Minimum Sum
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3047 Accepted Submission(s): 701
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
区间第k大的题目一般两种做法，，划分树 主席树，不过貌似划分树只支持静态区间第k大。
这道题由于内存限制 只能用划分树。。具体就是 先找出区间[l,r]内的中位数，（为什么是中位数，大白书貌似有介绍），然后直接求和。
MLE到死啊，，注释memset部分之后就过了，不知道为什么
划分树代码：
1 #include <set>
2 #include <map>
3 #include <cmath>
4 #include <ctime>
5 #include <queue>
6 #include <stack>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 typedef unsigned long long ull;
16 typedef long long ll;
17 const int inf = 0x3f3f3f3f;
18 const double eps = 1e-8;
19 const int maxn = 100010;
20 int sorted[maxn],tree[20][maxn],toleft[20][maxn]; //toleft[dep][i]表示第dep层1-i中进入左子树元素的个数
21 ll leftsum[20][maxn], sum[maxn]; //leftsum[dep][i]表示第dep层1-i中进入左子树元素的和
22
23 void build (int l,int r,int dep)
24 {
25 if (l == r)
26 return;
27 int mid = (l + r) >> 1;
28 int same = mid - l + 1;
29 for (int i = l; i <= r; i++)
30 if (tree[dep][i] < sorted[mid])
31 same--;
32 int lpos = l,rpos = mid + 1;
33 for (int i = l; i <= r; i++)
34 {
35 if (tree[dep][i] < sorted[mid])
36 {
37 tree[dep+1][lpos++] = tree[dep][i];
38 leftsum[dep][i] = leftsum[dep][i-1] + tree[dep][i];
39 }
40 else if (tree[dep][i] == sorted[mid] && same > 0)
41 {
42 tree[dep+1][lpos++] = tree[dep][i];
43 leftsum[dep][i] = leftsum[dep][i-1] + tree[dep][i];
44 same--;
45 }
46 else
47 {
48 tree[dep+1][rpos++] = tree[dep][i];
49 leftsum[dep][i] = leftsum[dep][i-1];
50 }
51 toleft[dep][i] = toleft[dep][l-1] + lpos - l;
52 }
53 build (l,mid,dep+1);
54 build (mid+1,r,dep+1);
55 }
56 int lnum,rnum;
57 ll lsum,rsum;
58 int query(int l,int r,int dep,int ua,int ub,int k)
59 {
60 if (ua == ub)
61 return tree[dep][ua];
62 int mid = (l + r) >> 1;
63 int cnt = toleft[dep][ub] - toleft[dep][ua-1];
64 if (cnt >= k)
65 {
66 int newl = l + toleft[dep][ua-1] - toleft[dep][l-1];
67 int newr = newl + cnt - 1;
68 return query(l,mid,dep+1,newl,newr,k);
69 }
70 else
71 {
72 int newr = ub + toleft[dep][r] - toleft[dep][ub];
73 int newl = newr - (ub - ua - cnt);
74 lnum += cnt;
75 lsum += leftsum[dep][ub] - leftsum[dep][ua-1];
76 return query(mid+1,r,dep+1,newl,newr,k-cnt);
77 }
78 }
79
80 int main(void)
81 {
82 #ifndef ONLINE_JUDGE
83 freopen("in.txt","r",stdin);
84 #endif
85 int T, cas = 1;
86 scanf ("%d",&T);
87 while (T--)
88 {
89 int n,m;
90 scanf ("%d",&n);
91 sum[0] = 0;
92 for (int i = 1; i <= n; i++)
93 {
94 scanf ("%d",&tree[0][i]);
95 sum[i] = sum[i-1] + tree[0][i];
96 sorted[i] = tree[0][i];
97 }
98 sort(sorted+1,sorted+n+1);
99 build (1,n,0);
100 scanf ("%d",&m);
101 printf ("Case #%d:\n",cas++);
102 while (m--)
103 {
104 int u, v;
105 scanf ("%d%d",&u,&v);
106 u++,v++;
107 lnum = 0;
108 lsum = 0;
109 int mid_num = query(1,n,0,u,v,(v-u)/2+1); //中位数
110 rnum = (v - u + 1 - lnum); // u~v 区间内大于mid_num的个数
111 rsum = (sum[v] - sum[u-1] - lsum); //u~v 区间内大于mid_num的数的和
112 ll ans = rsum - lsum + mid_num * (lnum - rnum);
113 printf("%I64d\n",ans);
114 }
115 printf("\n");
116 }
117 return 0;
118 }
主席树部分代码：（主席树会MLE，已经和划分树代码对拍过，应该没问题）
1 typedef long long ll;
2 const int maxn = 1e5+10;
3 int n,q,tot;
4
5 //主席树部分
6 int lson[maxn*20],rson[maxn*20],c[maxn*20],tree[maxn];
7
8 ll sum[maxn*20];
9 int build (int l,int r)
10 {
11 int root = tot++;
12 c[root] = 0;
13 sum[root] = 0;
14 if (l != r)
15 {
16 int mid = (l + r) >> 1;
17 lson[root] = build(l,mid);
18 rson[root] = build(mid+1,r);
19 }
20 return root;
21 }
22 int update(int root,int pos,int val,int k)
23 {
24 int newroot = tot++;
25 int tmp = newroot;
26 int l = 1, r = n;
27 c[newroot] = c[root] + val;
28 sum[newroot] = sum[root] + k;
29 while (l < r)
30 {
31 int mid = (l + r) >> 1;
32 if (pos <= mid)
33 {
34 rson[newroot] = rson[root];
35 root = lson[root];
36 lson[newroot] = tot++;
37 newroot = lson[newroot];
38 r = mid;
39 }
40 else
41 {
42 lson[newroot] = lson[root];
43 root = rson[root];
44 rson[newroot] = tot++;
45 newroot = rson[newroot];
46 l = mid + 1;
47 }
48 c[newroot] = c[root] + val;
49 sum[newroot] = sum[root] + k;
50 }
51 return tmp;
52 }
53 ll lsum,rsum; //lsum小于中位数的和，rsum大于中位数的和
54 ll query(int root,int l,int r,int ua,int ub) //查询1-root 大于ua小于ub的和
55 {
56 if (ub < ua)
57 return 0;
58 if (ua <= l && ub >= r)
59 {
60 return sum[root];
61 }
62 int mid = (l + r) >> 1;
63 ll t1 = 0,t2 = 0;
64 if (ua <= mid)
65 t1 = query(lson[root],l,mid,ua,ub);
66 if (ub > mid)
67 t2 = query(rson[root],mid+1,r,ua,ub);
68 return t1 + t2;
69 }
70 int query1(int root,int l,int r,int ua,int ub) //查询1-root 在ua ub 之间的数的个数
71 {
72 if (ub < ua)
73 return 0;
74 if (ua <= l && ub >= r)
75 {
76 return c[root];
77 }
78 int mid = (l + r) >> 1;
79 int t1 = 0,t2 = 0;
80 if (ua <= mid)
81 t1 = query1(lson[root],l,mid,ua,ub);
82 if (ub > mid)
83 t2 = query1(rson[root],mid+1,r,ua,ub);
84 return t1 + t2;
85 }
86 int query(int left,int right,int k) //查询left right区间第k大
87 {
88 int l_root = tree[left-1];
89 int r_root = tree[right];
90 int l = 1, r = n;
91 while (l < r)
92 {
93 int mid = (l + r) >> 1;
94 int tmp = c[lson[r_root]] - c[lson[l_root]];
95 if (tmp <= k)
96 {
97 k -= tmp;
98 r_root = rson[r_root];
99 l_root = rson[l_root];
100 l = mid + 1;
101 }
102 else
103 {
104 l_root = lson[l_root];
105 r_root = lson[r_root];
106 r = mid;
107 }
108 }
109 return l;
110 }
111 int vec[maxn],rel[maxn],idx;
112 //离散化
113 inline void init_hash()
114 {
115 sort(vec,vec+idx);
116 idx = unique(vec,vec+idx) - vec;
117 }
118 inline int _hash(ll x)
119 {
120 return lower_bound(vec,vec+idx,x) - vec + 1;
121 }
122
123
124 int main(void)
125 {
126 #ifndef ONLINE_JUDGE
127 freopen("in.txt","r",stdin);
128 freopen("wa.txt","w",stdout);
129 #endif
130 int t,cas = 1;
131 scanf ("%d",&t);
132 while (t--)
133 {
134 memset(sum,0,sizeof(sum));
135 scanf ("%d",&n);
136 idx = tot = 0;
137 for (int i = 1; i<= n; i++)
138 {
139 //scanf ("%d",a+i);
140 scanf ("%d",tree+i);
141 vec[idx++] = tree[i];
142 }
143 init_hash();
144 tree[0] = build(1,n);
145 for (int i = 1; i <= n; i++)
146 {
147 int tmp = _hash(tree[i]);
148 rel[tmp] = tree[i];
149
150 tree[i] = update(tree[i-1],tmp,1,tree[i]);
151 //tree[i] = tmp2;
152 }
153 scanf ("%d",&q);
154 printf("Case #%d:\n",cas++);
155 while (q--)
156 {
157 int u,v;
158 scanf ("%d%d",&u,&v);
159 u++,v++;
160 int vir_mid = query(u,v,(v-u)/2);
161 int mid_num = rel[vir_mid]; //中位数
162 lsum = query(tree[u-1],1,n,1,vir_mid-1);
163 rsum = query(tree[u-1],1,n,vir_mid+1,n);
164 ll x1 = lsum, x2 = rsum;
165
166 lsum = query(tree[v],1,n,1,vir_mid-1);
167 rsum = query(tree[v],1,n,vir_mid+1,n);
168 int lnum = query1(tree[v],1,n,1,vir_mid - 1) - query1(tree[u-1],1,n,1,vir_mid - 1);
169 int rnum = query1(tree[v],1,n,vir_mid + 1,n) - query1(tree[u-1],1,n,vir_mid + 1,n);
170 printf("%I64d\n",rsum - x2 - (lsum - x1) + (lnum - rnum) * mid_num);
171 }
172 printf("\n");
173 }
174 return 0;
175
176 }
时间： 2024-12-09 22:33:47