HDU 3473 Minimum Sum(划分树)
Minimum Sum
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2235 Accepted Submission(s): 512
Problem Description
You are given N positive integers, denoted as x0, x1 ... xN-1. Then give you some intervals [l, r]. For each interval, you need to find a number x to make as small as possible!
Input
The first line is an integer T (T <= 10), indicating the number of test cases. For each test case, an integer N (1 <= N <= 100,000) comes first. Then comes N positive integers x (1 <= x <= 1,000, 000,000) in the next line. Finally, comes an integer Q (1 <= Q <= 100,000), indicting there are Q queries. Each query consists of two integers l, r (0 <= l <= r < N), meaning the interval you should deal with.
Output
For the k-th test case, first output “Case #k:” in a separate line. Then output Q lines, each line is the minimum value of . Output a blank line after every test case.
Sample Input
2
5
3 6 2 2 4
2
1 4
0 2
2
7 7
2
0 1
1 1
Sample Output
Case #1:
6
4
Case #2:
0
0
Author
standy
Source
Recommend
zhengfeng
其实就是找到中间那个数。
然后需要记录和,右边的减掉左边的
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> using namespace std; const int MAXN = 100010; int tree[20][MAXN]; int sorted[MAXN]; int toleft[20][MAXN]; long long sum[20][MAXN]; void build(int l,int r,int dep) { if(l == r) { sum[dep][l] = sum[dep][l-1]+tree[dep][l]; return; } int mid = (l+r)>>1; int same = mid - l + 1; for(int i = l;i <= r;i++) { if(tree[dep][i] < sorted[mid]) same--; sum[dep][i] += sum[dep][i-1]+tree[dep][i]; } int lpos = l; int rpos = mid+1; for(int i = l;i <= r;i++) { if(tree[dep][i] < sorted[mid]) tree[dep+1][lpos++] = tree[dep][i]; else if(tree[dep][i] == sorted[mid] && same > 0) { tree[dep+1][lpos++] = tree[dep][i]; same--; } else tree[dep+1][rpos++] = tree[dep][i]; toleft[dep][i] = toleft[dep][l-1] + lpos - l; } build(l,mid,dep+1); build(mid+1,r,dep+1); } long long ans; int query(int L,int R,int l,int r,int dep,int k) { if(l == r)return tree[dep][l]; int mid = (L+R)>>1; int cnt = toleft[dep][r] - toleft[dep][l-1]; if(cnt >= k) { int ee = r-L+1-(toleft[dep][r]-toleft[dep][L-1])+mid; int ss = l-L-(toleft[dep][l-1]-toleft[dep][L-1])+mid; ans += sum[dep+1][ee]-sum[dep+1][ss]; int newl = L + toleft[dep][l-1]-toleft[dep][L-1]; int newr = newl + cnt -1; return query(L,mid,newl,newr,dep+1,k); } else { int s = L + toleft[dep][l-1] - toleft[dep][L-1]; int e = s + cnt - 1; ans -= sum[dep+1][e] - sum[dep+1][s-1]; int newr = r + toleft[dep][R] - toleft[dep][r]; int newl = newr - (r-l+1-cnt) + 1; return query(mid+1,R,newl,newr,dep+1,k-cnt); } } int main() { int T; int n; scanf("%d",&T); int iCase = 0; while(T--) { iCase++; scanf("%d",&n); memset(tree,0,sizeof(tree)); memset(sum,0,sizeof(sum)); for(int i = 1;i <= n;i++) { scanf("%d",&tree[0][i]); sorted[i] = tree[0][i]; } sort(sorted+1,sorted+n+1); build(1,n,0); printf("Case #%d:\n",iCase); int m,l,r; scanf("%d",&m); while(m--) { scanf("%d%d",&l,&r); l++;r++; ans = 0; int tmp = query(1,n,l,r,0,(l+r)/2-l+1); if((l+r)%2)ans-=tmp; printf("%I64d\n",ans); } printf("\n"); } return 0; }
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