Luogu P2880 [USACO07JAN]平衡的陣容Balanced Lineup (ST表模板)
阿新 • • 發佈:2018-12-04
傳送門(ST表裸題)
ST表是一種很優雅的演算法,用於求靜態RMQ
陣列l[i][j]表示從i開始,長度為2^j的序列中的最大值
注意事項:
1.核心部分:
for(int j = 1; (1<<j) <= n; j++)
for(int i = 1; i+(1<<j)-1 <= n; i++) {
l[i][j] = max(l[i][j-1],l[i+(1<<(j-1))][j-1]);
s[i][j] = min(s[i][j-1],s[i+(1<<(j-1 ))][j-1]);
}
因為i~j的位數是j-i+1位,所以迴圈的邊界需要-1,而所求的兩段區間是不相交的,所以迴圈內不用-1(或者說,-1又+1了)
2.位運算需要頻繁地打括號
程式碼如下
#include<cstdio>
#include<iostream>
using namespace std;
const int maxn = 50005;
int n,q;
int a[maxn],l[maxn][50],s[maxn][50];
int al,as,x,y;
int main() {
scanf("%d%d ",&n,&q);
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
l[i][0] = a[i];
s[i][0] = a[i];
}
for(int j = 1; (1<<j) <= n; j++)
for(int i = 1; i+(1<<j)-1 <= n; i++) {
l[i][j] = max(l[i][j-1],l[i+(1<<(j-1))][j-1 ]);
s[i][j] = min(s[i][j-1],s[i+(1<<(j-1))][j-1]);
}
while(q) {
q--;
scanf("%d%d",&x,&y);
int k = 0;
while(x+(1<<(k+1))<= y)k++;
al = max(l[x][k],l[y-(1<<k)+1][k]);
as = min(s[x][k],s[y-(1<<k)+1][k]);
printf("%d\n",al-as);
}
return 0;
}