HDU 3397 Sequence operation (线段树)

最新推荐文章于 2019-10-06 08:12:56 发布

原创最新推荐文章于 2019-10-06 08:12:56 发布 · 515 阅读

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本文深入探讨了段树区间合并的概念及其应用，并通过具体实例展示了如何利用线段树优化算法实现高效的区间操作。文章详细介绍了算法背后的逻辑、实现细节以及在实际编程中的应用，旨在为读者提供一种解决类似问题的强大工具。

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Sequence operation

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4362 Accepted Submission(s): 1258

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

线段树区间合并。每个节点维护当前区间1个总数sum,当前区间最长序列msum，左端点往右的最长序列lsum，右端点往左的最长序列rsum。注意把异或操作转换成成段更新即可，既当异或操作覆盖整个区间的时候，如果该区间有懒惰标记，把标记进行异或。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 111111
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

using namespace std;

int sum[SIZE<<2];
int msum[SIZE<<2],lsum[SIZE<<2],rsum[SIZE<<2],cv[SIZE<<2];

void pushUp(int rt,int itv)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
    lsum[rt] = lsum[rt<<1];
    rsum[rt] = rsum[rt<<1|1];
    if(lsum[rt] == (itv - (itv>>1)))
        lsum[rt] += lsum[rt<<1|1];
    if(rsum[rt] == (itv >> 1))
        rsum[rt] += rsum[rt<<1];
    msum[rt] = max(max(msum[rt<<1],msum[rt<<1|1]),lsum[rt<<1|1]+rsum[rt<<1]);
}

void pushDown(int rt,int itv)
{
    if(cv[rt] != -1)
    {
        cv[rt<<1] = cv[rt<<1|1] = cv[rt];
        sum[rt<<1] = cv[rt]*(itv - (itv>>1));
        sum[rt<<1|1] = cv[rt]*(itv >> 1);
        msum[rt<<1] = lsum[rt<<1] = rsum[rt<<1] = sum[rt<<1];
        msum[rt<<1|1] = lsum[rt<<1|1] = rsum[rt<<1|1] = sum[rt<<1|1];
        cv[rt] = -1;
    }
}

void build(int l,int r,int rt)
{
    if(l == r)
    {
        scanf("%d",&cv[rt]);
        msum[rt] = lsum[rt] = rsum[rt] = sum[rt] = cv[rt]?1:0;
        return ;
    }
    cv[rt] = -1;
    int mid = (l + r) >> 1;
    build(ls);
    build(rs);
    pushUp(rt,r-l+1);
}

void update(int l,int r,int rt,int L,int R,int w)
{
    if(L <= l && r <= R)
    {
        sum[rt] = w*(r-l+1);
        msum[rt] = lsum[rt] = rsum[rt] = sum[rt];
        cv[rt] = w;
        return ;
    }
    pushDown(rt,r-l+1);
    int mid = (l + r) >> 1;
    if(L <= mid) update(ls,L,R,w);
    if(R > mid) update(rs,L,R,w);
    pushUp(rt,r-l+1);
}

void do_XOR(int l,int r,int rt,int L,int R)
{
    if(L <= l && r <= R && cv[rt] != -1)
    {
        cv[rt] ^= 1;
        sum[rt] = msum[rt] = lsum[rt] = rsum[rt] = cv[rt]?(r-l+1):0;
        return ;
    }
    pushDown(rt,r-l+1);
    int mid = (l + r) >> 1;
    if(L <= mid) do_XOR(ls,L,R);
    if(R > mid) do_XOR(rs,L,R);
    pushUp(rt,r-l+1);
}

int querySum(int l,int r,int rt,int L,int R)
{
    if(L <= l && r <= R)
        return sum[rt];
    pushDown(rt,r-l+1);
    int mid = (l + r) >> 1;
    int ret = 0;
    if(L <= mid) ret += querySum(ls,L,R);
    if(R > mid) ret += querySum(rs,L,R);
    return ret;
}

int queryNum(int l,int r,int rt,int L,int R)
{
    if(L <= l && r <= R)
        return msum[rt];
    pushDown(rt,r-l+1);
    int mid = (l + r) >> 1;
    if(L > mid)
        return queryNum(rs,L,R);
    else if(R <= mid)
        return queryNum(ls,L,R);
    else
    {
        int ll = queryNum(ls,L,R);
        int rr = queryNum(rs,L,R);
        int Lt = min(rsum[rt<<1],mid-L+1);
        int Rt = min(lsum[rt<<1|1],R-mid);
        return max(max(ll,rr),Lt+Rt);
    }
}

int T;
int N,M;
int o,s,e;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&N,&M);
        build(1,N,1);
        while(M--)
        {
            scanf("%d%d%d",&o,&s,&e);
            s ++; e ++;
            if(o == 0 || o == 1)
                update(1,N,1,s,e,o);
            else if(o == 2)
                do_XOR(1,N,1,s,e);
            else if(o == 3)
                printf("%d\n",querySum(1,N,1,s,e));
            else
                printf("%d\n",queryNum(1,N,1,s,e));
        }
    }
    return 0;
}