题目链接
很多人写这道题都用的是二维RMQ,但是,我觉得这道题可以锻炼一下我二维线段树的思维,但是,无独有偶,这道题会卡一些二维线段树的模板,一开始我想也没想,直接敲了刚学的线段树,然后不停的RE,后来改了下,换成单点更新与区间更新二维线段树,还是不行TLE了,于是,就开始想,我们该如何把二维线段树双重区间写出来,然后,研究了下他的向下更新操作,我们不妨可以结合一下单点更新作出改进,譬如看向X轴与Y轴,我们先处理X轴的内容,然后对于Y轴,我们在X轴的单位1开始向上延伸,在再Y轴上做区间更新,这样问题就来了,如何判断是否是单位一,单位一时可以直接区间更新,还是easy的,然而不是单位一时,就是长度大于一了,那么我们建外树(X轴的建树)不妨考虑一下先建立完所对应的外树对应节点,然后推上给Y轴相匹配,这里怎么说呢:
if(l == r)
{
if(flag) tree_in[Xi][rt] = a[pos][l];
else tree_in[Xi][rt] = max(tree_in[Xi<<1][rt], tree_in[Xi<<1|1][rt]);
return;
}
在内树中,我们发现其长度为单位一(flag判断是否单位长度为1),于是就直接赋予其点值,否则就是两个X轴上的区间最大值。
建立外树也是有点东西的,我们不能简单的处理,不然岂不是有些内树的赋值不到位就会WA:
int mid = (l + r)>>1;
build_OutTree(rt<<1, l, mid);
build_OutTree(rt<<1|1, mid+1, r);
build_InTree(1, rt, 1, M, false, l);
我们先处理外树,然后再看向内树,毕竟内树的叶子节点是需要外树的赋值的,所以就是有序内外树的。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 305;
int N, M, Q;
ll a[maxN][maxN], tree_in[maxN<<2][maxN<<2];
void build_InTree(int rt, int Xi, int l, int r, bool flag, int pos)
{
if(l == r)
{
if(flag) tree_in[Xi][rt] = a[pos][l];
else tree_in[Xi][rt] = max(tree_in[Xi<<1][rt], tree_in[Xi<<1|1][rt]);
return;
}
int mid = (l + r)>>1;
build_InTree(rt<<1, Xi, l, mid, flag, pos);
build_InTree(rt<<1|1, Xi, mid+1, r, flag, pos);
tree_in[Xi][rt] = max(tree_in[Xi][rt<<1], tree_in[Xi][rt<<1|1]);
}
void build_OutTree(int rt, int l, int r)
{
if(l == r)
{
build_InTree(1, rt, 1, M, true, l);
return;
}
int mid = (l + r)>>1;
build_OutTree(rt<<1, l, mid);
build_OutTree(rt<<1|1, mid+1, r);
build_InTree(1, rt, 1, M, false, l);
}
ll query_in(int rt, int Xi, int l, int r, int ql, int qr)
{
if(ql<=l && qr>=r) return tree_in[Xi][rt];
int mid = (l + r)>>1;
if(ql>mid) return query_in(rt<<1|1, Xi, mid+1, r, ql, qr);
else if(qr<=mid) return query_in(rt<<1, Xi, l, mid, ql, qr);
else
{
ll ans = query_in(rt<<1|1, Xi, mid+1, r, ql, qr);
ans = max(ans, query_in(rt<<1, Xi, l, mid, ql, qr));
return ans;
}
}
ll query_out(int rt, int l, int r, int qlx, int qly, int qrx, int qry)
{
if(qlx<=l && qrx>=r) return query_in(1, rt, 1, M, qly, qry);
int mid = (l + r)>>1;
if(qlx>mid) return query_out(rt<<1|1, mid+1, r, qlx, qly, qrx, qry);
else if(qrx<=mid) return query_out(rt<<1, l, mid, qlx, qly, qrx, qry);
else
{
ll ans = query_out(rt<<1|1, mid+1, r, qlx, qly, qrx, qry);
ans = max(ans, query_out(rt<<1, l, mid, qlx, qly, qrx, qry));
return ans;
}
}
int main()
{
while(scanf("%d%d", &N, &M)!=EOF)
{
for(int i=1; i<=N; i++)
{
for(int j=1; j<=M; j++)
{
scanf("%lld", &a[i][j]);
}
}
build_OutTree(1, 1, N);
scanf("%d", &Q);
while(Q--)
{
int e1, e2, e3, e4;
scanf("%d%d%d%d", &e1, &e2, &e3, &e4);
bool flag = false;
ll tmp = query_out(1, 1, N, e1, e2, e3, e4);
if(a[e1][e2] == tmp || a[e1][e4] == tmp || a[e3][e2] == tmp || a[e3][e4] == tmp) flag = true;
printf("%lld ", tmp);
printf(flag?"yes\n":"no\n");
}
}
return 0;
}