九度OJ 1544 数字序列区间最小值
题目地址：http://ac.jobdu.com/problem.php?pid=1544
- 题目描述：
-
给定一个数字序列，查询任意给定区间内数字的最小值。
- 输入：
-
输入包含多组测试用例，每组测试用例的开头为一个整数n(1<=n<=100000)，代表数字序列的长度。
接下去一行给出n个数字，代表数字序列。数字在int范围内。
下一行为一个整数t(1<=t<=10000)，代表查询的次数。
最后t行，每行给出一个查询，由两个整数表示l、r(1<=l<=r<=n)。
- 输出：
-
对于每个查询，输出区间[l,r]内的最小值。
- 样例输入：
-
5 3 2 1 4 3 3 1 3 2 4 4 5
- 样例输出：
-
1 1 3
#include <stdio.h>
#include <math.h>
#include <limits.h>
void Preprocess (int arr[], int n, int loc[]){
int i, j, k;
int step;
int min;
step = (int)sqrt((double)n);
i = j = k = 0;
while (1){
min = k*step;
for (j=k*step+1; j<n && j<(k+1)*step; ++j){
if (arr[min] > arr[j])
min = j;
}
loc[k] = min;
++k;
if (j >= n)
break;
}
}
int main(void){
int n;
int arr[100001];
int loc[1000];
int t;
int i;
int left;
int right;
int min;
int step;
int index;
while (scanf ("%d", &n) != EOF){
step = (int)sqrt((double)n);
for (i=0; i<n; ++i){
scanf ("%d", &arr[i]);
}
Preprocess (arr, n, loc);
scanf ("%d", &t);
while (t-- != 0){
scanf ("%d%d", &left, &right);
--left;
--right;
min = INT_MAX;
while (left <= right && left % step != 0){
if (arr[left] < min){
min = arr[left];
}
++left;
}
while (left + step <= right){
index = loc[left/step];
if (arr[index] < min){
min = arr[index];
}
left += step;
}
while (left <= right){
if (arr[left] < min){
min = arr[left];
}
++left;
}
printf ("%d\n", min);
}
}
return 0;
}
/**************************************************************
Problem: 1544
User: 简简单单Plus
Language: C
Result: Accepted
Time:330 ms
Memory:1252 kb
****************************************************************/
Sparse Table (ST) algorithm
//Sparse Table (ST) algorithm
#include <stdio.h>
#include <math.h>
#include <limits.h>
#define MAX 100000
#define LOGMAX 20
void Preprocess (int M[MAX][LOGMAX], int arr[MAX], int n){
int i, j;
for (i = 0; i < n; ++i)
M[i][0] = i;
for (j = 1; 1 << j <= n; ++j){
for (i = 0; i + (1 << j) - 1 < n; ++i){
if (arr[M[i][j - 1]] < arr[M[i + (1 << (j - 1))][j - 1]])
M[i][j] = M[i][j - 1];
else
M[i][j] = M[i + (1 << (j - 1))][j - 1];
}
}
}
int main(void){
int n;
int arr[MAX];
int M[MAX][LOGMAX];
int test;
int left;
int right;
int i;
int min;
int k;
while (scanf ("%d", &n) != EOF){
for (i = 0; i < n; ++i){
scanf ("%d", &arr[i]);
}
Preprocess (M, arr, n);
scanf ("%d", &test);
while (test-- != 0){
scanf ("%d%d", &left, &right);
--left;
--right;
k = 0;
while ((1 << (k + 1)) < (right - left + 1))
++k;
if (arr[M[left][k]] <= arr[M[right - (1 << k) + 1][k]])
min = arr[M[left][k]];
else
min = arr[M[right - (1 << k) + 1][k]];
printf ("%d\n", min);
}
}
return 0;
}
/**************************************************************
Problem: 1544
User: 简简单单Plus
Language: C
Result: Accepted
Time:270 ms
Memory:9040 kb
****************************************************************/
Segment trees
#include <stdio.h>
#include <math.h>
#define MAX 100000
#define MAXN 200000
void Initialize (int node, int b, int e, int M[MAXN], int arr[MAX], int n){
if (b == e){
M[node] = b;
}
else{
//compute the values in the left and right subtrees
Initialize(2 * node, b, (b + e) / 2, M, arr, n);
Initialize(2 * node + 1, (b + e) / 2 + 1, e, M, arr, n);
//search for the minimum value in the first and
//second half of the interval
if (arr[M[2 * node]] <= arr[M[2 * node + 1]])
M[node] = M[2 * node];
else
M[node] = M[2 * node + 1];
}
}
int Query(int node, int b, int e, int M[MAXN], int arr[MAX], int i, int j)
{
int p1, p2;
//if the current interval doesn't intersect
//the query interval return -1
if (i > e || j < b)
return -1;
//if the current interval is included in
//the query interval return M[node]
if (b >= i && e <= j)
return M[node];
//compute the minimum position in the
//left and right part of the interval
p1 = Query(2 * node, b, (b + e) / 2, M, arr, i, j);
p2 = Query(2 * node + 1, (b + e) / 2 + 1, e, M, arr, i, j);
//return the position where the overall
//minimum is
if (p1 == -1)
//return M[node] = p2;
return p2;
if (p2 == -1)
//return M[node] = p1;
return p1;
if (arr[p1] <=arr[p2])
return p1;
return p2;
}
int main(void){
int n;
int arr[MAX];
int M[MAXN];
int t;
int left;
int right;
int i;
int index;
while (scanf ("%d", &n) != EOF){
for (i = 0; i < n; ++i){
scanf ("%d", &arr[i]);
}
for (i=0; i<MAXN; ++i){
M[i] = -1;
}
Initialize (1, 0, n - 1, M, arr, n);
scanf ("%d", &t);
while (t-- != 0){
scanf ("%d%d", &left, &right);
--left;
--right;
index = Query(1, 0, n - 1, M, arr, left, right);
printf ("%d\n", arr[index]);
}
}
return 0;
}
/**************************************************************
Problem: 1544
User: 简简单单Plus
Language: C
Result: Accepted
Time:290 ms
Memory:2016 kb
****************************************************************/
参考资料：http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor
http://dongxicheng.org/structure/lca-rmq/
http://dongxicheng.org/structure/segment-tree/