【解题报告】BZOJ1858 裸线段树

给定n个0/1，m个操作，操作1是区间清零，操作2是区间置1，操作3区间0/1互换，操作4查询有多少个1，操作5查询最多有多少个连续的1。

一看就是线段树，维护一段区间的前趋后继和最多连续1以及sum即可。

debug了好久发现自己pushdown写错了一句话。心痛，本来一小时的事情(敲代码还是慢啊)  我觉得我写的比网上的题解优雅多了

#include<iostream>#include<string>#include<string.h>#include<stdlib.h>#include<cstdio>#include<stdio.h>#include<set>#include<map>#include<deque>#include<stack>#include<vector>#include<queue>#include<algorithm>#include<cstring>#include<cmath>#include<sstream>#include<istream>#include<ostream>#include<sstream>using namespace std;#define lc n << 1#define rc n << 1 | 1const int maxn = 100010;struct segtree {    int l, r;    int pre[2], suf[2], sum, mx[2];    bool txor;    int lazy;}t[maxn << 2];inline int getlen(int n) { return t[n].r - t[n].l + 1; }inline void pushup(int n){    t[n].sum = t[lc].sum + t[rc].sum;    for (int i = 0;i < 2;i++)    {        t[n].pre[i] = t[lc].pre[i];        if (t[n].pre[i] == getlen(lc)) { t[n].pre[i] += t[rc].pre[i]; }        t[n].suf[i] = t[rc].suf[i];        if (t[n].suf[i] == getlen(rc)) { t[n].suf[i] += t[lc].suf[i]; }        t[n].mx[i] = max(t[lc].mx[i], t[rc].mx[i]);        t[n].mx[i] = max(t[n].mx[i], t[lc].suf[i] + t[rc].pre[i]);    }} inline void setstatus(int n, int cmd){    if (cmd == 0)    {        t[n].sum = t[n].pre[1] = t[n].suf[1] = t[n].mx[1] = 0;        t[n].pre[0] = t[n].suf[0] = t[n].mx[0] = getlen(n);        t[n].lazy = -1;t[n].txor = false;    }    else if (cmd == 1)    {        t[n].sum = t[n].pre[1] = t[n].suf[1] = t[n].mx[1] = getlen(n);        t[n].pre[0] = t[n].suf[0] = t[n].mx[0] = 0;        t[n].lazy = 1;t[n].txor = false;    }    else if (cmd == 2)    {        t[n].sum = getlen(n) - t[n].sum;        swap(t[n].pre[0], t[n].pre[1]);        swap(t[n].suf[0], t[n].suf[1]);        swap(t[n].mx[0], t[n].mx[1]);        if (t[n].lazy != 0)            t[n].lazy = -t[n].lazy;        else            t[n].txor = !t[n].txor;    }}inline void pushdown(int n){    if (t[n].lazy != 0)    {        t[lc].lazy = t[rc].lazy = t[n].lazy;        t[lc].txor = t[rc].txor = false;        int x = t[n].lazy == 1;        t[lc].sum = x * getlen(lc);        t[rc].sum = x * getlen(rc);        t[lc].pre[x] = t[lc].suf[x] = t[lc].mx[x] = getlen(lc);        t[rc].pre[x] = t[rc].suf[x] = t[rc].mx[x] = getlen(rc);        t[lc].pre[x ^ 1] = t[lc].suf[x ^ 1] = t[lc].mx[x ^ 1]            = t[rc].pre[x ^ 1] = t[rc].suf[x ^ 1] = t[rc].mx[x ^ 1] = 0;        t[n].lazy = 0;    }    else if (t[n].txor)    {        setstatus(lc, 2);        setstatus(rc, 2);        t[n].txor = false;    }}void build(int n, int l, int r){    t[n].l = l;t[n].r = r;    if (l == r)    {        scanf("%d", &t[n].sum);        t[n].pre[1] = t[n].suf[1] = t[n].sum;        t[n].pre[0] = t[n].suf[0] = !t[n].pre[1];        t[n].mx[1] = t[n].sum;        t[n].mx[0] = !t[n].mx[1];        return;    }    int mid = l + ((r - l) >> 1);    build(lc, l, mid);    build(rc, mid + 1, r);    pushup(n);}void update(int n, int l, int r, int cmd){    if (l > t[n].r || r < t[n].l)return;    if (l <= t[n].l && r >= t[n].r)    {        setstatus(n, cmd);        return;    }    pushdown(n);    update(lc, l, r, cmd);    update(rc, l, r, cmd);    pushup(n);}int qsum(int n, int l, int r){    if (l > t[n].r || r < t[n].l)return 0;    if (l <= t[n].l && r >= t[n].r)return t[n].sum;    pushdown(n);    return qsum(lc, l, r) + qsum(rc, l, r);}int qpre(int n, int l, int r){    if (l > t[n].r || r < t[n].l)return 0;    if (l <= t[n].l && r >= t[n].r)return t[n].pre[1];    pushdown(n);    if (r <= t[lc].r)return qpre(lc, l, r);    if (l >= t[rc].l)return qpre(rc, l, r);    int ret = qpre(lc, l, r);    if (ret == getlen(lc))        return ret + qpre(rc, l, r);    return ret;}int qsuf(int n, int l, int r){    if (l > t[n].r || r < t[n].l)return 0;    if (l <= t[n].l && r >= t[n].r)return t[n].suf[1];    pushdown(n);    if (l >= t[rc].l)return qsuf(rc, l, r);    if (r <= t[lc].r)return qsuf(lc, l, r);    int ret = qsuf(rc, l, r);    if (ret == getlen(rc))        return ret + qsuf(lc, l, r);    return ret;}int qmax(int n, int l, int r){    if (l > t[n].r || r < t[n].l)return 0;    if (l <= t[n].l && r >= t[n].r)return t[n].mx[1];    pushdown(n);    if (r <= t[lc].r)return qmax(lc, l, r);    if (l >= t[rc].l)return qmax(rc, l, r);    int ret = max(qmax(lc, l, r), qmax(rc, l, r));    return max(ret, qsuf(lc, l, r) + qpre(rc, l, r));}int main(){    int n, m, c, l, r;    scanf("%d%d", &n, &m);    memset(t, 0, sizeof t);    build(1, 1, n);    while (m--)    {        scanf("%d%d%d", &c, &l, &r);        if (c < 3)update(1, l + 1, r + 1, c);        else if (c == 3)printf("%d\n", qsum(1, l + 1, r + 1));        else if (c == 4)printf("%d\n", qmax(1, l + 1, r + 1));      }    return 0;}