1) {if (!ch[x][0] && !ch[x][1]) x = 0;else if (!ch[x][0] || !ch[x][1]) x = ch[x][0]?ch[x][0]:ch[x][1];else {int d2 = key[ch[x][0]] > key[ch[x][1]]?0:1;rotate(x,d2);Remove(ch[x][d2^1],w);maintain(x);if (key[ch[x][d2^1]] > key[x])rotate(x,d2^1);}return;}Remove(ch[x][d],w);maintain(x);if (key[ch[x][d]] > key[x])rotate(x,d);}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifcin >> n >> k;if (k == 1) {cout << 0;return 0;}for (int i = 1; i <= n; i++) scanf("%d",&a[i]);for (int i = 1; i <= k; i++)Insert(Root,a[i]);for (int i = k; i <= n; i++) {if (i > k) {Remove(Root,a[i-k]);Insert(Root,a[i]);}if (k&1) {int mid = (k + 1) >> 1;LL Now = Rank(Root,mid),tot = 0;LL z = mid-1,sum = Sum(Root,mid-1);tot += (z*Now - sum);sum = Sum(Root,k) - Sum(Root,mid);tot += (sum - z*Now);Ans = min(Ans,tot);}else {int mid1 = k>>1,mid2 = mid1+1;int Now1 = Rank(Root,mid1);int Now2 = Rank(Root,mid2);int Now = (Now1 + Now2) >> 1;Solve(Now,mid1,mid2);if (Now1 != Now2) Solve(Now + 1,mid1,mid2);}}cout << Ans;return 0;}

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