HDU-1540 Tunnel Warfare 【線段樹+單點修改+區間更新】


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
 

 
思路:
一道感覺並不是很難的線段樹,菜雞看着別人的題解按自己的習慣勉強寫完的,真的菜啊。
lm[o], rm[o], sum[o],分別代表區間o左、右端點開始的最長連續區間,該區間內的最長連續區間。
 
Code:
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define INF 0x3f3f3f3f
 4 #define M(a, b) memset(a, b, sizeof(a))
 5 #define lson o<<1
 6 #define rson o<<1|1
 7 const int maxn = 50000 + 10;
 8 int lm[maxn<<2], rm[maxn<<2], sum[maxn<<2], v, p;
 9 int S[maxn], top;
10 
11 void build(int o, int L, int R) {
12     lm[o] = rm[o] = sum[o] = R - L + 1;
13     if (L == R) return;
14     else {
15         int M = (L + R) >> 1;
16         build(lson, L, M);
17         build(rson, M+1, R);
18     }
19 }
20 
21 void update(int o, int L, int R) {
22     int M = (L + R) >> 1;
23     if (L == R) {
24         lm[o] = rm[o] = sum[o] = v;
25         return;
26     }
27     else {
28         if (p <= M) update(lson, L, M);
29         else update(rson, M+1, R);
30     }
31     lm[o] = lm[lson]; rm[o] = rm[rson];
32     sum[o] = max(rm[lson]+lm[rson], max(sum[lson], sum[rson]));
33     if (lm[lson] == M - L + 1) lm[o] += lm[rson];
34     if (rm[rson] == R - M) rm[o] += rm[lson];
35 }
36 
37 int query(int o, int L, int R, int x) {
38     if (L == R || sum[o] == 0 || sum[o] == R - L + 1) return sum[o];
39     else {
40         int M = (L + R) >> 1;
41         if (x <= M) {
42             if (x >= M - rm[lson] + 1) 
43                 return query(lson, L, M, x) + query(rson, M+1, R, M+1);
44             else return query(lson, L, M, x);
45         }
46         else {
47             if (x <= M + lm[rson]) 
48                 return query(rson, M+1, R, x) + query(lson, L, M, M);
49             else return query(rson, M+1, R, x);
50         }
51     }
52 }
53 
54 int main() {
55     int n, m;
56     while(~scanf("%d%d", &n, &m)) {
57         build(1, 1, n);
58         char op[3];
59         top = 0;
60         while(m--) {
61             scanf("%s", op);
62             if (op[0] == 'D') {
63                 scanf("%d", &p);
64                 S[top++] = p;
65                 v = 0;
66                 update(1, 1, n);
67             }
68             else if (op[0] == 'R') {
69                 if (top == 0) continue;
70                 p = S[--top];
71                 v = 1;
72                 update(1, 1, n);
73             }
74             else {
75                 scanf("%d", &p);
76                 printf("%d\n", query(1, 1, n, p));
77             }
78         }
79     }
80 
81     return 0;
82 }

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