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poj3264RMQ
区间最值。学了下 st算法,o(1)的查询,这个要比线段树犀利。而且线段树的log(n)前面的常数也比较大。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n, q;const int maxn = 111111;int dp[maxn][20];int dp1[maxn][20];int a[maxn];void gao(){ for (int i = 0; i < n; i++) dp[i][0] = a[i]; for (int j = 1; (1 << j) <= n; j++){ for (int i = 0; i + (1 << j) - 1 < n; i++){ dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j-1))][j - 1]); } }}int ask(int l, int r){ int k = 0; while ((1 << (k + 1)) <= r - l + 1) k++; return min(dp[l][k], dp[r-(1<<k)+1][k]);}void gao1(){ memset(dp1,0,sizeof(dp1)); for (int i = 0; i < n; i++) dp1[i][0] = a[i]; for (int j = 1; (1 << j) <= n; j++){ for (int i = 0; i + (1 << j) - 1 < n; i++){ dp1[i][j] = max(dp1[i][j - 1], dp1[i + (1 << (j-1))][j - 1]); } }}int ask1(int l, int r){ int k = 0; while ((1 << (k + 1)) <= r - l + 1) k++; return max(dp1[l][k], dp1[r + 1 - (1 << k)][k]);}int main(){ int l, r; while (cin >> n >> q){ for (int i = 0; i < n; i++) scanf("%d",&a[i]); gao(); gao1(); for (int i = 0; i < q; i++){ scanf("%d%d",&l,&r); l--;r--; printf("%d\n",ask1(l,r)- ask(l,r)); } } return 0;}
poj3264RMQ
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