POJ 2892-Tunnel Warfare【树状数组+二分】

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本文链接： https://blog.csdn.net/qq_41785863/article/details/101864437

题意：

有几个村庄，D a表示a村庄被摧毁，R表示修复上一个村庄，Q a表示包含a的连续村庄有多少个，如果a被摧毁则输出0。

1. D x: The x-th village was destroyed.
2. 第x个村庄被摧毁
3. Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.
4. 输出包含x的连续村庄有多少个。
5. R: The village destroyed last was rebuilt.
6. 上一个摧毁的村庄被修复了。

思路：这题可以用树状数组来实现，首先删除不用说，我们只需要记录一下方便以后恢复，主要是查询，我们可以考虑二分往左右最远能延伸的边界，然后就可以得到答案了，也是比较水的题

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define ls rt << 1
#define rs rt << 1|1
#define mid ((l + r) >> 1)
#define lson l, mid, ls
#define rson mid + 1, r, rs
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
int c[maxn], d[maxn], vis[maxn];
int n;

int lowbit(int i)
{
    return (i & (-i));
}
void add(int i, int x)
{
    while(i <= n)
    {
        c[i] += x;
        i += lowbit(i);
    }
}
int ask(int i)
{
    int ret = 0;
    while(i > 0)
    {
        ret += c[i];
        i -= lowbit(i);
    }
    return ret;
}

int main()
{
    int m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i)
        add(i, 1);
    int tot = 0;
    while(m--)
    {
        char s[2];
        int x;
        scanf("%s", s);
        if(s[0] == 'D')
        {
            scanf("%d", &x);
            add(x, -1);
            d[++tot] = x; //记录第几次修改的内容
            vis[x] = 1;
        }
        else if(s[0] == 'R')
        {
            add(d[tot], 1);
            vis[d[tot--]] = 0;
        }
        else if(s[0] == 'Q')
        {
            scanf("%d", &x);
            if(vis[x])
            {
                printf("0\n");
                continue;
            }
            int l = 1, r = x;
            int L = r;
            while(l <= r)
            {
                if(ask(r) - ask(mid - 1) >= r - mid + 1)
                    L = mid, r = mid - 1;
                else
                    l = mid + 1;
            }
            l = x, r = n;
            int R = l;
            while(l <= r)
            {
                if(ask(mid) - ask(l - 1) >= mid - l + 1)
                    R = mid, l = mid + 1;
                else
                    r = mid - 1;
            }
            int ans = R - L + 1;
            printf("%d\n", ans);
        }
    }
    return 0;
}