hdu Tunnel Warfare(线段树)

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3789 Accepted Submission(s): 1427

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
线段树记录每个点的前缀和，以及后缀和包含其本身！
这题有几个地方比较坑爹：
1、在修复点的时候记得先判断一下点是否已经修复，如果已经修复过了就不用再修复了；
2、在摧毁一个点的时候，也要先判断一下这个点是否已经被摧毁，若已摧毁就不用再摧毁；
3、如果一个点被连续摧毁，比如：
D 2
D 1
D 1
D 1
1被摧毁3次，那么接着：
R
R
R
这3个R都是在修复1！！
4、用cin 还超时！我用了scanf！！
#include <iostream>#include <cstdio>#include <stack>using namespace std;const int maxn = 50010;struct tree{int l , r , sl , sr;tree(int a = 0 , int b = 0 , int c = 0 , int d = 0){l = a , r = b , sl = c , sr = d;}}a[4*maxn];int n , m , SL , SR , sta[maxn];stack<int> D;void pushdown(int k){if(a[k].l < a[k].r){a[2*k].sl += a[k].sl;a[2*k].sr += a[k].sr;a[2*k+1].sl += a[k].sl;a[2*k+1].sr += a[k].sr;a[k].sl = 0;a[k].sr = 0;}}void build(int l , int r , int k){a[k].l = l;a[k].r = r;a[k].sl = 0;a[k].sr = 0;if(l == r){a[k].sl = l;a[k].sr = n-l+1;return;}int mid = (l+r)/2;build(l , mid , 2*k);build(mid+1 , r , 2*k+1);}void add(int l , int r , int k , int sl , int sr){if(l <= a[k].l && a[k].r <= r){a[k].sl += sl;a[k].sr += sr;}else{int mid = (a[k].l+a[k].r)/2;if(mid >= r){add(l , r , 2*k , sl , sr);}else if(mid < l){add(l , r , 2*k+1 , sl , sr);}else{add(l , mid , 2*k , sl , sr);add(mid+1 , r , 2*k+1, sl , sr);}}}void querry(int l , int r , int k){if(l <= a[k].l && a[k].r <= r){SL = a[k].sl;SR = a[k].sr;}else{pushdown(k);int mid = (a[k].l+a[k].r)/2;if(mid >= r){querry(l , r , 2*k);}else if(mid < l){querry(l , r , 2*k+1);}else{querry(l , mid , 2*k);querry(mid , r , 2*k+1);}}}void initial(){build(0 , n+1 , 1);SL = 0;SR = 0;while(!D.empty()) D.pop();for(int i = 0; i < maxn; i++){sta[i] = 1;}sta[0] = 0;}void computing(){char op;sta[n+1] = 0;while(m--){getchar();/*for(int i = 1; i <= n; i++){querry(i , i , 1);cout << SL+SR-1 << " ";}cout <<endl;*///cin >> op;scanf("%c" , &op);if(op == 'D'){int d;//cin >> d;scanf("%d" , &d);D.push(d);if(sta[d] == 1){//D.push(d);querry(d , d , 1);add(d-SL+1 , d , 1 , 0 , -1*SR);add(d , d+SR-1 , 1 , -1*SL , 0);sta[d] = 0;}}if(op == 'R'){if(!D.empty()){int R = D.top();D.pop();if(sta[R] == 0){sta[R] = 1;int l1 , l2 , r1 , r2;querry(R-1 , R-1 , 1);l1 = SL;r1 = SR;querry(R+1 , R+1 , 1);l2 = SL;r2 = SR;add(R-l1 , R , 1 , 0 , r2+1);add(R , R+r2 , 1 , l1+1 , 0);}}}if(op == 'Q'){int q;//cin >> q;scanf("%d" , &q);if(sta[q] == 0) cout << 0 <<endl;else{querry(q , q , 1);cout << SL+SR-1 <<endl;}}}}int main(){while(scanf("%d%d" , &n , &m) != EOF){initial();computing();}return 0;}