• CodeForces 46DParking Lot线段树


     对于100的查询,用线段树搞。。

    还是队友帮我调的,= =!错的很2b。同hotel搞法,只不过把他变长了,好处理头和为的停车位置。

    #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>
    #include<vector>
    using namespace std;
    
    const int maxn = 111111;
    int sum[maxn << 2]; int lsum[maxn << 2], rsum[maxn << 2], color[maxn << 2];
    int q[maxn], q1[maxn];
    #define lson l,mid,rt<<1
    #define rson mid+1,r,rt<<1|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(max(sum[rt << 1], sum[rt << 1 | 1]), rsum[rt << 1] + lsum[rt << 1 | 1]);
    }
    
    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 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){color[rt] = add;
            sum[rt] = lsum[rt] = rsum[rt] = color[rt] ? 0 : r - l + 1;
             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);
        if (rsum[rt << 1] + lsum[rt << 1 | 1] >= key) return mid - rsum[rt << 1] + 1;
        return ask(key, rson);
    }
    
    int main()
    {
        int n, a, b, c, d, m;
        scanf("%d%d%d", &n, &c, &d);
        build(1, c + n + d, 1);
        scanf("%d", &m);
        for (int i = 1; i <= m; i++){
            scanf("%d%d", &a, &b);
            if (a == 1){
                if (c + b + d > sum[1]){
                    printf("-1
    "); continue;
                }
                int t = ask(c + b + d, 1, c + n + d, 1);
                printf("%d
    ",t - 1);
                update(c + t, c + t + b - 1, 1, 1, c + n + d, 1);
                q[i] = t; q1[i] = b;
            }
            else{
                update(c + q[b], c + q[b] + q1[b] - 1, 0, 1, c + n + d, 1);
            }
        }
        return 0;
    
    }
  • 相关阅读:
    C++多态深入分析!
    字符编码总结
    算法:并查集
    树的非递归遍历:一种很好的算法
    算法:快速排序
    算法:堆排序
    字符串匹配:KMP算法, Boyer-Moore算法理解与总结
    shodan搜索
    google hacking 语法
    FOFA的搜索语句
  • 原文地址:https://www.cnblogs.com/yigexigua/p/3928297.html
Copyright © 2020-2023  润新知