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]
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.
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
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; }