一、题目

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

顺便附上原题链接→_→Problem-1540

二、题目分析

和普通的线段树没什么太大差别，只是每个区间加了两个变量——lsum和rsum，分别表示这个区间左起向右连续“1”串的长度和右起向左连续“1”串的长度。在维护树回溯时，一个区间的lsum等于其左区间的lsum，rsum等于其右区间的rsum。特别的，当一个区间左区间的lsum等于其左区间长度，即该区间左区间是一个连续的“1”串时，该区间lsum需额外加上其右区间的lsum；对于其rsum同理。

三、代码实现

这题有个天坑，题目完全没提到，但这题有多组数据_(:з」∠)_

  1 #include<cstdio>
  2 #include<cstring>
  3 const int MAXN=5e4+10;
  4 int n,m;
  5 struct node
  6 {
  7     int l,r;
  8     int lsum,rsum;
  9 }tr[MAXN<<2];
 10 int stack[MAXN],top;
 11 void build(int x,int y,int i)
 12 {
 13     tr[i].l=x,tr[i].r=y;
 14     if(x==y)
 15     {
 16         tr[i].lsum=1;
 17         tr[i].rsum=1;
 18         return;
 19     }
 20     int mid=(tr[i].l+tr[i].r)>>1;
 21     build(x,mid,i<<1);
 22     build(mid+1,y,i<<1|1);
 23     tr[i].lsum=y-x+1,tr[i].rsum=y-x+1;
 24     return;
 25 }
 26 void update(int x,int i,int val)
 27 {
 28     if(tr[i].l==x&&tr[i].r==x)
 29     {
 30         tr[i].lsum=val;
 31         tr[i].rsum=val;
 32         return;
 33     }
 34     int mid=(tr[i].l+tr[i].r)>>1;
 35     if(x<=mid)
 36     {
 37         update(x,i<<1,val);
 38
 39     }
 40     else
 41     {
 42         update(x,i<<1|1,val);
 43     }
 44     tr[i].lsum=tr[i<<1].lsum;
 45     tr[i].rsum=tr[i<<1|1].rsum;
 46     if(tr[i<<1].lsum==tr[i<<1].r-tr[i<<1].l+1)tr[i].lsum+=tr[i<<1|1].lsum;
 47     if(tr[i<<1|1].rsum==tr[i<<1|1].r-tr[i<<1|1].l+1)tr[i].rsum+=tr[i<<1].rsum;
 48 }
 49 int query_left(int x,int i)
 50 {
 51     if(tr[i].r==x)
 52     {
 53         if(tr[i].rsum==tr[i].r-tr[i].l+1&&tr[i].l!=1)
 54         {
 55             return tr[i].rsum+query_left(tr[i].l-1,1);
 56         }
 57         return tr[i].rsum;
 58     }
 59     int mid=(tr[i].l+tr[i].r)>>1;
 60     if(x<=mid)
 61     {
 62         return query_left(x,i<<1);
 63     }
 64     else
 65     {
 66         return query_left(x,i<<1|1);
 67     }
 68 }
 69 int query_right(int x,int i)
 70 {
 71
 72     if(tr[i].l==x)
 73     {
 74         if(tr[i].lsum==tr[i].r-tr[i].l+1&&tr[i].r!=n)
 75         {
 76             return tr[i].lsum+query_right(tr[i].r+1,1);
 77         }
 78         return tr[i].lsum;
 79     }
 80     int mid=(tr[i].l+tr[i].r)>>1;
 81     if(x<=mid)
 82     {
 83         return query_right(x,i<<1);
 84     }
 85     else
 86     {
 87         return query_right(x,i<<1|1);
 88     }
 89 }
 90 int main()
 91 {
 92     while(scanf("%d%d",&n,&m)!=EOF)
 93     {
 94         memset(tr,0,sizeof(tr));
 95         memset(stack,0,sizeof(stack));
 96         top=0;
 97         build(1,n,1);
 98         for(int i=1;i<=m;++i)
 99         {
100             char c;
101             int x;
102             scanf("\n%c",&c);
103             if(c==‘D‘)
104             {
105                 scanf("%d",&x);
106                 stack[++top]=x;
107                 update(x,1,0);
108             }
109             else if(c==‘R‘)
110             {
111                 update(stack[top--],1,1);
112             }
113             else
114             {
115                 scanf("%d",&x);
116                 int ans=query_left(x,1)+query_right(x,1)-1;
117                 if(ans<=0)printf("0\n");
118                 else printf("%d\n",ans);
119             }
120         }
121     }
122     return 0;
123 }

HDU1540-Tunnel Warfare

弱弱地说一句，本蒟蒻码字也不容易，转载请注明出处http://www.cnblogs.com/Maki-Nishikino/p/6230606.html