HDU 3397（线段树，较难~）

最新推荐文章于 2021-11-15 22:13:06 发布

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#tree #output #build #struct #string #input

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本文介绍了一种涉及字符序列的编程问题，通过五种操作实现对序列中字符的修改和查询，重点在于使用线段树进行高效处理。文章详细展示了如何通过线段树结构来完成这些操作，并提供了一个具体的样例输入输出。

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Sequence operation

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 927 Accepted Submission(s): 259

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
/*看了好几个大牛的代码，学习思路，然后自己敲  手都敲酸了~~~~  @-@  线段树恐惧症
前几次不是WA就是TLE~~ 最后全删了又继续敲，总共花了1天半的时间~~~*/
#include <iostream>
using namespace std;
#define  N 100010
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)

struct LineTree 
{
	int l,r,mid,len;
	int count0,left0,right0,max0;
	int count1,left1,right1,max1;
	bool fill0,fill1,change;
}tree[N*3];

int val[N];

void Update(int root)
{
	tree[root].count0 = tree[root<<1].count0 + tree[(root<<1)+1].count0;
	tree[root].count1 = tree[root<<1].count1 + tree[(root<<1)+1].count1;

	tree[root].max0 = max( max(tree[root<<1].max0,tree[(root<<1)+1].max0),tree[root<<1].right0 + tree[(root<<1)+1].left0 );
	tree[root].max1 = max( max(tree[root<<1].max1,tree[(root<<1)+1].max1),tree[root<<1].right1 + tree[(root<<1)+1].left1 );

	tree[root].left0 = tree[root<<1].left0;
	tree[root].right0 = tree[(root<<1)+1].right0;

	if(tree[root<<1].left0 == tree[root<<1].len)
		tree[root].left0 += tree[(root<<1)+1].left0;

	if(tree[(root<<1)+1].right0 == tree[(root<<1)+1].len)
		tree[root].right0 += tree[root<<1].right0;

	tree[root].left1 = tree[root<<1].left1;
	tree[root].right1 = tree[(root<<1)+1].right1;

	if(tree[root<<1].left1 == tree[root<<1].len)
		tree[root].left1 += tree[(root<<1)+1].left1;

	if(tree[(root<<1)+1].right1 == tree[(root<<1)+1].len)
		tree[root].right1 += tree[root<<1].right1;
}

void Build(int root,int a,int b)
{
	tree[root].l=a;
	tree[root].r=b;
	tree[root].mid=(a+b)>>1;
	tree[root].len=b-a;
	tree[root].fill0 = tree[root].fill1 = tree[root].change = 0;
	if(a+1 == b) 
	{
		tree[root].left0 = tree[root].right0 = tree[root].count0 = tree[root].max0 = 0;
		tree[root].left1 = tree[root].right1 = tree[root].count1 = tree[root].max1 = 0;
		if(val[a])
			tree[root].left1 = tree[root].right1 = tree[root].count1 = tree[root].max1 = 1;
		else
			tree[root].left0 = tree[root].right0 = tree[root].count0 = tree[root].max0 = 1;
		return;
	}
	Build(root<<1,a,tree[root].mid);
	Build((root<<1)+1,tree[root].mid,b);
	Update(root);
}

void Fill0(int root)
{
	tree[root].count0 = tree[root].max0 = tree[root].left0 = tree[root].right0 = tree[root].len;
	tree[root].count1 = tree[root].max1 = tree[root].left1 = tree[root].right1 = 0;
	tree[root].fill0 = 0;
	if(tree[root].len == 1) return;
	tree[root<<1].fill0 = tree[(root<<1)+1].fill0 = 1;
	tree[root<<1].fill1 = tree[(root<<1)+1].fill1 = 0;
	tree[root<<1].change = tree[(root<<1)+1].change = 0;
}

void Fill1(int root)
{
	tree[root].count1 = tree[root].max1 = tree[root].left1 = tree[root].right1 = tree[root].len;
	tree[root].count0 = tree[root].max0 = tree[root].left0 = tree[root].right0 = 0;
	tree[root].fill1 = 0;
	if(tree[root].len == 1) return;
	tree[root<<1].fill0 = tree[(root<<1)+1].fill0 = 0;
	tree[root<<1].fill1 = tree[(root<<1)+1].fill1 = 1;
	tree[root<<1].change = tree[(root<<1)+1].change = 0;
}

void Change(int root)
{
	swap(tree[root].count0,tree[root].count1);
	swap(tree[root].left0,tree[root].left1);
	swap(tree[root].right0,tree[root].right1);
	swap(tree[root].max0,tree[root].max1);
	tree[root].change = 0;
	if(tree[root].len == 1) return;
	tree[root<<1].change ^= 1;
	tree[(root<<1)+1].change ^= 1;
}

void Down(int root)
{
	if(tree[root].fill0) Fill0(root);
	if(tree[root].fill1) Fill1(root);
	if(tree[root].change) Change(root);
}

void Oper(int root,int a,int b,int x)
{
	Down(root);
	if(a == tree[root].l && b == tree[root].r)
	{
		if(x == 0) tree[root].fill0 = 1;
		else if(x == 1) tree[root].fill1 = 1;
		else tree[root].change ^= 1;
		Down(root);
		return;
	}
	if( b <= tree[root].mid ) Oper(root<<1,a,b,x);
	else if( tree[root].mid <= a) Oper((root<<1)+1,a,b,x);
	else
	{
		Oper(root<<1,a,tree[root].mid,x);
		Oper((root<<1)+1,tree[root].mid,b,x);
	}
	Down(root<<1);
	Down((root<<1)+1);
	Update(root);
}

int Onenum(int root,int a,int b)
{
	Down(root);
	if(a == tree[root].l && b == tree[root].r)
	{
		return tree[root].count1;
	}
	if(b <= tree[root].mid) return Onenum(root<<1,a,b);
	else if( tree[root].mid <= a) return Onenum((root<<1)+1,a,b);
	else
	{
		return Onenum(root<<1,a,tree[root].mid) + Onenum((root<<1)+1,tree[root].mid,b);
	}
}

int Conenum(int root,int a,int b)
{
	Down(root);
	if(a == tree[root].l && b == tree[root].r)
	{
		return tree[root].max1;
	}
	if(b <= tree[root].mid) return Conenum(root<<1,a,b);
	else if(tree[root].mid <= a) return Conenum((root<<1)+1,a,b);
	else
	{
		int m1,m2,res;
		m1 = Conenum(root<<1,a,tree[root].mid);
		m2 = Conenum((root<<1)+1,tree[root].mid,b);
		res = min(tree[root<<1].right1,tree[root].mid-a) + min(tree[(root<<1)+1].left1,b-tree[root].mid);
		return max(res,max(m1,m2));
	}
}

int main()
{
	int t,n,m,i,x,a,b;
	scanf("%d",&t);
	while (t--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
			scanf("%d",&val[i]);
		Build(1,0,n);
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&x,&a,&b);
			if(x<=2) Oper(1,a,b+1,x);
			else if(x==3) printf("%d/n",Onenum(1,a,b+1));
			else printf("%d/n",Conenum(1,a,b+1));
		}
	}
	return 0;
}