Tunnel Warfare
Time Limit: 2000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 154064-bit integer IO format: %I64d Java class name: Main
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!
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.
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
Source
解题：线段树的区间合并。。。
关键是合并的时候看怎么合并
如果在左儿子的右连续区间中，那么一定要加上右儿子的左连续区间
如果在右儿子的左连续区间中，那么一定要加上左儿子的右连续区间
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 50010; 4 struct node { 5 int lsum,rsum; 6 } tree[maxn<<2]; 7 void build(int L,int R,int v) { 8 tree[v].lsum = tree[v].rsum = R - L + 1; 9 if(L == R) return; 10 int mid = (L + R)>>1; 11 build(L,mid,v<<1); 12 build(mid+1,R,v<<1|1); 13 } 14 void pushup(int v,int k) { 15 tree[v].lsum = tree[v<<1].lsum; 16 tree[v].rsum = tree[v<<1|1].rsum; 17 if(tree[v].lsum == k - (k>>1)) tree[v].lsum += tree[v<<1|1].lsum; 18 if(tree[v].rsum == (k>>1)) tree[v].rsum += tree[v<<1].rsum; 19 } 20 void update(int L,int R,int id,int val,int v) { 21 if(L == R) { 22 tree[v].lsum = tree[v].rsum = val; 23 return; 24 } 25 int mid = (L + R)>>1; 26 if(id <= mid) update(L,mid,id,val,v<<1); 27 if(id > mid) update(mid+1,R,id,val,v<<1|1); 28 pushup(v,R - L + 1); 29 } 30 int ans,tot,stk[maxn]; 31 void query(int L,int R,int id,int v) { 32 if(L == R) { 33 ans = tree[v].lsum; 34 return; 35 } 36 int mid = (L + R)>>1; 37 if(id <= mid) { 38 query(L,mid,id,v<<1); 39 if(id >= mid - tree[v<<1].rsum + 1) 40 ans += tree[v<<1|1].lsum; 41 } 42 if(id > mid) { 43 query(mid+1,R,id,v<<1|1); 44 if(id <= mid + tree[v<<1|1].lsum) 45 ans += tree[v<<1].rsum; 46 } 47 } 48 int main() { 49 int n,m,x; 50 char op[10]; 51 while(~scanf("%d %d",&n,&m)) { 52 build(1,n,1); 53 while(m--) { 54 scanf("%s",op); 55 if(op[0] == 'Q') { 56 ans = 0; 57 scanf("%d",&x); 58 query(1,n,x,1); 59 printf("%d ",ans); 60 } else if(op[0] == 'D') { 61 scanf("%d",&x); 62 stk[tot++] = x; 63 update(1,n,x,0,1); 64 } else if(op[0] == 'R' && tot) 65 update(1,n,stk[--tot],1,1); 66 } 67 } 68 return 0; 69 }