hdu3397 Sequence operation(线段树成段更新)
Sequence operation
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4699 Accepted Submission(s): 1368
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
Author
lxhgww&&shǎ崽
Source
HDOJ Monthly Contest – 2010.05.01
题目大意：给一个01串，给3种操作，分别是将一个区间置0，置1和取反。给2种查询，分别是查询某个区间1的个数和查询某个区间最长连续1的个数。
题目分析：线段树好题。
对区间的3种操作中，置0和置1操作很简单，直接将区间lazy标记，强制覆盖。如果是对区间取反的话，就要考虑区间原来的状态了。如果原来该区间已经取反了，再取反相当于还原，如果该区间没有标记，标记上取反直接返回，如果有标记，则当前区间lazy标记取反更新相应值返回即可。
可以看出，每个区间每次操作只可能有一种逻辑结果，所以只需要一个lazy标记即可。lazy有4个取值：-1，0，1，2，分别表示当前区间无标记，当前区间覆盖0，当前区间覆盖1意见当前区间取反。
其他的就是简单的线段树成段更新了。
不过查询操作有个查询某个区间最大连续1的个数。这稍微复杂些，不过也好办。如果没有取反操作，那就和这道Hotel一样了，直接标记左右连续1的个数即可。但是有了取反操作，那么还要维护当前区间左右连续0的个数，因为取反的时候，直接交换当前区间连续0和连续1的个数即可。
详情请见代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1000005;
int n,m;
struct node
{
int sum;//区间1的个数
int lazy;//区间更新延迟标记和翻转标记
int long1,long0;//最长连续的1和0
int l[2],r[2];//记录区间左右连续的1，0长度
}tree[N];
int Max(int a,int b)
{
return a > b?a:b;
}
int Min(int a,int b)
{
return a > b?b:a;
}
void pushup(int num,int s,int e)
{
int ls = num<<1;
int rs = num<<1|1;
int mid = (s + e)>>1;
tree[num].sum = tree[ls].sum + tree[rs].sum;
tree[num].l[0] = tree[ls].l[0];
if(tree[ls].sum == 0)
tree[num].l[0] += tree[rs].l[0];
tree[num].l[1] = tree[ls].l[1];
if(tree[ls].sum == mid - s + 1)
tree[num].l[1] += tree[rs].l[1];
tree[num].r[0] = tree[rs].r[0];
if(tree[rs].sum == 0)
tree[num].r[0] += tree[ls].r[0];
tree[num].r[1] = tree[rs].r[1];
if(tree[rs].sum == e - mid)
tree[num].r[1] += tree[ls].r[1];
tree[num].long0 = Max(tree[ls].long0,Max(tree[rs].long0,tree[ls].r[0] + tree[rs].l[0]));
tree[num].long1 = Max(tree[ls].long1,Max(tree[rs].long1,tree[ls].r[1] + tree[rs].l[1]));
}
void pushdown(int num,int s,int e)
{
if(s == e || tree[num].lazy == -1)
return;
int ls = num<<1;
int rs = num<<1|1;
int mid = (s + e)>>1;
if(tree[num].lazy == 0)
{
tree[ls].lazy = tree[rs].lazy = 0;
tree[ls].sum = tree[rs].sum = 0;
tree[ls].long1 = tree[rs].long1 = 0;
tree[ls].l[1] = tree[ls].r[1] = tree[rs].l[1] = tree[rs].r[1] = 0;
tree[ls].l[0] = tree[ls].r[0] = tree[ls].long0 = mid - s + 1;
tree[rs].l[0] = tree[rs].r[0] = tree[rs].long0 = e - mid;
}
if(tree[num].lazy == 1)
{
tree[ls].lazy = tree[rs].lazy = 1;
tree[ls].sum = mid - s + 1;tree[rs].sum = e - mid;
tree[ls].long0 = tree[rs].long0 = 0;
tree[ls].l[0] = tree[ls].r[0] = tree[rs].l[0] = tree[rs].r[0] = 0;
tree[ls].l[1] = tree[ls].r[1] = tree[ls].long1 = mid - s + 1;
tree[rs].l[1] = tree[rs].r[1] = tree[rs].long1 = e - mid;
}
if(tree[num].lazy == 2)
{
if(tree[ls].lazy == -1 || tree[ls].lazy == 2)
{
tree[ls].lazy = 1 - tree[ls].lazy;
tree[ls].sum = mid - s + 1 - tree[ls].sum;
swap(tree[ls].long0,tree[ls].long1);
swap(tree[ls].l[0],tree[ls].l[1]);
swap(tree[ls].r[0],tree[ls].r[1]);
}
else if(tree[ls].lazy == 0)
{
tree[ls].lazy = 1;
tree[ls].sum = tree[ls].long1 = mid - s + 1;
tree[ls].long0 = 0;
tree[ls].l[0] = tree[ls].r[0] = 0;
tree[ls].l[1] = tree[ls].r[1] = mid - s + 1;
}
else if(tree[ls].lazy == 1)//....少了2个else 跪了这么久。。。
{
tree[ls].lazy = 0;
tree[ls].sum = tree[ls].long1 = 0;
tree[ls].long0 = tree[ls].l[0] = tree[ls].r[0] = mid - s + 1;
tree[ls].l[1] = tree[ls].r[1] = 0;
}
if(tree[rs].lazy == -1 || tree[rs].lazy == 2)
{
tree[rs].lazy = 1 - tree[rs].lazy;
tree[rs].sum = e - mid - tree[rs].sum;
swap(tree[rs].long0,tree[rs].long1);
swap(tree[rs].l[0],tree[rs].l[1]);
swap(tree[rs].r[0],tree[rs].r[1]);
}
else if(tree[rs].lazy == 0)//....sb
{
tree[rs].lazy = 1;
tree[rs].sum = tree[rs].long1 = e - mid;
tree[rs].long0 = 0;
tree[rs].l[0] = tree[rs].r[0] = 0;
tree[rs].l[1] = tree[rs].r[1] = e - mid;
}
else if(tree[rs].lazy == 1)
{
tree[rs].lazy = 0;
tree[rs].sum = tree[rs].long1 = 0;
tree[rs].long0 = tree[rs].l[0] = tree[rs].r[0] = e - mid;
tree[rs].l[1] = tree[rs].r[1] = 0;
}
}
tree[num].lazy = -1;
}
void build(int num,int s,int e)
{
tree[num].lazy = -1;tree[num].sum = 0;
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = 0;
tree[num].long1 = tree[num].l[1] = tree[num].r[1] = 0;
if(s == e)
{
scanf("%d",&tree[num].sum);
if(tree[num].sum)
tree[num].long1 = tree[num].l[1] = tree[num].r[1] = 1;
else
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = 1;
return;
}
int mid = (s + e)>>1;
build(num<<1,s,mid);
build(num<<1|1,mid + 1,e);
pushup(num,s,e);
}
void update(int num,int s,int e,int l,int r,int op)
{
if(s == l && e == r)
{
if(op == 0)
{
tree[num].sum = tree[num].long1 = tree[num].lazy = 0;
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = e - s + 1;
tree[num].l[1] = tree[num].r[1] = 0;
}
if(op == 1)
{
tree[num].lazy = 1;
tree[num].sum = tree[num].long1 = e - s + 1;
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = 0;
tree[num].l[1] = tree[num].r[1] = e - s + 1;
}
if(op == 2)
{
switch(tree[num].lazy)
{
case 0:
tree[num].lazy = 1;
tree[num].long1 = tree[num].sum = e - s + 1;
tree[num].l[1] = tree[num].r[1] = e - s + 1;
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = 0;
break;
case 1:
tree[num].lazy = 0;
tree[num].long0 = tree[num].l[0] = tree[num].r[0] = e - s + 1;
tree[num].long1 = tree[num].sum = 0;
tree[num].l[1] = tree[num].r[1] = 0;
break;
case 2:
case -1:
tree[num].lazy = 1 - tree[num].lazy;
tree[num].sum = e - s + 1 - tree[num].sum;
swap(tree[num].long0,tree[num].long1);
swap(tree[num].l[0],tree[num].l[1]);
swap(tree[num].r[0],tree[num].r[1]);
}
}
return;
}
pushdown(num,s,e);
int mid = (s + e)>>1;
if(r <= mid)
update(num<<1,s,mid,l,r,op);
else
{
if(l > mid)
update(num<<1|1,mid + 1,e,l,r,op);
else
{
update(num<<1,s,mid,l,mid,op);
update(num<<1|1,mid + 1,e,mid + 1,r,op);
}
}
pushup(num,s,e);
}
int query(int num,int s,int e,int l,int r,int op)
{
if(s == l && r == e)
{
if(op == 3)
return tree[num].sum;
else
return tree[num].long1;
}
pushdown(num,s,e);
int mid = (s + e)>>1;
if(r <= mid)
return query(num<<1,s,mid,l,r,op);
else
{
if(l > mid)
return query(num<<1|1,mid + 1,e,l,r,op);
else
{
if(op == 3)
return query(num<<1,s,mid,l,mid,op) + query(num<<1|1,mid + 1,e,mid + 1,r,op);
else
{
int ret = Min(tree[num<<1].r[1],mid - l + 1) + Min(tree[num<<1|1].l[1],r - mid);
return Max(query(num<<1,s,mid,l,mid,op),Max(query(num<<1|1,mid + 1,e,mid + 1,r,op),ret));
}
}
}
}
int main()
{
int i,t;
int op,a,b;
scanf("%d",&t);
while(t --)
{
scanf("%d%d",&n,&m);
build(1,1,n);
while(m --)
{
scanf("%d%d%d",&op,&a,&b);
a ++;b ++;
if(op <= 2)
update(1,1,n,a,b,op);
else
printf("%d\n",query(1,1,n,a,b,op));
}
}
return 0;
}
//843MS 8840K
这题调了好久，原来是pushdown的时候犯了一个sb错误，if-else结构写成了3个并列的if。。
引以为戒。。。