HDU 3397 Sequence operation (线段树)

Sequence operation

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4357    Accepted Submission(s): 1256

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

 

Author

lxhgww&&shǎ崽

 

Source

HDOJ Monthly Contest – 2010.05.01

 

Recommend

lcy

题意：给个一只有0，1 的序列，有以下5种操作
0 a b change all characters into '0's in [a , b] //把a，b间的都改为0
1 a b change all characters into '1's in [a , b] //把a，b间都改为1
2 a b change all '0's into '1's and change all '1's into '0's in [a, b] //1的改为0，0的改为1
Output operations:
3 a b output the number of '1's in [a, b]  //a，b间1 的数量
4 a b output the length of the longest continuous '1' string in [a , b]//a，b间最大连续的1的个数

思路：经典的线段树，只是操作过多，但其实很简单，相当于把以前做过的题的操作合在一块 什么值的更改，区间和的查询，最开连续的个数 。所以没什么特殊的

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

const int N=100010;

int num[N];

struct node{
    int l,r,len;    //len 是该区间的长度
    int sum,cover;  //sum 是该区间1的数量，cover 有3个值-1，0，1分别表示该区间是否被0，1完全覆盖
    int lm,rm,ma;   //lm rm ma 分别表示 左连续 右连续 和最大的连续1的个数
}tree[N<<2];

void PushUp(int rt){    //区间合并，由子结点修改父结点的信息
    tree[rt].sum=tree[L(rt)].sum+tree[R(rt)].sum;
    if(tree[L(rt)].len==tree[L(rt)].lm)
        tree[rt].lm=tree[L(rt)].len+tree[R(rt)].lm;
    else
        tree[rt].lm=tree[L(rt)].lm;

    if(tree[R(rt)].len==tree[R(rt)].rm)
        tree[rt].rm=tree[R(rt)].len+tree[L(rt)].rm;
    else
        tree[rt].rm=tree[R(rt)].rm;

    int a=max(tree[L(rt)].ma,tree[R(rt)].ma);
    int b=max(tree[rt].lm,tree[rt].rm);
    tree[rt].ma=max(max(a,b),tree[L(rt)].rm+tree[R(rt)].lm);
}

void build(int l,int r,int rt){
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].len=r-l+1;
    tree[rt].cover=-1;
    if(l==r){
        if(num[l]){
            tree[rt].sum=1;
            tree[rt].cover=1;
            tree[rt].lm=tree[rt].rm=tree[rt].ma=1;
        }else{
            tree[rt].sum=0;
            tree[rt].cover=0;
            tree[rt].lm=tree[rt].rm=tree[rt].ma=0;
        }
        return ;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void PushDown(int rt){  //延迟操作，由父结点修改子结点
    tree[L(rt)].cover=tree[rt].cover;
    tree[R(rt)].cover=tree[rt].cover;
    if(tree[rt].cover){
        tree[L(rt)].lm=tree[L(rt)].rm=tree[L(rt)].ma=tree[L(rt)].len;
        tree[L(rt)].sum=tree[L(rt)].len;
        tree[R(rt)].lm=tree[R(rt)].rm=tree[R(rt)].ma=tree[R(rt)].len;
        tree[R(rt)].sum=tree[R(rt)].len;
    }else{
        tree[L(rt)].lm=tree[L(rt)].rm=tree[L(rt)].ma=0;
        tree[L(rt)].sum=0;
        tree[R(rt)].lm=tree[R(rt)].rm=tree[R(rt)].ma=0;
        tree[R(rt)].sum=0;
    }
    tree[rt].cover=-1;
}

void Update1(int flag,int l,int r,int rt){  //都是一个套路 左边，右边，或一起更新
    if(tree[rt].l==l && tree[rt].r==r){
        tree[rt].cover=flag;
        if(flag){
            tree[rt].lm=tree[rt].rm=tree[rt].ma=tree[rt].r-tree[rt].l+1;
            tree[rt].sum=tree[rt].ma;
        }else{
            tree[rt].lm=tree[rt].rm=tree[rt].ma=0;
            tree[rt].sum=0;
        }
        return ;
    }
    if(tree[rt].cover!=-1)
        PushDown(rt);
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(r<=mid)
        Update1(flag,l,r,L(rt));
    else if(l>=mid+1)
        Update1(flag,l,r,R(rt));
    else{
        Update1(flag,l,mid,L(rt));
        Update1(flag,mid+1,r,R(rt));
    }
    PushUp(rt);
}

void Update2(int l,int r,int rt){
    if(tree[rt].l==l && tree[rt].r==r && (tree[rt].sum==0 || tree[rt].sum==tree[rt].len)){
        if(tree[rt].sum>0){
            tree[rt].lm=tree[rt].rm=tree[rt].ma=0;
            tree[rt].sum=0;
            tree[rt].cover=0;
        }else{
            tree[rt].lm=tree[rt].rm=tree[rt].ma=tree[rt].len;
            tree[rt].sum=tree[rt].len;
            tree[rt].cover=1;
        }
        return ;
    }
    if(tree[rt].cover!=-1)
        PushDown(rt);
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(r<=mid)
        Update2(l,r,L(rt));
    else if(l>=mid+1)
        Update2(l,r,R(rt));
    else{
        Update2(l,mid,L(rt));
        Update2(mid+1,r,R(rt));
    }
    PushUp(rt);
}

int Query(int flag,int l,int r,int rt){ //查询操作 因为根据flag不同返回不同的值
    if(tree[rt].l==l && tree[rt].r==r){
        if(flag)
            return tree[rt].ma;
        else
            return tree[rt].sum;
    }
    if(tree[rt].cover!=-1)
        PushDown(rt);
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(r<=mid)
        return Query(flag,l,r,L(rt));
    else if(l>=mid+1)
        return Query(flag,l,r,R(rt));
    else{   //这是的return得注意，当返回 连续1的个数时 注意中间断开的情况
        if(flag){   //min (node[L(u)].rm,node[L(u)].r-left+1) 这是求左子结点右边最大的连续 右子结点同理
            int a=max(Query(flag,l,mid,L(rt)),Query(flag,mid+1,r,R(rt)));
            return max(a,min(tree[L(rt)].rm,tree[L(rt)].r-l+1)+min(tree[R(rt)].lm,r-tree[R(rt)].l+1));
        }else
            return (Query(flag,l,mid,L(rt))+Query(flag,mid+1,r,R(rt)));
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        build(1,n,1);
        int op,a,b;
        while(m--){
            scanf("%d%d%d",&op,&a,&b);
            a++;b++;
            if(op==0)
                Update1(0,a,b,1);
            else if(op==1)
                Update1(1,a,b,1);
            else if(op==2)
                Update2(a,b,1);
            else if(op==3)
                printf ("%d\n",Query(0,a,b,1));
            else
                printf ("%d\n",Query(1,a,b,1));
        }
    }
    return 0;
}