hdu 3397 线段树双标记
Sequence operation
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7492 Accepted Submission(s): 2230
Problem Description
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 , b]
1 a b change all characters into '1's in [a , b]
2 a b change all '0's into '1's and change all '1's into '0's in [a, b]
Output operations:
3 a b output the number of '1's in [a, b]
4 a b output the length of the longest continuous '1' string in [a , b]
We have five operations here:
Change operations:
0 a b change all characters into '0's in [a , b]
1 a b change all characters into '1's in [a , b]
2 a b change all '0's into '1's and change all '1's into '0's in [a, b]
Output operations:
3 a b output the number of '1's in [a, b]
4 a b output the length of the longest continuous '1' string in [a , b]
Input
T(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.
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.
Output
For 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
/*
hdu 3397 线段树双标记
给你5个操作：
0:将区间[a,b]之间的数全部置为0
1:将区间[a,b]之间的数全部置为1
2:将区间[a,b]之间的 1->0 0->1
3:求区间[a,b]之间1的个数
4:求区间[a,b]之间1的最长连续长度
先设定ls1,rs1,ms1,num1 ls0,rs0,ms0,num0分别记录1,0的情况
然后是rev和same标记,区间合并这些到是没什么问题,主要是在标记下放
最开始没有注意标记之间的互相影响的问题,假设[a,b]上即有rev又有same
我们应该怎么处理之.所以最开始WR了几次,然后想值记录same标记,当进行
rev操作时 要么更新到点,要么遇到same对标记进行修改，尝试了很久，一直
TLE
然后考虑same和rev之间的关系，
假设same在[a,b]上遇到一个rev,那么把该区间上原先有的rev删除.
如果rev在[a,b]上遇见一个same标记,那么只需要对same进行修改即可,否则
将0 1的数据进行交换
hhh-2016-04-01 21:52:19
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
const int maxn = 200050;
struct node
{
int l,r;
int same,rev;
int ls1,rs1,ms1;
int ls0,rs0,ms0;
int num1,num0;
int mid()
{
return (l+r)>>1;
}
int len()
{
return (r-l+1);
}
} tree[maxn<<2];
void push_up(int i)
{
tree[i].ls0 = tree[lson].ls0,tree[i].ls1 = tree[lson].ls1;
tree[i].rs0 = tree[rson].rs0,tree[i].rs1 = tree[rson].rs1;
tree[i].num1 = tree[lson].num1 + tree[rson].num1;
tree[i].num0 = tree[lson].num0 + tree[rson].num0;
if(tree[i].ls1 == tree[lson].len() )
tree[i].ls1 += tree[rson].ls1;
if(tree[i].ls0 == tree[lson].len() )
tree[i].ls0 += tree[rson].ls0;
if(tree[i].rs1 == tree[rson].len() )
tree[i].rs1 += tree[lson].rs1;
if(tree[i].rs0 == tree[rson].len() )
tree[i].rs0 += tree[lson].rs0;
tree[i].ms1 = max(tree[lson].ms1,tree[rson].ms1);
tree[i].ms0 = max(tree[lson].ms0,tree[rson].ms0);
tree[i].ms1 = max(tree[i].ms1,tree[lson].rs1+tree[rson].ls1);
tree[i].ms0 = max(tree[i].ms0,tree[lson].rs0+tree[rson].ls0);
}
void ini1(int i,int val)
{
tree[i].num1=tree[i].ls1 = tree[i].rs1 = tree[i].ms1 = val;
}
void ini0(int i,int val)
{
tree[i].num0=tree[i].ls0 = tree[i].rs0 = tree[i].ms0 = val;
}
void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
ini1(i,0),ini0(i,0);
tree[i].same = -1;
tree[i].rev = 0;
if(l == r)
{
int x;
scanf("%d",&x);
if(x)
ini1(i,1);
else
ini0(i,1);
return ;
}
int mid = tree[i].mid();
build(lson,l,mid);
build(rson,mid+1,r);
push_up(i);
}
void exchange(int i)
{
swap(tree[i].ls1,tree[i].ls0);
swap(tree[i].rs1,tree[i].rs0);
swap(tree[i].ms0,tree[i].ms1);
swap(tree[i].num0,tree[i].num1);
}
void solve(int i)
{
tree[i].same ^= 1;
if(tree[i].same)
{
ini1(i,tree[i].len());
ini0(i,0);
}
else
{
ini1(i,0);
ini0(i,tree[i].len());
}
}
void push_down(int i)
{
if(tree[i].same != -1)
{
tree[lson].rev = tree[rson].rev = 0;
tree[lson].same = tree[i].same;
tree[rson].same = tree[i].same;
if(tree[i].same)
{
ini1(lson,tree[lson].len()),ini0(lson,0);
ini1(rson,tree[rson].len()),ini0(rson,0);
}
else
{
ini1(lson,0),ini1(rson,0);
ini0(lson,tree[lson].len()),ini0(rson,tree[rson].len());
}
tree[i].same = -1;
}
if(tree[i].rev)
{
if(tree[lson].same != -1)
{
solve(lson);
}
else
{
tree[lson].rev ^= 1;
exchange(lson);
}
if(tree[rson].same != -1)
{
solve(rson);
}
else
{
tree[rson].rev ^= 1;
exchange(rson);
}
tree[i].rev = 0;
}
}
void update_same(int i,int l,int r,int va)
{
if(tree[i].l >= l && tree[i].r <= r )
{
tree[i].rev = 0;
if(va)
{
ini1(i,tree[i].len()),ini0(i,0);
}
else
{
ini1(i,0),ini0(i,tree[i].len());
}
tree[i].same = va;
return;
}
push_down(i);
int mid = tree[i].mid();
if(l <= mid)
update_same(lson,l,r,va);
if(r > mid)
update_same(rson,l,r,va);
push_up(i);
}
void update_rev(int i,int l,int r)
{
if(tree[i].l >= l && tree[i].r <= r)
{
if(tree[i].same != -1)
{
solve(i);
return ;
}
tree[i].rev ^= 1;
exchange(i);
return;
}
push_down(i);
int mid = tree[i].mid();
if(l <= mid)
update_rev(lson,l,r);
if(r > mid)
update_rev(rson,l,r);
push_up(i);
}
int query1(int i,int l,int r)
{
if(tree[i].l >= l && tree[i].r <= r)
{
return tree[i].ms1;
}
int mid = tree[i].mid();
push_down(i);
if(r <= mid)
return query1(lson,l,r);
else if(l > mid)
return query1(rson,l,r);
else
{
int ans1 = query1(lson,l,mid);
int ans2 = query1(rson,mid+1,r);
return max(max(ans1,ans2),min(tree[lson].rs1,mid-l+1)+min(tree[rson].ls1,r-mid));
}
}
int query2(int i,int l,int r)
{
if(l <= tree[i].l && tree[i].r <= r)
{
return tree[i].num1;
}
push_down(i);
int mid = tree[i].mid();
int num = 0;
if(l <= mid)
num += query2(lson,l,r);
if(r > mid)
num += query2(rson,l,r);
return num;
}
int op;
int x,y;
int T,n,m;
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
build(1,0,n-1);
while(m--)
{
scanf("%d",&op);
scanf("%d%d",&x,&y);
if(op == 0)
update_same(1,x,y,0);
else if(op == 1)
update_same(1,x,y,1);
else if(op == 2)
update_rev(1,x,y);
else if(op == 3)
printf("%d\n",query2(1,x,y));
else
printf("%d\n",query1(1,x,y));
}
}
return 0;
}
