hdu3397 Sequence operation

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]

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.

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

这道题是前面线段树知识的综合题，也和之前有些不同，是个很好的训练题，做了很长时间啊，还需努力。这里的难点是异或运算，即把区间中的0变成1，1变成0.这里我维护每个区间的llen0,rlen0,tlen0,llen1,rlen1,tlen1,即区间从左端点起的向右最大连续0（1）区间长度，从右端起的最大向左连续0（1）区间长度，以及区间最大的连续0（1）区间长度，然后异或操作的时候，只要把0,1换一下，然后用pushup操作就可以了，再记录一个异或操作符x_or.
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100005
int a[maxn];
struct node{
	int l,r,sum,color,x_or;
	int llen0,rlen0,tlen0;
	int llen1,rlen1,tlen1;
}b[4*maxn];
int max(int a,int b){
	return a>b?a:b;
}
int min(int a,int b){
	return a<b?a:b;
}

void XOR(int i)
{
	int mid;
	swap(b[i].llen0,b[i].llen1);
	swap(b[i].rlen0,b[i].rlen1);
	swap(b[i].tlen0,b[i].tlen1);
	b[i].sum=b[i].r-b[i].l+1-b[i].sum;
}
void pushup(int i)
{
	if(b[i].l==b[i].r)return;
	b[i].tlen0=max(b[i*2].tlen0,b[i*2+1].tlen0);
	b[i].tlen0=max(b[i].tlen0,b[i*2].rlen0+b[i*2+1].llen0);

	b[i].llen0=b[i*2].llen0;b[i].rlen0=b[i*2+1].rlen0;
	if(b[i*2].llen0==b[i*2].r-b[i*2].l+1)b[i].llen0+=b[i*2+1].llen0;
	if(b[i*2+1].rlen0==b[i*2+1].r-b[i*2+1].l+1)b[i].rlen0+=b[i*2].rlen0;

	b[i].tlen1=max(b[i*2].tlen1,b[i*2+1].tlen1);
	b[i].tlen1=max(b[i].tlen1,b[i*2].rlen1+b[i*2+1].llen1);

	b[i].llen1=b[i*2].llen1;b[i].rlen1=b[i*2+1].rlen1;
	if(b[i*2].llen1==b[i*2].r-b[i*2].l+1)b[i].llen1+=b[i*2+1].llen1;
	if(b[i*2+1].rlen1==b[i*2+1].r-b[i*2+1].l+1)b[i].rlen1+=b[i*2].rlen1;

	b[i].sum=b[i*2].sum+b[i*2+1].sum;
}
void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;b[i].color=-1;b[i].x_or=0;
	if(l==r){
		b[i].llen0=b[i].rlen0=b[i].tlen0=a[l]?0:1;
		b[i].llen1=b[i].rlen1=b[i].tlen1=a[l]?1:0;
		b[i].color=a[l];
		b[i].sum=a[l];return;
	}
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
	pushup(i);
}

void pushdown(int i)
{
	int mid;
	if(b[i].l==b[i].r)return;
	if(b[i].color!=-1){   //如果是纯色，那么需要下传
		b[i*2].color=b[i*2+1].color=b[i].color;
		b[i*2].x_or=b[i*2+1].x_or=0; //这里因为左右子树的值和父亲一样，所以如果父亲是纯色且翻转一次，那么传到左右子树也是一样的，不管子树记录多少次（先记为0，再后面一步变成1） 

		b[i*2].llen0=b[i*2].rlen0=b[i*2].tlen0=b[i].color?0:(b[i*2].r-b[i*2].l+1);
		b[i*2].llen1=b[i*2].rlen1=b[i*2].tlen1=b[i].color?(b[i*2].r-b[i*2].l+1):0;
		b[i*2].sum=b[i].color?(b[i*2].r-b[i*2].l+1):0;

		b[i*2+1].llen0=b[i*2+1].rlen0=b[i*2+1].tlen0=b[i].color?0:(b[i*2+1].r-b[i*2+1].l+1);
		b[i*2+1].llen1=b[i*2+1].rlen1=b[i*2+1].tlen1=b[i].color?(b[i*2+1].r-b[i*2+1].l+1):0;
		b[i*2+1].sum=b[i].color?(b[i*2+1].r-b[i*2+1].l+1):0;
		b[i].color=-1;
	}
	if(b[i].x_or){
		b[i*2].x_or^=1;
		b[i*2+1].x_or^=1;
		b[i].x_or=0;
		XOR(i*2);
		XOR(i*2+1);
	}
}

void update(int l,int r,int flag,int i)
{
	int mid;
	//if(b[i].color==flag)return;  这里和之前做过的染色题不同，颜色相同不能直接返回，因为可能这段上有翻转标记，比如整段区间都是0，但是x_or=1,那么其实这段是1，当flag=0时是要更新的，如果返回就错了。
	pushdown(i); //这里也和之前的题目不同，之前都是在第一个if下面写的，这里提前写是因为要把翻转信息提前传给子树，如果没有的话，可能下面第一个if中翻转标记就消失了，那么子树就没有得到翻转信息。
	if(b[i].l==l && b[i].r==r){
		if(flag<2){
			b[i].color=flag;
			b[i].x_or=0;
			b[i].llen0=b[i].rlen0=b[i].tlen0=flag?0:(b[i].r-b[i].l+1);
			b[i].llen1=b[i].rlen1=b[i].tlen1=flag?(b[i].r-b[i].l+1):0;
			b[i].sum=flag?(b[i].r-b[i].l+1):0;
		}
		else{
			b[i].x_or=1;
			XOR(i);
		}
		return;
	}

	mid=(b[i].l+b[i].r)/2;
	if(r<=mid)update(l,r,flag,i*2);
	else if(l>mid)update(l,r,flag,i*2+1);
	else {
		update(l,mid,flag,i*2);
		update(mid+1,r,flag,i*2+1);
	}
	pushup(i);
}

int question(int l,int r,int flag,int i)
{
	int mid,r1,l1,ans1,ans2;
	if(b[i].l==l && b[i].r==r){
		if(flag==3)return b[i].sum;
		else return b[i].tlen1;
	}
	pushdown(i);  //询问操作可以在if后面的
	mid=(b[i].l+b[i].r)/2;
	if(r<=mid) return question(l,r,flag,i*2);
	else if(l>mid)return question(l,r,flag,i*2+1);
	if(flag==3)return question(l,mid,flag,i*2)+question(mid+1,r,flag,i*2+1);

	if(b[i*2+1].l+b[i*2+1].llen1-1>r)r1=r;
	else r1=b[i*2+1].l+b[i*2+1].llen1-1;

	if(b[i*2].r-b[i*2].rlen1+1<l)l1=l;
	else l1=b[i*2].r-b[i*2].rlen1+1;

	ans1=r1-l1+1;
	ans2=max(question(l,mid,flag,i*2),question(mid+1,r,flag,i*2+1));
	return max(ans1,ans2);
}

int main()
{
	int n,m,i,j,T,flag,c,d;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		build(1,n,1);
		for(i=1;i<=m;i++){
			scanf("%d%d%d",&flag,&c,&d);
			c++;d++;
			if(flag<=2){
				update(c,d,flag,1);
			}
			else printf("%d\n",question(c,d,flag,1));
		}
	}
	return 0;
}