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hdoj 1540 Tunnel Warfare 【线段树 区间合并】

2016-01-09 02:51 393 查看

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6106    Accepted Submission(s): 2361


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

 
题意：给定n个排成一行的村庄（编号从1到n）和m次操作。

D x 表示破坏x村庄，Q x 表示查询x村庄所能联络的村庄数，R 表示重建最后一个被破坏的村庄。

思路：查询单点所在连续块的最大长度，以前做过区间的，点和区间也就查询不同。

点只有两种情况，区间内的连续长度 或者 左区间右连续+右区间左连续

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (50000+10)
#define MAXM (500000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
struct Tree
{
int l, r, len;
int lsum, rsum, sum;
};
Tree tree[MAXN<<2];
void PushUp(int o)
{
tree[o].lsum = tree[ll].lsum;
tree[o].rsum = tree[rr].rsum;
tree[o].sum = max(tree[ll].sum, tree[rr].sum);
tree[o].sum = max(tree[o].sum, tree[ll].rsum + tree[rr].lsum);
if(tree[o].lsum == tree[ll].len)
tree[o].lsum += tree[rr].lsum;
if(tree[o].rsum == tree[rr].len)
tree[o].rsum += tree[ll].rsum;
}
void Build(int o, int l, int r)
{
tree[o].l = l; tree[o].r = r;
tree[o].len = tree[o].lsum = tree[o].rsum = tree[o].sum = r-l+1;
if(l == r)
return ;
int mid = (l + r) >> 1;
Build(lson); Build(rson);
}
void Update(int o, int pos, int val)
{
if(tree[o].l == tree[o].r)
{
tree[o].lsum = tree[o].rsum = tree[o].sum = val;
return ;
}
int mid = (tree[o].l + tree[o].r) >> 1;
if(pos <= mid)
Update(ll, pos, val);
else
Update(rr, pos, val);
PushUp(o);
}
int Query(int o, int pos)
{
if(tree[o].l == tree[o].r)
return tree[o].sum;
int mid = (tree[o].l + tree[o].r) >> 1;
if(pos <= mid)
{
int ans1 = Query(ll, pos), ans2 = 0;
if(pos >= tree[ll].r - tree[ll].rsum + 1)
ans2 = tree[ll].rsum + tree[rr].lsum;
return max(ans1, ans2);
}
else
{
int ans1 = Query(rr, pos), ans2 = 0;
if(pos <= tree[rr].l + tree[rr].lsum - 1)
ans2 = tree[ll].rsum + tree[rr].lsum;
return max(ans1, ans2);
}
}
int main()
{
int n, m;
while(scanf("%d%d", &n, &m) != EOF)
{
Build(1, 1, n);
stack<int> S;
W(m)
{
char op[2]; int x;
Rs(op);
if(op[0] == 'D')
{
Ri(x);
Update(1, x, 0);
S.push(x);
}
else if(op[0] == 'Q')
{
Ri(x);
Pi(Query(1, x));
}
else
{
if(!S.empty())
{
x = S.top(); S.pop();
Update(1, x, 1);
}
}
}
}
return 0;
}

                                            

                                        

                        
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