Sequence operation
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3114 Accepted Submission(s): 909
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]
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.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 100003
using namespace std;
struct node
{
int lm,rm,m;//lm代表从区间左端开始以1开始的最长连续序列，rm类同，m代表区间最长的连续1了
int len,cover,sum;//sum就是成段更新的附带水货了
};
node st[N<<2];
void up(int &k)
{
int ls=k<<1,rs=k<<1|1;
st[k].sum=st[ls].sum+st[rs].sum;
st[k].lm=st[ls].lm==st[ls].len?st[ls].len+st[rs].lm:st[ls].lm;
st[k].rm=st[rs].rm==st[rs].len?st[rs].len+st[ls].rm:st[rs].rm;
st[k].m=max(max(st[ls].m,st[rs].m),st[ls].rm+st[rs].lm);
st[k].m=max(max(st[k].lm,st[k].rm),st[k].m);//经研究、发现这句可以不用
}
void qup(node &rt,node &ls,node &rs)
{
rt.lm=ls.lm==ls.len?ls.len+rs.lm:ls.lm;
rt.rm=rs.rm==rs.len?rs.len+ls.rm:rs.rm;
rt.m=max(max(ls.m,rs.m),ls.rm+rs.lm);
rt.m=max(max(rt.lm,rt.rm),rt.m);//这里就类同了，可以不要这句、是多余的比较了
}
void down(int &k)
{
st[k<<1].cover=st[k<<1|1].cover=st[k].cover;
st[k<<1].lm=st[k<<1].rm=st[k<<1].m=st[k<<1].sum=st[k].cover?st[k<<1].len:0;
st[k<<1|1].lm=st[k<<1|1].rm=st[k<<1|1].m=st[k<<1|1].sum=st[k].cover?st[k<<1|1].len:0;
st[k].cover=-1;
}
void build(int l,int r,int k)
{
st[k].len=r-l+1;
if(l==r)
{
scanf("%d",&st[k].cover);
st[k].lm=st[k].rm=st[k].m=st[k].sum=st[k].cover?1:0;
return ;
}
st[k].cover=-1;
int m=(l+r)>>1;
build(lson);
build(rson);
up(k);
}
bool flag;
void update(int &L,int &R,int l,int r,int k)
{ //printf("%d ",flag);
if(L<=l&&R>=r)
{
st[k].lm=st[k].rm=st[k].m=st[k].sum=flag?st[k].len:0;
st[k].cover=flag;
return ;
}
if(st[k].cover!=-1)
down(k);
int m=(l+r)>>1;
if(L<=m) update(L,R,lson);
if(R>m) update(L,R,rson);
up(k);
}
void Xor(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r&&st[k].cover!=-1)
{ // printf("k=%d %d %d\n",l,r,st[k].cover);
st[k].cover^=1;
st[k].lm=st[k].rm=st[k].m=st[k].sum=st[k].cover?st[k].len:0;
return ;
}
if(st[k].cover!=-1)
down(k);
int m=(l+r)>>1;
if(L<=m) Xor(L,R,lson);
if(R>m) Xor(L,R,rson);
up(k);
}
int Qsum(int L,int R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
return st[k].sum;
}
if(st[k].cover!=-1)
down(k);
int m=(l+r)>>1;
int t1=0,t2=0;
if(L<=m) t1=Qsum(L,R,lson);
if(R>m) t2=Qsum(L,R,rson);
return t1+t2;
}
node Qconti(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
return st[k];
}
if(st[k].cover!=-1)
down(k);
int m=(l+r)>>1;
node t1,t2;
t1.len=t2.len=0;
if(L<=m) t1=Qconti(L,R,lson);
if(R>m) t2=Qconti(L,R,rson);
if(t1.len&&t2.len)
{ node te;
qup(te,t1,t2);
te.len=t1.len+t2.len;
return te;
}
if(t1.len)
return t1;
return t2;
}
int main()
{
int T;
scanf("%d",&T);
int n,m;
int op,a,b;
node t;
while(T--)
{
scanf("%d%d",&n,&m);
n--;
build(0,n,1);
while(m--)
{
scanf("%d%d%d",&op,&a,&b);
switch(op)
{
case 0:flag=0;update(a,b,0,n,1);break;
case 1:flag=1;update(a,b,0,n,1);break;
case 2:Xor(a,b,0,n,1);break;
case 3:printf("%d\n",Qsum(a,b,0,n,1));break;
case 4:t=Qconti(a,b,0,n,1);printf("%d\n",t.m);break;
}
}
}
return 0;
}