51Nod-1571-最近等对

ACM模版

描述

题解

挺常见的一种线段树问题，首先对 a[] 进行离散化，然后询问离线处理，处理时需要根据右区间进行排序，然后不断维护左区间就好了……每次添加离散化后的 a[i] 时都要处理完询问中右区间端点为 i 的查询，避免后续添加对它的影响。

区间更新，区间查询，注意延迟标记。

代码

#include <iostream>
#include <cstdio>
#include <algorithm>

#define lson rt << 1
#define rson rt << 1 | 1

using namespace std;

const int MAXN = 5e5 + 10;
const int INF = 0x3f3f3f3f;

struct node
{
int l, r, id;
} q[MAXN];  //  查询

int n, m;
int a[MAXN];
int b[MAXN];
int pre[MAXN];
int ans[MAXN];
int minv[MAXN << 2];
int lazy[MAXN << 2];

int cmp(node a, node b)
{
return a.r < b.r;
}

void build(int rt, int l, int r)
{
minv[rt] = n, lazy[rt] = INF;
if (l == r)
{
return ;
}
int m = (l + r) >> 1;
build(lson, l, m);
build(rson, m + 1, r);
}

void pushup(int rt)
{
minv[rt] = min(minv[lson], minv[rson]);
}

void pushdown(int rt)
{
if (lazy[rt] != INF)
{
minv[lson] = min(minv[lson], lazy[rt]);
minv[rson] = min(minv[rson], lazy[rt]);
lazy[lson] = min(lazy[lson], lazy[rt]);
lazy[rson] = min(lazy[rson], lazy[rt]);
lazy[rt] = INF;
}
}

void update(int rt, int l, int r, int l_, int r_, int v)
{
if (l_ <= l && r_ >= r)
{
minv[rt] = min(minv[rt], v);
lazy[rt] = min(lazy[rt], v);
return ;
}

pushdown(rt);
int m = (l + r) >> 1;
if (l_ <= m)
{
update(lson, l, m, l_, r_, v);
}
if (r_ > m)
{
update(rson, m + 1, r, l_, r_, v);
}
pushup(rt);
}

int query(int rt, int l, int r, int l_, int r_)
{
if (l_ <= l && r_ >= r)
{
return minv[rt];
}

pushdown(rt);
int ans = INF, m = (l + r) >> 1;
if (l_ <= m)
{
ans = min(ans, query(lson, l, m, l_, r_));
}
if (r_ > m)
{
ans = min(ans, query(rson, m + 1, r, l_, r_));
}
pushup(rt);

return ans;
}

void slove()
{
build(1, 1, n);

for (int i = 1; i <= n; i++)
{
pre[i] = n;
}

for (int i = 1, j = 1; i <= n && j <= m; i++)
{
if (pre[a[i]] == n) //  记录离散化后的 a[i] 的前一个位置
{
pre[a[i]] = i;
}
else
{
update(1, 1, n, 1, pre[a[i]], i - pre[a[i]]);
pre[a[i]] = i;
}
while (j <= m && q[j].r == i)
{
ans[q[j].id] = query(1, 1, n, q[j].l, n);
j++;
}
}
}

template <class T>
inline bool scan_d(T &ret)
{
char c;
int sgn;
if (c = getchar(), c == EOF)
{
return 0; //EOF
}
while (c != '-' && (c < '0' || c > '9'))
{
c = getchar();
}
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0');
}
ret *= sgn;
return 1;
}