Sequence operation
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4952 Accepted Submission(s): 1452
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
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
2
6
5
Author
lxhgww&&shǎ崽
Source
Recommend
lcy
题意：
给你一个0,1数组。你可以对数组进行以下操作。
0 x y 把[x,y]之间的元素置为0。
1 x y 把[x,y]之间的元素置为1。
2 x y 把[x,y]之间的元素置为1变成0,0变成1。
3 x y 求[x,y]之间元素1的个数。
4 x y 求[x,y]之间最长连续1的长度。
思路：
对于0,1,3,4操作都很传统。区间更新区间统计。但是由于多了2操作所以就要多维护关于0的信息。这样进行2操作的时候之间交换1和0的信息就行了。还有查询的时候也有注意的地方。开始没注意那个地方导致wa数次。反复检查冗长的代码无数次。。。。可怜我的时间呀。。。哎。。。。。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100100;
int sta[maxn<<2],ml0[maxn<<2],mr0[maxn<<2],ml1[maxn<<2];//sta为标记。ml1为1左端连续。ml0为0左端连续
int mr1[maxn<<2],ma1[maxn<<2],ma0[maxn<<2],one[maxn<<2];//ma1为最大1连续数.one记录1的个数
void opp(int L,int R,int k)//对于2操作。0和1互换
{
swap(ma0[k],ma1[k]);
swap(ml0[k],ml1[k]);
swap(mr0[k],mr1[k]);
one[k]=R-L-one[k]+1;
}
void pushup(int L,int R,int k)//下传标记
{
int ls,rs,mid;
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
ml0[k]=ml0[ls];
mr0[k]=mr0[rs];
ml1[k]=ml1[ls];
mr1[k]=mr1[rs];
if(ml0[ls]==mid-L+1)
ml0[k]+=ml0[rs];
if(mr0[rs]==R-mid)
mr0[k]+=mr0[ls];
if(ml1[ls]==mid-L+1)
ml1[k]+=ml1[rs];
if(mr1[rs]==R-mid)
mr1[k]+=mr1[ls];
ma1[k]=max(ma1[ls],ma1[rs]);
ma1[k]=max(ma1[k],mr1[ls]+ml1[rs]);
ma0[k]=max(ma0[ls],ma0[rs]);
ma0[k]=max(ma0[k],mr0[ls]+ml0[rs]);
one[k]=one[ls]+one[rs];
}
void pushdown(int L,int R,int k)//上传
{
int ls,rs,mid;
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
//printf("mark %d->%d and %d->%d %d\n",L,mid,mid+1,R,sta[k]);
if(sta[k]==0)
{
sta[ls]=sta[rs]=0;
ma0[ls]=ml0[ls]=mr0[ls]=mid-L+1;
ma1[ls]=ml1[ls]=mr1[ls]=one[ls]=ma1[rs]=ml1[rs]=mr1[rs]=one[rs]=0;
ma0[rs]=ml0[rs]=mr0[rs]=R-mid; }
else if(sta[k]==1)
{
sta[ls]=sta[rs]=1;
ma0[ls]=ml0[ls]=mr0[ls]=ma0[rs]=ml0[rs]=mr0[rs]=0;
ma1[ls]=ml1[ls]=mr1[ls]=one[ls]=mid-L+1;
ma1[rs]=ml1[rs]=mr1[rs]=one[rs]=R-mid;
}
else
{
if(sta[ls]!=-1)//2操作对于0,1标记直接0,1标记互换
{
if(sta[ls]==2)//原先有2直接变-1
sta[ls]=-1;
else
sta[ls]^=1;
}
else
sta[ls]=2;
if(sta[rs]!=-1)
{
if(sta[rs]==2)
sta[rs]=-1;
else
sta[rs]^=1;
}
else
sta[rs]=2;
opp(L,mid,ls);
opp(mid+1,R,rs);
}
sta[k]=-1;
}
void btree(int L,int R,int k)
{
int ls,rs,mid;
sta[k]=-1;
if(L==R)
{
scanf("%d",&one[k]);
if(one[k])
{
ma1[k]=ml1[k]=mr1[k]=1;
ma0[k]=ml0[k]=mr0[k]=0;
}
else
{
ma1[k]=ml1[k]=mr1[k]=0;
ma0[k]=ml0[k]=mr0[k]=1;
}
return ;
}
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
btree(L,mid,ls);
btree(mid+1,R,rs);
pushup(L,R,k);
//printf("%d->%d\n",L,R);
//printf("%d---%d----%d\n",ml1[k],mr1[k],one[k]);
}
void update(int L,int R,int l,int r,int k,int op)
{
int ls,rs,mid;
if(l==L&&r==R)
{
if(op==0)
{
sta[k]=0;
ma0[k]=ml0[k]=mr0[k]=R-L+1;
ma1[k]=ml1[k]=mr1[k]=one[k]=0;
}
else if(op==1)
{
sta[k]=1;
ma0[k]=ml0[k]=mr0[k]=0;
ma1[k]=ml1[k]=mr1[k]=one[k]=R-L+1;
}
else
{
if(sta[k]==-1)
sta[k]=2;
else if(sta[k]==2)
sta[k]=-1;
else
sta[k]^=1;
opp(L,R,k);
}
//printf("mark %d->%d %d\n",L,R,op);
//printf("%d->%d\n",L,R);
//printf("%d---%d----%d\n",ml1[k],mr1[k],one[k]);
return ;
}
if(sta[k]!=-1)
pushdown(L,R,k);
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(l>mid)
update(mid+1,R,l,r,rs,op);
else if(r<=mid)
update(L,mid,l,r,ls,op);
else
{
update(L,mid,l,mid,ls,op);
update(mid+1,R,mid+1,r,rs,op);
}
pushup(L,R,k);
//printf("%d->%d\n",L,R);
//printf("%d---%d----%d\n",ml1[k],mr1[k],one[k]);
}
int qu(int L,int R,int l,int r,int k,int op)
{
int ls,rs,mid,tmp,ll,rr;
if(sta[k]==0)
return 0;
if(sta[k]==1)
return r-l+1;
if(l==L&&r==R)
{
if(op==3)
return one[k];
else
return ma1[k];
}
if(sta[k]!=-1)
pushdown(L,R,k);
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(l>mid)
return qu(mid+1,R,l,r,rs,op);
else if(r<=mid)
return qu(L,mid,l,r,ls,op);
else
{
if(op==3)
return qu(L,mid,l,mid,ls,op)+qu(mid+1,R,mid+1,r,rs,op);
else//4对于分离操作尤其注意！！最大值只能是以下几种情况。
{
tmp=max(qu(L,mid,l,mid,ls,op),qu(mid+1,R,mid+1,r,rs,op));//最大值在左儿子或右儿子中
ll=max(mid-mr1[ls]+1,l);//最大值在中间区域
rr=min(mid+ml1[rs],r);//注意范围
tmp=max(tmp,rr-ll+1);
return tmp;
}
} }
int main()
{
int t,n,m,op,a,b; scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
btree(1,n,1);
while(m--)
{
scanf("%d%d%d",&op,&a,&b);
a++,b++;
if(op<3)
update(1,n,a,b,1,op);
else
printf("%d\n",qu(1,n,a,b,1,op));
}
}
return 0;
}