（线段树）hdoj 3397-Sequence operation

lxhgww got a sequence contains n characters which are all '0's or '1's. 
We have five operations here: 
Change operations: 
0 a b change all characters into '0's in a,ba,b 
1 a b change all characters into '1's in a,ba,b 
2 a b change all '0's into '1's and change all '1's into '0's in a,ba,b 
Output operations: 
3 a b output the number of '1's in a,ba,b 
4 a b output the length of the longest continuous '1' string in a,ba,b

InputT(T<=10) in the first line is the case number. 
Each case has two integers in the first line: n and m (1 <= n , m <= 100000). 
The next line contains n characters, '0' or '1' separated by spaces. 
Then m lines are the operations: 
op a b: 0 <= op <= 4 , 0 <= a <= b < n.OutputFor each output operation , output the result.Sample Input

1
10 10
0 0 0 1 1 0 1 0 1 1
1 0 2
3 0 5
2 2 2
4 0 4
0 3 6
2 3 7
4 2 8
1 0 5
0 5 6
3 3 9

Sample Output

5
2
6
5

感觉HOTEL一类的线段树题目已经练的差不多了，接下来开始练习扫描线。

  1 #include <iostream>
  2 //#include<bits/stdc++.h>
  3 #include <stack>
  4 #include <queue>
  5 #include <map>
  6 #include <set>
  7 #include <cstdio>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <math.h>
 11 using namespace std;
 12 int t,n,m;
 13 const int MAX=1e5+5;
 14 struct node
 15 {
 16     int left,mid,right;
 17     int l,r;
 18     int status;
 19     int num;
 20     int len()
 21     {
 22         return r-l+1;
 23     }
 24     void actl()
 25     {
 26         if(status!=-1)
 27         num=left=mid=right=(status?len():0);
 28     }
 29 }st[10*MAX];
 30 void renew(int k)
 31 {
 32     st[k].num=st[2*k].num+st[2*k+1].num;
 33     st[k].mid=max(st[2*k].mid,st[2*k+1].mid);
 34     st[k].left=st[2*k].left;
 35     st[k].right=st[2*k+1].right;
 36     st[k].mid=max(st[k].mid,st[2*k].right+st[2*k+1].left);
 37     if(st[2*k].left==st[2*k].len())
 38         st[k].left+=st[2*k+1].left;
 39     if(st[2*k+1].right==st[2*k+1].len())
 40         st[k].right+=st[2*k].right;
 41 }
 42 void init(int l,int r,int k)//allright
 43 {
 44     st[k].l=l;st[k].r=r;
 45     st[k].num=0;
 46     st[k].status=-1;
 47     if(l==r)
 48     {
 49         scanf("%d",&st[k].status);
 50         st[k].actl();
 51 //        st[k].num=1;
 52         return;
 53     }
 54     init(l,(l+r)/2,2*k);
 55     init((l+r)/2+1,r,2*k+1);
 56     renew(k);
 57 }
 58 void pushdown(int k)//allright
 59 {
 60     if(st[k].status!=-1)
 61     {
 62         st[2*k].status=st[2*k+1].status=st[k].status;
 63         st[2*k].actl();
 64         st[2*k+1].actl();
 65         st[k].status=-1;
 66     }
 67 }
 68 int query(int ql,int qr,int k)//返回1的个数
 69 {
 70     if(st[k].l==ql&&st[k].r==qr)
 71         return st[k].num;
 72     int mid=(st[k].l+st[k].r)/2;
 73     pushdown(k);
 74     if(qr<=mid)
 75     {
 76         return query(ql,qr,2*k);
 77     }
 78     else if(ql>mid)
 79         return query(ql,qr,2*k+1);
 80     else
 81         return query(ql,mid,2*k)+query(mid+1,qr,2*k+1);
 82 }
 83 int query_1(int ql,int qr,int k)//返回连续的1的个数
 84 {
 85     if(ql==st[k].l&&qr==st[k].r)
 86     {
 87         return st[k].mid;
 88     }
 89     int mid=(st[k].l+st[k].r)/2;
 90     pushdown(k);
 91     if(qr<=mid)
 92     {
 93         return query_1(ql,qr,2*k);
 94     }
 95     else if(ql>mid)
 96         return query_1(ql,qr,2*k+1);
 97     else
 98     {
 99         int an;
100         int zuo,you;
101         an=min(st[2*k].right,st[2*k].r-ql+1)+min(st[2*k+1].left,qr-st[2*k+1].l+1);
102 
103         zuo=query_1(ql,mid,2*k);an=max(an,zuo);
104         you=query_1(mid+1,qr,2*k+1);an=max(an,you);
105         return an;
106     }
107 }
108 void update(int ul,int ur,int k,int val)
109 {
110     if(ul==st[k].l&&ur==st[k].r)
111     {
112         if(val<=1)
113         {
114             st[k].status=val;
115             st[k].actl();
116             return;
117         }
118         else
119         {
120             if(st[k].status==0||st[k].status==1)
121             {
122                 st[k].status=(st[k].status?0:1);
123                 st[k].actl();
124                 return;
125             }
126             else
127             {
128                 pushdown(k);
129                 update(ul,(ul+ur)/2,2*k,val);
130                 update((ul+ur)/2+1,ur,2*k+1,val);
131                 renew(k);
132                 return;
133             }
134         }
135     }
136     pushdown(k);
137     int mid=(st[k].l+st[k].r)/2;
138     if(ur<=mid)
139         update(ul,ur,2*k,val);
140     else if(ul>mid)
141         update(ul,ur,2*k+1,val);
142     else
143     {
144         update(ul,mid,2*k,val);
145         update(mid+1,ur,2*k+1,val);
146     }
147     renew(k);
148 }
149 int main()
150 {
151     scanf("%d",&t);
152     int x,y;
153     int opt;
154     while(t--)
155     {
156         scanf("%d %d",&n,&m);
157         init(0,n-1,1);
158         while(m--)
159         {
160             scanf("%d %d %d",&opt,&x,&y);
161             if(opt<3)
162             {
163                 update(x,y,1,opt);
164             }
165             else if(opt==3)
166                 printf("%d\n",query(x,y,1));
167             else printf("%d\n",query_1(x,y,1));
168         }
169     }
170 }

 

转载于:https://www.cnblogs.com/quintessence/p/6485850.html