HDU 1540 Tunnel Warfare （线段树，区间合并）

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9270    Accepted Submission(s): 3609

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 9D 3D 6D 5Q 4Q 5RQ 4RQ 4

 

Sample Output

1024

题意：D表示摧毁某个村庄，R表示修复最后一个被摧毁的村庄，Q表示询问包含x在内的最大连续区间

思路：利用线段树求最大连续区间，我最近刚学线段树，这种题目对我来说还是有难度的，参考了别人的博客，才理解

上代码

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<algorithm>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define ls l,m,rt<<1#define rs m+1,r,rt<<1|1#define maxn 50010int len, R, L;struct node{int llen, rlen;//左右最大连续区间}tree[maxn << 2];void build(int l, int r, int rt){tree[rt].llen = tree[rt].rlen = r - l + 1;//一开始左右最大连续区间都是满的if (l == r)return;int m = (l + r) / 2;build(ls);build(rs);}void updata(int l, int r, int rt, int id, int a){if (l == r){tree[rt].llen = tree[rt].rlen = a;//摧毁或者回复return;}int m = (l + r) / 2;if (id <= m)updata(ls, id, a);elseupdata(rs, id, a);//以下是代码的核心//父区间的最大左右区间tree[rt].llen = tree[rt << 1].llen;tree[rt].rlen = tree[rt << 1 | 1].rlen;if (tree[rt].llen == m - l + 1)//如果左子树区间满了，父左区间要加上右子树的左区间tree[rt].llen += tree[rt << 1 | 1].llen;if (tree[rt].rlen == r - m)//如果右子树右区间满了，父右区间要加上左子树的右区间tree[rt].rlen += tree[rt << 1].rlen;}//查询的话，就从那个数开始，看能不能和自己左（右）区间的合并void query(int l, int r, int rt, int id){if (l == r){if (tree[rt].llen)R = L = 1;elseR = L = 0;len = tree[rt].llen;return;}int m = (l + r) / 2;//左区间满向右，右区间满向左if (id <= m){query(ls, id);if (L)len += tree[rt << 1 | 1].llen;//能和右区间的合并if (tree[rt << 1 | 1].llen<r - m)//父区间不满，不能再和右区间合并了L = 0;}else{query(rs, id);if (R)len += tree[rt << 1].rlen;//能和左区间合并if (tree[rt << 1].rlen<m - l + 1)//父区间不满，不能再和左区间合并了R = 0;}}int main(){//freopen("Text.txt", "r", stdin);int n, m, s[maxn], sn, id;char ch[2];while (scanf("%d%d", &n, &m) != EOF){build(1, n, 1);sn = 0;while (m--){scanf("%s", ch);if (ch[0] == 'D'){scanf("%d", &id);s[++sn] = id;updata(1, n, 1, id, 0);}else if (ch[0] == 'R'){if (sn)updata(1, n, 1, s[sn], 1), sn--;//修复最后一个被破坏的村庄}else{scanf("%d", &id);query(1, n, 1, id);printf("%d\n", len);}}}}