RMQ問題模板
阿新 • • 發佈：2019-02-24
ring turn string 兩種 cmp eof 最大值 cstring print
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define MAXN 10010 #define INF 1000000000 using namespace std; int A[MAXN]; int N, M; int Amax[MAXN][50];//Amax[i][j]表示從i開始的，長度為2的j次方的區間裏面的最大值 int Amin[MAXN][50];//Amin[i][j]表示從i開始的，長度為2的j次方的區間裏面的最小值 void RMQ_init() {for(int i = 1; i <= N; i++) Amax[i][0] = Amin[i][0] = A[i]; for(int j = 1; (1<<j) <= N; j++) { for(int i = 1; i + (1<<j)-1 <= N; i++) { Amax[i][j] = max(Amax[i][j-1], Amax[i+(1<<(j-1))][j-1]); Amin[i][j] = min(Amin[i][j-1], Amin[i+(1<<(j-1))][j-1]); } } } int query(char *op, int L, int R) { //k值求法有兩種 int k = 0; while((1<<(k+1)) <= R-L+1) k++; //int k = (int)(log(double(R - L + 1))/log((double)2)); if(strcmp(op, "MAX") == 0) return max(Amax[L][k], Amax[R-(1<<k)+1][k]); else return min(Amin[L][k], Amin[R-(1<<k)+1][k]); } int main() { while(scanf("%d%d", &N, &M), N||M) { for(int i = 1; i <= N; i++) scanf("%d", &A[i]); RMQ_init(); char op[10]; int x, y; while(M--) { scanf("%s%d%d", op, &x, &y); printf("%d\n", query(op, x, y)); } } return 0; }
優化版：
#include <cstdio> #include <cstring> #include <algorithm> #define MAXN 1000000+10 using namespace std; int N, M, k; int A[MAXN]; int Amax[MAXN]; int Amin[MAXN]; void RMQ_init() { for(int i = 1; i <= N; i++) Amax[i] = Amin[i] = A[i]; for(int j = 1; (1<<j) <= k; j++) { for(int i = 1; i + (1<<j) - 1 <= N; i++) { Amax[i] = max(Amax[i], Amax[i + (1<<(j-1))]);//壓縮 Amin[i] = min(Amin[i], Amin[i + (1<<(j-1))]); } } } int query(char *op, int L, int R) { int len = 0; while(1<<(len+1) <= R-L+1) len++;//求len值 if(strcmp(op, "MAX") == 0) return max(Amax[L], Amax[R-(1<<len)+1]); else return min(Amin[L], Amin[R-(1<<len)+1]); } int main() { while(scanf("%d%d%d", &N, &M, &k) != EOF) { for(int i = 1; i <= N; i++) scanf("%d", &A[i]); RMQ_init(); char op[10]; int x; while(M--) { scanf("%s%d", op, &x); printf("%d\n", query(op, x, x+k-1)); } } return 0; }
RMQ問題模板