POJ - 3264 Balanced Lineup（线段树或RMQ）

题意：求区间最大值-最小值。

分析：

1、线段树

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 50000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int minv[MAXN << 2], maxv[MAXN << 2];
int _min, _max;
void build(int id, int L, int R){
    if(L == R){
        minv[id] = maxv[id] = a[L];
    }
    else{
        int mid = L + (R - L) / 2;
        build(id << 1, L, mid);
        build(id << 1 | 1, mid + 1, R);
        minv[id] = min(minv[id << 1], minv[id << 1 | 1]);
        maxv[id] = max(maxv[id << 1], maxv[id << 1 | 1]);
    }
}
void query(int l, int r, int id, int L, int R){
    if(l <= L && R <= r){
        _max = max(_max, maxv[id]);
        _min = min(_min, minv[id]);
        return;
    }
    int mid = L + (R - L) / 2;
    if(l <= mid){
        query(l, r, id << 1, L, mid);
    }
    if(r > mid){
        query(l, r, id << 1 | 1, mid + 1, R);
    }
}
int main(){
    int N, Q;
    scanf("%d%d", &N, &Q);
    for(int i = 1; i <= N; ++i){
        scanf("%d", &a[i]);
    }
    build(1, 1, N);
    while(Q--){
        int A, B;
        scanf("%d%d", &A, &B);
        _min = INT_INF;
        _max = 0;
        query(A, B, 1, 1, N);
        printf("%d\n", _max - _min);
    }
    return 0;
}

2、RMQ

Sparse-Table算法，预处理时间O(nlogn),查询O(1)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 50000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int N, Q;
int minv[MAXN][20];
int maxv[MAXN][20];
void RMQ_init(){
    for(int i = 1; i <= N; ++i){
        minv[i][0] = maxv[i][0] = a[i];
    }
    for(int j = 1; (1 << j) <= N; ++j){
        for(int i = 1; (i + (1 << j) - 1) <= N; ++i){
            minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]);
            maxv[i][j] = max(maxv[i][j - 1], maxv[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int RMQ(int L, int R){
    int k = 0;
    while((1 << (k + 1)) <= (R - L + 1)) ++k;
    return max(maxv[L][k], maxv[R - (1 << k) + 1][k]) - min(minv[L][k], minv[R - (1 << k) + 1][k]);
}
int main(){
    scanf("%d%d", &N, &Q);
    for(int i = 1; i <= N; ++i){
        scanf("%d", &a[i]);
    }
    RMQ_init();
    while(Q--){
        int A, B;
        scanf("%d%d", &A, &B);
        printf("%d\n", RMQ(A, B));
    }
    return 0;
}