POJ 3667 & 1823 Hotel (线段树区间合并)

来源：互联网发布：视频文件加密软件编辑：程序博客网时间：2024/06/07 07:58

两个题目都是用同一个模板，询问最长的连续未覆盖的区间。

lazy代表是否有人，msum代表区间内最大的连续长度，lsum是从左结点往右的连续长度，rsum是从右结点往左的连续长度。

区间合并很恶心啊，各种左左右右左右左右........

#include <cstdio>#include <iostream>#include <cstring>#include <cmath># define MAX 211111# define ll(x) x << 1# define rr(x) x << 1 | 1using namespace std;struct node {    int l,r,mid,lazy;    int msum,lsum,rsum;} tree[MAX*4];void up(int x) {    int p = tree[x].r - tree[x].l + 1;    tree[x].msum = max(tree[rr(x)].lsum + tree[ll(x)].rsum, max(tree[ll(x)].msum, tree[rr(x)].msum));    tree[x].lsum = tree[ll(x)].lsum;    tree[x].rsum = tree[rr(x)].rsum;    if(tree[x].lsum == p - (p >> 1)) tree[x].lsum += tree[rr(x)].lsum;    if(tree[x].rsum == (p >> 1)) tree[x].rsum += tree[ll(x)].rsum;}void down (int x) {    int p = tree[x].r - tree[x].l + 1;    if(tree[x].lazy != -1) {        tree[ll(x)].lazy = tree[rr(x)].lazy = tree[x].lazy;        tree[ll(x)].lsum = tree[ll(x)].rsum = tree[ll(x)].msum = tree[x].lazy ? 0 : p - (p>>1);        tree[rr(x)].lsum = tree[rr(x)].rsum = tree[rr(x)].msum = tree[x].lazy ? 0 : (p>>1);        tree[x].lazy = -1;    }}void build(int l,int r,int x) {    tree[x].l = l;    tree[x].r = r;    tree[x].mid = (l+r) >> 1;    tree[x].lazy = -1;    tree[x].lsum = tree[x].rsum = tree[x].msum = r - l + 1;    if(l == r) {        return ;    }    build(l,tree[x].mid,ll(x));    build(tree[x].mid+1,r,rr(x));}void update(int l,int r,int v,int x) {    if(l == tree[x].l && r == tree[x].r) {        tree[x].msum = tree[x].lsum = tree[x].rsum = v ? 0: tree[x].r - tree[x].l + 1;        tree[x].lazy = v;        return ;    }    down(x);    if(r <= tree[x].mid ) update(l,r,v,ll(x));    else if(l > tree[x].mid) update(l,r,v,rr(x));    else {        update(l,tree[x].mid,v,ll(x));        update(tree[x].mid + 1,r,v,rr(x));    }    up(x);}int query(int v,int x) {    if(tree[x].l == tree[x].r) return tree[x].l;    down(x);    if(tree[ll(x)].msum >= v) return query(v,ll(x));    else if(tree[ll(x)].rsum + tree[rr(x)].lsum >= v) return tree[x].mid - tree[ll(x)].rsum + 1;    else {        return query(v,rr(x));    }}int main() {       return 0;}