HDU - 1540 Tunnel Warfare（線段樹最大連續區間）

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9770    Accepted Submission(s): 3820

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

 

Source

POJ Monthly

 

Recommend

LL

題意：有一排村莊，村莊之間由地道相連，現有3個操作 D：破壞一座村莊，R：修復上一座被破壞的村莊，Q：詢問當前村莊與幾個村莊聯通

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int n, m;
struct xx{
    int l, r;
    int ls, rs, ms;
} T[N<<2];

void Build(int l, int r, int k){
    T[k].l = l, T[k].r = r;
    T[k].ls = T[k].rs = T[k].ms = r-l+1;
    if(l == r) return;
    int mid = (l+r)>>1;
    Build(l, mid, k<<1);
    Build(mid+1, r, k<<1|1);
}

void Update(int c, int k, int f){
    if(T[k].l == T[k].r){
        if(f){
            T[k].ls = T[k].rs = T[k].ms = 1;
        }
        else{
            T[k].ls = T[k].rs = T[k].ms = 0;
        }
        return;
    }
    int mid = (T[k].l+T[k].r)>>1;
    if(c <= mid) Update(c, k<<1, f);
    else Update(c, k<<1|1, f);
    T[k].ls = T[k<<1].ls, T[k].rs = T[k<<1|1].rs;
    T[k].ms = max(max(T[k<<1].ms, T[k<<1|1].ms), T[k<<1].rs+T[k<<1|1].ls);
    if(T[k<<1].ls == T[k<<1].r-T[k<<1].l+1) T[k].ls += T[k<<1|1].ls;
    if(T[k<<1|1].rs == T[k<<1|1].r-T[k<<1|1].l+1) T[k].rs += T[k<<1].rs;
}

int Query(int c, int k){
    if(T[k].l == T[k].r || !T[k].ms || T[k].r-T[k].l+1 == T[k].ms) return T[k].ms;
    int mid = (T[k].l+T[k].r)>>1;
    if(c <= mid){
        if(c >= T[k<<1].r-T[k<<1].rs+1){
            return Query(c, k<<1) + Query(mid+1, k<<1|1);
        }
        else return Query(c, k<<1);
    }
    else{
        if(c <= T[k<<1|1].l+T[k<<1|1].ls-1){
            return Query(c, k<<1|1) + Query(mid, k<<1);
        }
        else return Query(c, k<<1|1);
    }
}

int main(){
    while(scanf("%d%d", &n, &m) == 2){
        Build(1, n, 1);
        stack<int> k;
        for(int i = 0; i < m; i++){
            char op[3];
            scanf("%s", op);
            if(op[0] == 'D'){
                int x;
                scanf("%d", &x);
                k.push(x);
                Update(x, 1, 0);
            }
            else if(op[0] == 'Q'){
                int x;
                scanf("%d", &x);
                printf("%d\n", Query(x, 1));
            }
            else{
                Update(k.top(), 1, 1);
                k.pop();
            }
        }
    }
}