Tunnel Warfare（线段树 + 区间合并）

最新推荐文章于 2020-04-21 16:15:14 发布

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最新推荐文章于 2020-04-21 16:15:14 发布

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分类专栏： HDOJ 文章标签： java

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本文深入探讨了中国东北平原上，抗日战争时期隧道战的广泛运用。文章详细介绍了村庄通过隧道相连的布局特点，以及敌军攻击村庄、破坏隧道后，八路军司令部对当前隧道连接状态的迫切需求。通过对不同格式事件的处理，包括摧毁村庄、请求查询连接村庄数量及重建村庄等，展示了线段树和区间合并技术在解决实际问题中的应用。

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Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4070 Accepted Submission(s): 1526

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

Sample Input

7 9

D 3

D 6

D 5

Q 4

Q 5

R

Q 4

R

Q 4

Sample Output

1

0

2

4

题意：

给出 n,m（1 ~ 50000），代表有 n 个连续的城市和 m 个操作，操作有 3 类，D ans，代表要把编号为 ans 的城市毁灭，R 代表要重建最后毁灭的那么城市，Q ans 代表询问包含 ans 的完好的连续城市有几个。每次 Q 给出一个结果。

思路：

线段树，区间合并。注意 updata 的时候要判断往左还是往右，如果不判断的话，则会 TLE。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>

using namespace std;

const int MAX = 500050;

int lmax[MAX], rmax[MAX], mmax[MAX];

void push_up(int node, int l, int r) {
        lmax[node] = lmax[node << 1];
        rmax[node] = rmax[node << 1 | 1];

        int mid = (r + l) >> 1;
        if (lmax[node] == mid - l + 1) lmax[node] += lmax[node << 1 | 1];
        if (rmax[node] == r - mid) rmax[node] += rmax[node << 1];

        mmax[node] = max(rmax[node << 1] + lmax[node << 1 | 1],
                         max(mmax[node << 1], mmax[node << 1 | 1]));
}

void build (int node, int l, int r) {
        if (l == r) {
                lmax[node] = rmax[node] = mmax[node] = 1;
        } else {
                int mid = (r + l) >> 1;
                build(node << 1, l, mid);
                build(node << 1 | 1, mid + 1, r);
                push_up(node, l, r);
        }
}

void update(int node, int l, int r, int loc, int s) {
        if (l == r) {
                if (l == loc) {
                        mmax[node] = lmax[node] =
                        rmax[node] = (s ? 0 : 1);
                }
                return;
        }

        int mid = (r + l) >> 1;
        if (loc <= mid) update(node << 1, l, mid, loc, s);
        else update(node << 1 | 1, mid + 1, r, loc, s);
        push_up(node, l, r);
}

int query (int node, int l, int r, int loc) {
        int mid = (r + l) >> 1;

        if (mmax[node] == r - l + 1 || !mmax[node])
            return mmax[node];

        if (loc <= mid) {
                if (loc >= mid - rmax[node << 1] + 1)
                        return query(node << 1, l, mid, loc) +
                               query(node << 1 | 1, mid + 1, r, mid + 1);
                else return query(node << 1, l, mid, loc);
        } else {
                if (loc <= mid + 1 + lmax[node << 1 | 1] - 1)
                        return query(node << 1, l, mid, mid) +
                               query(node << 1 | 1, mid + 1, r, loc);
                else return query(node << 1 | 1, mid + 1, r, loc);
        }
}

int main() {
        int n, m;

        while(~scanf("%d%d", &n, &m)) {
                stack<int> s;

                build (1, 1, n);

                while (m--) {
                        char op;
                        scanf(" %c", &op);

                        if (op == 'D') {
                                int loc;
                                scanf("%d", &loc);
                                update(1, 1, n, loc, 1);
                                s.push(loc);
                        } else if (op == 'Q') {
                                int loc;
                                scanf("%d", &loc);

                                printf("%d\n", query(1, 1, n, loc));

                        } else {
                                int loc = s.top();
                                s.pop();
                                update(1, 1, n, loc, 0);
                        }
                }

        }

        return 0;
}