【HDOJ】4251 The Famous ICPC Team Again

划分树模板题目,主席树也可解。
划分树。

  1 /* 4251 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <algorithm>
 12 #include <cstdio>
 13 #include <cmath>
 14 #include <ctime>
 15 #include <cstring>
 16 #include <climits>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <functional>
 20 #include <iterator>
 21 #include <iomanip>
 22 using namespace std;
 23 //#pragma comment(linker,"/STACK:102400000,1024000")
 24
 25 #define sti                set<int>
 26 #define stpii            set<pair<int, int> >
 27 #define mpii            map<int,int>
 28 #define vi                vector<int>
 29 #define pii                pair<int,int>
 30 #define vpii            vector<pair<int,int> >
 31 #define rep(i, a, n)     for (int i=a;i<n;++i)
 32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 33 #define clr                clear
 34 #define pb                 push_back
 35 #define mp                 make_pair
 36 #define fir                first
 37 #define sec                second
 38 #define all(x)             (x).begin(),(x).end()
 39 #define SZ(x)             ((int)(x).size())
 40 #define lson            l, mid, rt<<1
 41 #define rson            mid+1, r, rt<<1|1
 42
 43 const int maxn = 1e5+5;
 44 int order[maxn];
 45 int toLeft[18][maxn];
 46 int val[18][maxn];
 47 int n;
 48
 49 void Build(int l, int r, int dep) {
 50     if (l == r) {
 51         toLeft[dep][l] = toLeft[dep][l-1] + 1;
 52         return ;
 53     }
 54
 55     int mid = (l + r) >> 1;
 56     int same = mid - l + 1;
 57
 58     rep(i, l, r+1) {
 59         if (val[dep][i] < order[mid])
 60             --same;
 61     }
 62
 63     int lpos = l, rpos = mid + 1;
 64
 65     rep(i, l, r+1) {
 66         if (val[dep][i] < order[mid]) {
 67             val[dep+1][lpos++] = val[dep][i];
 68         } else if (val[dep][i]==order[mid] && same>0) {
 69             val[dep+1][lpos++] = val[dep][i];
 70             --same;
 71         } else {
 72             val[dep+1][rpos++] = val[dep][i];
 73         }
 74         toLeft[dep][i] = toLeft[dep][l-1] + lpos - l;
 75     }
 76
 77     Build(l, mid, dep+1);
 78     Build(mid+1, r, dep+1);
 79 }
 80
 81 int Query(int l, int r, int k, int L, int R, int dep) {
 82     if (l == r) {
 83         return val[dep][l];
 84     }
 85
 86     int mid = (L + R) >> 1;
 87     int tmp = toLeft[dep][r] - toLeft[dep][l-1];
 88
 89     if (tmp >= k) {
 90         int ll = L + toLeft[dep][l-1] - toLeft[dep][L-1];
 91         int rr = ll + tmp - 1;
 92
 93         return Query(ll, rr, k, L, mid, dep+1);
 94     } else {
 95         k -= tmp;
 96         int ll = mid+1 + l-L - (toLeft[dep][l-1] - toLeft[dep][L-1]);
 97         int rr = ll + (r-l+1-tmp) - 1;
 98
 99         return Query(ll, rr, k, mid+1, R, dep+1);
100     }
101 }
102
103 void solve() {
104     sort(order+1, order+1+n);
105     Build(1, n, 0);
106
107     int q;
108     int l, r, kth;
109     int ans;
110
111     scanf("%d", &q);
112     while (q--) {
113         scanf("%d %d", &l, &r);
114         kth = ((l+r)>>1) - l + 1;
115         ans = Query(l, r, kth, 1, n, 0);
116         printf("%d\n", ans);
117     }
118 }
119
120 int main() {
121     ios::sync_with_stdio(false);
122     #ifndef ONLINE_JUDGE
123         freopen("data.in", "r", stdin);
124         freopen("data.out", "w", stdout);
125     #endif
126
127     int t = 0;
128
129     while (scanf("%d", &n)!=EOF) {
130         rep(i, 1, n+1) {
131             scanf("%d", &val[0][i]);
132             order[i] = val[0][i];
133         }
134         printf("Case %d:\n", ++t);
135         solve();
136     }
137
138     #ifndef ONLINE_JUDGE
139         printf("time = %d.\n", (int)clock());
140     #endif
141
142     return 0;
143 }

主席树。

  1 /* 4251 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <algorithm>
 12 #include <cstdio>
 13 #include <cmath>
 14 #include <ctime>
 15 #include <cstring>
 16 #include <climits>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <functional>
 20 #include <iterator>
 21 #include <iomanip>
 22 using namespace std;
 23 //#pragma comment(linker,"/STACK:102400000,1024000")
 24
 25 #define sti                set<int>
 26 #define stpii            set<pair<int, int> >
 27 #define mpii            map<int,int>
 28 #define vi                vector<int>
 29 #define pii                pair<int,int>
 30 #define vpii            vector<pair<int,int> >
 31 #define rep(i, a, n)     for (int i=a;i<n;++i)
 32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 33 #define clr                clear
 34 #define pb                 push_back
 35 #define mp                 make_pair
 36 #define fir                first
 37 #define sec                second
 38 #define all(x)             (x).begin(),(x).end()
 39 #define SZ(x)             ((int)(x).size())
 40 // #define lson            l, mid, rt<<1
 41 // #define rson            mid+1, r, rt<<1|1
 42
 43 const int maxn = 1e5+5;
 44 const int maxm = maxn * 20;
 45 int a[maxn], b[maxn];
 46 int T[maxn];
 47 int lson[maxm], rson[maxm], c[maxm];
 48 int n, m, tot;
 49
 50 void init() {
 51     m = tot = 0;
 52 }
 53
 54 int Build(int l, int r) {
 55     int rt = tot++;
 56
 57     c[rt] = 0;
 58     if (l == r)
 59         return rt;
 60
 61     int mid = (l + r) >> 1;
 62
 63     lson[rt] = Build(l, mid);
 64     rson[rt] = Build(mid+1, r);
 65     return rt;
 66 }
 67
 68 int Insert(int rt, int p, int delta) {
 69     int nrt = tot++, ret = nrt;
 70     int l = 0, r = m - 1, mid;
 71
 72     c[nrt] = c[rt] + delta;
 73     while (l < r) {
 74         mid = (l + r) >> 1;
 75         if (p <= mid) {
 76             lson[nrt] = tot++;
 77             rson[nrt] = rson[rt];
 78             nrt = lson[nrt];
 79             rt = lson[rt];
 80             r = mid;
 81         } else {
 82             lson[nrt] = lson[rt];
 83             rson[nrt] = tot++;
 84             nrt = rson[nrt];
 85             rt = rson[rt];
 86             l = mid + 1;
 87         }
 88         c[nrt] = c[rt] + delta;
 89     }
 90
 91     return ret;
 92 }
 93
 94 int Query(int lrt, int rrt, int k, int l, int r) {
 95     if (l == r)
 96         return l;
 97
 98     int mid = (l + r) >> 1;
 99     int tmp = c[lson[rrt]] - c[lson[lrt]];
100
101     if (tmp >= k) {
102         return Query(lson[lrt], lson[rrt], k, l, mid);
103     } else {
104         k -= tmp;
105         return Query(rson[lrt], rson[rrt], k, mid+1, r);
106     }
107 }
108
109 void solve() {
110     init();
111     sort(b, b+n);
112     m = unique(b, b+n) - b;
113     T[0] = Build(0, m-1);
114     rep(i, 0, n) {
115         int id = lower_bound(b, b+m, a[i]) - b;
116         T[i+1] = Insert(T[i], id, 1);
117     }
118
119     int q;
120     int l, r, kth;
121     int ans;
122
123     scanf("%d", &q);
124     while (q--) {
125         scanf("%d %d", &l, &r);
126         kth = ((l+r)>>1)-l+1;
127         int id = Query(T[l-1], T[r], kth, 0, m-1);
128         ans = b[id];
129         printf("%d\n", ans);
130     }
131 }
132
133 int main() {
134     ios::sync_with_stdio(false);
135     #ifndef ONLINE_JUDGE
136         freopen("data.in", "r", stdin);
137         freopen("data.out", "w", stdout);
138     #endif
139
140     int t = 0;
141
142     while (scanf("%d", &n)!=EOF) {
143         rep(i, 0, n) {
144             scanf("%d", &a[i]);
145             b[i] = a[i];
146         }
147         printf("Case %d:\n", ++t);
148         solve();
149     }
150
151     #ifndef ONLINE_JUDGE
152         printf("time = %d.\n", (int)clock());
153     #endif
154
155     return 0;
156 }
时间: 2024-10-05 19:11:55

【HDOJ】4251 The Famous ICPC Team Again的相关文章

HDOJ 4251 The Famous ICPC Team Again

划分树水题..... The Famous ICPC Team Again Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 859    Accepted Submission(s): 415 Problem Description When Mr. B, Mr. G and Mr. M were preparing for the

【HDOJ】4956 Poor Hanamichi

基本数学题一道,看错位数,当成大数减做了,而且还把方向看反了.所求为最接近l的值. 1 #include <cstdio> 2 3 int f(__int64 x) { 4 int i, sum; 5 6 i = sum = 0; 7 while (x) { 8 if (i & 1) 9 sum -= x%10; 10 else 11 sum += x%10; 12 ++i; 13 x/=10; 14 } 15 return sum; 16 } 17 18 int main() { 1

【HDOJ】1099 Lottery

题意超难懂,实则一道概率论的题目.求P(n).P(n) = n*(1+1/2+1/3+1/4+...+1/n).结果如果可以除尽则表示为整数,否则表示为假分数. 1 #include <cstdio> 2 #include <cstring> 3 4 #define MAXN 25 5 6 __int64 buf[MAXN]; 7 8 __int64 gcd(__int64 a, __int64 b) { 9 if (b == 0) return a; 10 else return

【HDOJ】2844 Coins

完全背包. 1 #include <stdio.h> 2 #include <string.h> 3 4 int a[105], c[105]; 5 int n, m; 6 int dp[100005]; 7 8 int mymax(int a, int b) { 9 return a>b ? a:b; 10 } 11 12 void CompletePack(int c) { 13 int i; 14 15 for (i=c; i<=m; ++i) 16 dp[i]

【HDOJ】3509 Buge&#39;s Fibonacci Number Problem

快速矩阵幂,系数矩阵由多个二项分布组成.第1列是(0,(a+b)^k)第2列是(0,(a+b)^(k-1),0)第3列是(0,(a+b)^(k-2),0,0)以此类推. 1 /* 3509 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #incl

【HDOJ】1818 It&#39;s not a Bug, It&#39;s a Feature!

状态压缩+优先级bfs. 1 /* 1818 */ 2 #include <iostream> 3 #include <queue> 4 #include <cstdio> 5 #include <cstring> 6 #include <cstdlib> 7 #include <algorithm> 8 using namespace std; 9 10 #define MAXM 105 11 12 typedef struct {

【HDOJ】2424 Gary&#39;s Calculator

大数乘法加法,直接java A了. 1 import java.util.Scanner; 2 import java.math.BigInteger; 3 4 public class Main { 5 public static void main(String[] args) { 6 Scanner cin = new Scanner(System.in); 7 int n; 8 int i, j, k, tmp; 9 int top; 10 boolean flag; 11 int t

【HDOJ】2425 Hiking Trip

优先级队列+BFS. 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 7 #define MAXN 25 8 9 typedef struct node_st { 10 int x, y, t; 11 node_st() {} 12 node_st(int xx, int yy, int tt)

【HDOJ】1686 Oulipo

kmp算法. 1 #include <cstdio> 2 #include <cstring> 3 4 char src[10005], des[1000005]; 5 int next[10005], total; 6 7 void kmp(char des[], char src[]){ 8 int ld = strlen(des); 9 int ls = strlen(src); 10 int i, j; 11 12 total = i = j = 0; 13 while (