线段树解法
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 100010;
struct node
{
int l, r, mid, minn;
}tree[maxn<<2];
int a[maxn];
void build(int l, int r, int o)
{
tree[o].l = l;
tree[o].r = r;
int m = (l+r) >> 1;
tree[o].mid = m;
if (l == r)
{
tree[o].minn = a[l];
return ;
}
build(l, m, o<<1);
build(m+1, r, (o<<1)+1);
tree[o].minn = min(tree[o<<1].minn, tree[(o<<1)+1].minn);
}
int query(int l, int r, int o)
{
if (tree[o].l == l && tree[o].r == r)
return tree[o].minn;
if (r <= tree[o].mid)
return query(l, r, o<<1);
else if (l > tree[o].mid)
return query(l, r, (o<<1)+1);
else
return min(query(l, tree[o].mid, o<<1), query(tree[o].mid+1, r, (o<<1)+1));
}
int main()
{
int t, n, m, l, r;
scanf("%d",&t);
for (int k = 1; k <= t; k++)
{
scanf("%d%d",&n, &m);
for (int i = 1; i <= n; i++)
scanf("%d",&a[i]);
build(1, n, 1);
printf("Case %d:\n",k);
while (m--)
{
scanf("%d %d",&l, &r);
printf("%d\n",query(l, r, 1));
}
}
return 0;
}
Sparse-Table 算法 刘汝佳 训练指南 p197
ST算法:先是预处理部分(构造RMQ数组),DP处理。假设b是所求区间最值的数列,dp[i][j] 表示从i到i+2^j -1中最值(从i开始持续2^j个数)。即dp[i][j]=min{dp[i][j-1],dp[i+2^(j-1)][j-1]},或者dp[i][j]=max{dp[i][j-1],dp[i+2^(j-1)][j-1]},这个过程的复杂度为:O(n(longn))
接着就是查询最值了,可以通过在O(1)完成查询。就是将查询区间[s,v],分成两个2^k的区间。
这里只要知道这种算法即可,因为数据量过大,都编译不通过,不过思想算法没有任何问题。
解题代码
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int a[maxn];
int d[maxn][maxn];
void rmqinit(int n)
{
for (int i = 1; i <= n; i++)
d[i][0] = a[i];
for (int j = 1; (1<<j) <= n; j++)
{
for (int i = 1; i+j-1 <= n; i++)
d[i][j] = min(d[i][j-1], d[i+(1<<(j-1))][j-1]);
}
}
int rmq(int l, int r)
{
int k = 0;
while ((1<<(k+1)) <= r-l+1) k++;
return min(d[l][k], d[r-(1<<k)+1][k]);
}
int main()
{
int t, n, m, l, r;
scanf("%d",&t);
for (int k = 1; k <= t; k++)
{
scanf("%d%d",&n, &m);
for (int i = 1; i <= n; i++)
scanf("%d",&a[i]);
rmqinit(n);
printf("Case %d:\n",k);
while (m--)
{
scanf("%d %d",&l, &r);
printf("%d\n",rmq(l, r));
}
}
return 0;
}