HDU 1540 Tunnel Warfare

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1540

解题思路：线段树节点增加一个ld----区间左端点起始的最大长度，rd----区间右端点起始的最大长度，md----覆盖区间中点的最大长度，那么对于每次更新，更新到点然后更新父节点值；对于每次查询，若为中点，则返回md，否则从左右孩子中选择。需要注意的是ld，rd与md可能会有重叠部分，因此更新与查询的时候需要判断一下。

代码：

const int maxn = 5e4 + 5;

int ld[maxn * 4], rd[maxn * 4], md[maxn * 4];
int ok[maxn];
int n, m;
stack<int> s;

void build(int l, int r, int k){
    ld[k] = rd[k] = md[k] = (r - l + 1);
    if(l == r) return;
    int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;
    build(l, mid, lc);
    build(mid + 1, r, rc);
}
void update(int u, int x, int l, int r, int k){
    if(u == l && l == r){
        if(x == 0)
            ld[k] = rd[k] = md[k] = 0;
        else
            ld[k] = rd[k] = md[k] = 1;
        return;
    }
    int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;
    if(u <= mid) update(u, x, l, mid, lc);
    else update(u, x, mid + 1, r, rc);
    ld[k] = (ld[lc] >= mid - l + 1)? ld[lc] + ld[rc]: ld[lc];
    rd[k] = (rd[rc] >= r - mid)? rd[rc] + rd[lc]: rd[rc];
    md[k] = rd[lc] == 0? 0: ld[rc] + rd[lc];
}
int query(int u, int l, int r, int k){
    int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;
    if(u == mid) return md[k];
    if(u <= mid) {
        int q1 = query(u, l, mid, lc);
        if(rd[lc] >= mid - u + 1) q1 += ld[rc];
        return q1;
    }
    else{
        int q2 = query(u, mid + 1, r, rc);
        if(ld[rc] >= u - mid) q2 += rd[lc];
        return q2;
    }
}
int main(){
    while(scanf("%d %d", &n, &m) != EOF){
        while(!s.empty()) s.pop();
        build(1, n, 1);
        for(int i = 1; i <= n; i++) ok[i] = 1;
        while(m--){
            char ch;
            scanf(" %c", &ch);
            if(ch == 'D'){
                int u;
                scanf("%d", &u);
                if(ok[u]) update(u, 0, 1, n, 1);
                ok[u] = 0;
                s.push(u);
            }
            else if(ch == 'Q'){
                int u;
                scanf("%d", &u);
                int tmans = query(u, 1, n, 1);
                printf("%d
", tmans);
            }
            else{
                if(!s.empty()){
                    int u = s.top();
                    s.pop();
                    if(!ok[u]) update(u, 1, 1, n, 1);
                    ok[u] = 1;
                }
            }
        }
    }
}

题目：

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9320    Accepted Submission(s): 3633

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