hdu3437 划分树 区间内小于第K大的值得和

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3769 Accepted Submission(s): 872


[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
5
3 6 2 2 4
2
1 4
0 2
2
7 7
2
0 1
1 1

[align=left]Sample Output[/align]

Case #1:
6
4
Case #2:
0
0

题意: 有n个元素的数组，有q次查询，对于每次询问，希望得到一个值x，使区间[L,R]内， 的值最小。

思路:
既然要让这个值最小，那么这个区间内的中位数一定满足。不过这里还要处理这个区间里面小于中位数的值得和。
这时候，可以在建树的时候同时处理。

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 100010;
int num[20][MAXN],cnt[20][MAXN],sor[MAXN],n,leftnum;
ll sum[20][MAXN],leftsum,all[MAXN];//sum记录第d层 第i个数之前小于sor[m]的和
void build(int l,int r,int d)
{
if(l == r){
return ;
}
int m = (l + r) >> 1;
int same_m = m - l + 1;
for(int i = l; i <= r; i++){
if(num[d][i] < sor[m])same_m --;
}
int cnt_small = 0;
int pl,pr;
ll val = 0;
pl = l,pr = m + 1;
for(int i = l; i <= r; i++){
if(num[d][i] < sor[m]){
cnt_small ++;
val += num[d][i];
sum[d][i] = val;
cnt[d][i] = cnt_small;
num[d+1][pl++] = num[d][i];
}
else if(num[d][i] == sor[m] && same_m){
same_m --;
cnt_small ++;
val += num[d][i];
sum[d][i] = val;
cnt[d][i] = cnt_small;
num[d+1][pl++] = num[d][i];
}
else {
sum[d][i] = val;
cnt[d][i] = cnt_small;
num[d+1][pr++] = num[d][i];
}
}
build(l,m,d+1);
build(m+1,r,d+1);
}
ll query(int L,int R,int k,int l,int r,int d)
{
if(l == r){
return num[d][l];
}
int m = (l + r) >> 1;
int s,ss;
ll val = 0;
if(l == L)s = 0, val = sum[d][R];
else s = cnt[d][L-1], val = sum[d][R] - sum[d][L-1];
ss = cnt[d][R] - s;
if(ss >= k){
int newl = l + s;
int newr = l + s + ss - 1;
return query(newl,newr,k,l,m,d+1);
}
else {
leftnum += ss;//进入左孩子不用处理，进入右孩子时 就要加上左孩子的值
leftsum += val;
int a = L - l - s;
int b = R - L + 1 - ss;
int newl = m + 1 + a;
int newr = m + 1 + a + b - 1;
return query(newl,newr,k - ss,m+1,r,d+1);
}
}
int main()
{
int t,ff = 0;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
memset(all,0,sizeof(all));
memset(num,0,sizeof(num));
for(int i = 1; i <= n; i++){
scanf("%d",&num[1][i]);
sor[i] = num[1][i];
all[i] = all[i-1] + sor[i];
}
sort(sor + 1,sor + n + 1);
build(1,n,1);
int q,x,y;
scanf("%d",&q);
printf("Case #%d:\n",++ff);
while(q--){
scanf("%d%d",&x,&y);
x += 1;
y += 1;
int len = (y - x + 1);
ll tp;
leftnum = 0;
leftsum = 0;
if(len % 2){
int k = (len + 1) >> 1;
tp = query(x,y,k,1,n,1);
}
else {
int k = len >> 1;
tp = query(x,y,k,1,n,1);
}
//cout<<tp<<' '<<leftnum<<' '<<leftsum<<' '<<all[y]<<' '<<all[x+leftnum]<<endl;
ll ans = tp * (leftnum + 1) - (leftsum + tp) + (all[y] - all[x - 1] - (leftsum + tp)) - (y - x + 1 - (leftnum + 1)) * tp;
printf("%lld\n",ans);
}
printf("\n");
}
return 0;
}