Poj3667Hotel线段树

 区间合并。

需要记录三个东西,sum[rt] 表示这个区间最长连续空集的长度,lsum[rt]=表示从左边开始最长连续空集的长度,rum[rt]表示从右边最长连续空集的长度。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn = 55555;
int color[maxn << 2];
int sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];

void down(int rt, int m)
{
    if (~color[rt]){
        color[rt << 1] = color[rt << 1 | 1] = color[rt];
        sum[rt << 1] = lsum[rt << 1] = rsum[rt << 1] = color[rt] ? 0 : (m - (m >> 1));
        sum[rt << 1 | 1] = lsum[rt << 1 | 1] = rsum[rt << 1 | 1] = color[rt] ? 0 : (m >> 1);
        color[rt] = -1;
    }
}


void up(int rt, int m)
{
    lsum[rt] = lsum[rt << 1];
    rsum[rt] = rsum[rt << 1 | 1];
    if (lsum[rt] == (m - (m >> 1))) lsum[rt] += lsum[rt << 1 | 1];
    if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt << 1];
    sum[rt] = max(rsum[rt << 1] + lsum[rt << 1 | 1], max(sum[rt << 1], sum[rt << 1 | 1]));//这里就是很神,不知道怎么想到的,用这三个量就可以记录更新区间最长连续0。sum的大小只和左右儿子的sum,和横跨左右儿子的那一部分有关。
}

void build(int l, int r, int rt)
{
    color[rt] = -1; sum[rt] = lsum[rt] = rsum[rt] = r - l + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}

void update(int L, int R, int add, int l, int r, int rt)
{
    if (L <= l&&r <= R){
        sum[rt] = lsum[rt] = rsum[rt] = add ? 0 : r - l + 1;
        color[rt] = add; return;
    }
    down(rt, r - l + 1);
    int mid = (l + r) >> 1;
    if (L <= mid) update(L, R, add, lson);
    if (R>mid) update(L, R, add, rson);
    up(rt, r - l + 1);
}

int ask(int key, int l, int r, int rt)
{
    if (l == r) return l;
    int mid = (l + r) >> 1;
    down(rt, r - l + 1);
    if (sum[rt << 1] >= key) return ask(key, lson);
    else if (rsum[rt << 1] + lsum[rt << 1 | 1] >= key) return mid - rsum[rt << 1] + 1;
    return ask(key, rson);
}

int main()
{
    int n, m;
    int a, b, c;
    while (cin >> n >> m){
        build(1, n, 1);
        for (int i = 0; i < m; i++){
            scanf("%d", &a);
            if (a == 1){
                scanf("%d", &b);
                if (b>sum[1]){
                    printf("0
"); continue;
                }
                int pos = ask(b, 1, n, 1);
                printf("%d
", pos);
                update(pos, pos + b - 1, 1, 1, n, 1);
            }
            else{
                scanf("%d%d", &b, &c);
                update(b, b + c - 1, 0, 1, n, 1);
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/yigexigua/p/3928287.html