Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6931 Accepted Submission(s): 2680
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 #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <set> #include <stack> using namespace std; #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 const int maxn = 50010; int lsum[maxn<<2], rsum[maxn<<2]; void PushUp(int rt, int m) { lsum[rt] = lsum[rt<<1]; rsum[rt] = rsum[rt<<1|1]; if(lsum[rt] == m - (m >> 1)) lsum[rt] += lsum[rt<<1|1]; if(rsum[rt] == m >> 1) rsum[rt] += rsum[rt<<1]; } void build(int l, int r, int rt) { lsum[rt] = rsum[rt] = r - l + 1; if(l == r) return; int m = (l + r) >> 1; build(lson); build(rson); } void update(int p, int c, int l, int r, int rt) { if(l == r) { lsum[rt] = rsum[rt] = c; return; } int m = (l + r) >> 1; if(p <= m) update(p, c, lson); else update(p, c, rson); PushUp(rt, r - l + 1); } int flag; int query(int p, int l, int r, int rt) { if(rsum[rt] >= r - p + 1) { flag = 1; return rsum[rt]; } if(lsum[rt] >= p - l + 1) { flag = 1; return lsum[rt]; } int t; if(l == r) return 0; int m = (l + r) >> 1; if(p > m) { t = query(p, rson); if(flag) { flag = 0; t += rsum[rt<<1]; } } else { t = query(p, lson); if(flag) { flag = 0; t += lsum[rt<<1|1]; } } return t; } int main() { int n, m; while(~scanf("%d %d", &n, &m)) { flag = 0; build(1, n, 1); stack<int> stk; for(int i = 1; i <= m; i++) { char cmd[2]; scanf("%s", cmd); if(cmd[0] == 'D') { int tmp; scanf("%d", &tmp); stk.push(tmp); update(tmp, 0, 1, n, 1); } else if(cmd[0] == 'Q') { int tmp; scanf("%d", &tmp); flag = 0; printf("%d\n", query(tmp, 1, n, 1)); } else if(cmd[0] == 'R') { update(stk.top(), 1, 1, n, 1); stk.pop(); } } } }
转载于:https://www.cnblogs.com/lonewanderer/p/5648136.html
相关资源：JAVA上百实例源码以及开源项目