kuangbin专题七 HDU1540 Tunnel Warfare （前缀后缀线段树）-白红宇

kuangbin专题七 HDU1540 Tunnel Warfare （前缀后缀线段树）

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发布时间：2019-06-25

本文共 4109 字，大约阅读时间需要 13 分钟。

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!

InputThe 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.

OutputOutput the answer to each of the Army commanders’ request in order on a separate line.

Sample Input

7 9D 3D 6D 5Q 4Q 5RQ 4RQ 4

Sample Output

1024 表示初识线段树，对区间维护前缀后缀表示无力，不过最后还是卡过去了。注释应该比较清晰。。。 清楚线段树的结构，和区间前缀后缀 左孩子右孩子的拼接情况。。。说不清楚。。。。

1 #include 
     
        2 #include 
      
         3 #include 
       
          4 #include 
        
           5 #include 
         
            6 #include 
          
            7 #include 
           
             8 #include 
            
              9 #include 
              10 #include 
              
                11 #include 
               
                 12 #include 
                
                  13 #include 
                 
                   14 using namespace std; 15 #define FO freopen("in.txt","r",stdin); 16 #define rep(i,a,n) for (int i=a;i
                  
                   =a;i--) 18 #define pb push_back 19 #define mp make_pair 20 #define all(x) (x).begin(),(x).end() 21 #define fi first 22 #define se second 23 #define SZ(x) ((int)(x).size()) 24 #define debug(x) cout << "&&" << x << "&&" << endl; 25 #define lowbit(x) (x&-x) 26 #define mem(a,b) memset(a, b, sizeof(a)); 27 typedef vector
                   
                     VI; 28 typedef long long ll; 29 typedef pair
                    
                      PII; 30 const ll mod=1000000007; 31 const int inf = 0x3f3f3f3f; 32 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 34 //head 35 36 const int maxn=50010; 37 int pre[maxn<<2],suf[maxn<<2],maxx[maxn<<2],n,m,st[maxn];//维护区间前缀1，区间后缀1，st模拟栈 38 39 void pushup(int rt,int x) { 40 pre[rt]=pre[rt<<1];//rt的前缀1就是左孩子前缀1 41 suf[rt]=suf[rt<<1|1];//rt后缀1就是右孩子后缀1 42 maxx[rt]=max(maxx[rt<<1],maxx[rt<<1|1]);//rt的最大两种情况 43 maxx[rt]=max(maxx[rt],pre[rt<<1|1]+suf[rt<<1]); 44 if(suf[rt<<1|1]==(x>>1)) suf[rt]+=suf[rt<<1];// 如果右孩子的后缀全是1，可以拼接左孩子的后缀 45 if(pre[rt<<1]==(x-(x>>1))) pre[rt]+=pre[rt<<1|1];//如果左孩子前缀全是1 可以拼接右孩子的前缀 46 } 47 48 void build(int rt,int L,int R) { 49 pre[rt]=suf[rt]=maxx[rt]=R-L+1;//这里置为R-L+1 和 常规用pushup效果一样 50 int mid=(L+R)>>1; 51 if(L!=R) { 52 build(rt<<1,L,mid); 53 build(rt<<1|1,mid+1,R); 54 } 55 } 56 57 void updata(int rt,int L,int R,int pos,int val) { 58 if(L==R) { //单点修改 59 pre[rt]=suf[rt]=maxx[rt]=val; 60 return; 61 } 62 int mid=(L+R)>>1; 63 if(pos<=mid) updata(rt<<1,L,mid,pos,val); 64 else updata(rt<<1|1,mid+1,R,pos,val); 65 pushup(rt,R-L+1); 66 } 67 68 int query(int rt,int L,int R,int pos) { 69 if(L==R||maxx[rt]==0||maxx[rt]==(R-L+1))//断点 || maxx为0 || maxx最大 70 return maxx[rt]; 71 int mid=(L+R)>>1; 72 if(pos<=mid) { //如果在左子树 73 if(pos>=mid-suf[rt<<1]+1) return pre[rt<<1|1]+suf[rt<<1];//如果大于左孩子的后缀1 == 包含两部分 74 else return query(rt<<1,L,mid,pos); //否则搜左子树 75 } else { //如果在右子树 76 if(pos<=mid+1+pre[rt<<1|1]-1) return pre[rt<<1|1]+suf[rt<<1];//如果小于右孩子的前缀1 == 包含两部分 77 else return query(rt<<1|1,mid+1,R,pos);//否则右子树 78 } 79 } 80 81 int main() { 82 while(~scanf("%d%d",&n,&m)) { 83 int cur=0; 84 build(1,1,n); 85 int pos; 86 while(m--) { 87 char s[3]; 88 scanf("%s",s); 89 if(s[0]=='D') { 90 scanf("%d",&pos); 91 st[cur++]=pos; 92 updata(1,1,n,pos,0); 93 } else if(s[0]=='Q') { 94 scanf("%d",&pos); 95 printf("%d\n",query(1,1,n,pos)); 96 } else { 97 pos=st[--cur]; 98 updata(1,1,n,pos,1); 99 }100 }101 }102 }

转载于:https://www.cnblogs.com/ACMerszl/p/9886650.html

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