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 | 1using 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;}