[luoguP1816] 忠誠(RMQ || 線段樹)
阿新 • • 發佈:2017-06-11
onclick spl ide fin pri nbsp log urn def
傳送門
其實我就是想練練 rmq
本以為學了線段樹可以省點事不學 rmq 了
但是後綴數組中用 rmq 貌似很方便
所以還是學了吧,反正也不難
——代碼
1 #include <cstdio> 2 #define N 100001 3 #define min(x, y) ((x) < (y) ? (x) : (y)) 4 5 int n, m; 6 int a[N], d[N][21]; 7 8 int main() 9 { 10 int i, j, k, x, y; 11 scanf("View Code%d %d", &n, &m); 12 for(i = 1; i <= n; i++) scanf("%d", &a[i]); 13 for(i = 1; i <= n; i++) d[i][0] = a[i]; 14 for(j = 1; (1 << j) <= n; j++) 15 for(i = 1; i + (1 << j) - 1 <= n; i++) 16 d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]); 17 for(i = 1; i <= m; i++) 18 { 19 scanf("%d %d", &x, &y); 20 k = 0; 21 while((1 << (k + 1)) <= y - x + 1) k++; 22 printf("%d ", min(d[x][k], d[y - (1 << k) + 1][k])); 23 } 24 return 0; 25 }
[luoguP1816] 忠誠(RMQ || 線段樹)