Black And White
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5583 Accepted Submission(s): 1649
Problem Description
There are a bunch of stones on the beach; Stone color is white or black. Little Sheep has a magic brush, she can change the color of a continuous stone, black to white, white to black. Little Sheep like black very much, so she want to know the longest period of consecutive black stones in a range [i, j].
Input
There are multiple cases, the first line of each case is an integer n(1<= n <= 10^5), followed by n integer 1 or 0(1 indicates black stone and 0 indicates white stone), then is an integer M(1<=M<=10^5) followed by M operations formatted as x i j(x = 0 or 1) , x=1 means change the color of stones in range[i,j], and x=0 means ask the longest period of consecutive black stones in range[i,j]
Output
When x=0 output a number means the longest length of black stones in range [i,j].
Sample Input
4 1 0 1 0 5 0 1 4 1 2 3 0 1 4 1 3 3 0 4 4
Sample Output
1 2 0
Source
Recommend
其实就是建两颗树 SWAP一下 维护左边最长连续 右边最长连续 中间最长连续 得一个最大值连续
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX_N = 100005;
struct node {
int len;
int len_l;
int len_r;
int down;
}arr1[4*MAX_N],arr0[4*MAX_N];
void up(int p,int l,int r){
int mid = l+(r-l)/2,len_black,len_white;
if(arr1[p*2].len_l==(mid-l+1)){
arr1[p].len_l = arr1[p*2+1].len_l+arr1[p*2].len_l;
}
else {
arr1[p].len_l = arr1[p*2].len_l;
}
if(arr1[p*2+1].len_r==r-mid){
arr1[p].len_r = arr1[p*2+1].len_r+arr1[p*2].len_r;
}
else {
arr1[p].len_r = arr1[p*2+1].len_r;
}
len_black = arr1[p*2].len_r+arr1[p*2+1].len_l;
arr1[p].len=max(max(max(arr1[p*2].len_l,arr1[p*2+1].len_r),max(arr1[p*2+1].len,arr1[p*2].len)),len_black);
if(arr0[p*2].len_l==(mid-l+1)){
arr0[p].len_l = arr0[p*2+1].len_l+arr0[p*2].len_l;
}
else {
arr0[p].len_l = arr0[p*2].len_l;
}
if(arr0[p*2+1].len_r==r-mid){
arr0[p].len_r = arr0[p*2+1].len_r+arr0[p*2].len_r;
}
else {
arr0[p].len_r = arr0[p*2+1].len_r;
}
len_white = arr0[p*2].len_r+arr0[p*2+1].len_l;
arr0[p].len=max(max(max(arr0[p*2].len_l,arr0[p*2+1].len_r),max(arr0[p*2+1].len,arr0[p*2].len)),len_white);
}
void SWAP(int p){
int len_l = arr1[p].len_l;
int len_r = arr1[p].len_r;
int len_len =arr1[p].len;
arr1[p].len_l = arr0[p].len_l;
arr1[p].len_r = arr0[p].len_r;
arr1[p].len = arr0[p].len;
arr0[p].len_l = len_l;
arr0[p].len_r = len_r;
arr0[p].len = len_len;
}
void down(int p,int l,int r){
if(arr1[p].down){
arr1[p].down^=1;
arr1[p*2].down^=1;
arr1[p*2+1].down^=1;
SWAP(p);
SWAP(p*2);
SWAP(p*2+1);
}
}
void build(int p,int l,int r){
if(l==r){
scanf("%d",&arr1[p].len);
if(arr1[p].len){
arr1[p].len_l = 1;
arr1[p].len_r = 1;
}
else {
arr0[p].len = 1;
arr0[p].len_l = 1;
arr0[p].len_r = 1;
}
return;
}
down(p,l,r);
int mid =l+(r-l)/2;
build(p*2,l,mid);
build(p*2+1,mid+1,r);
up(p,l,r);
}
void modify(int p,int l,int r,int x,int y){
if(x<=l&&r<=y) {
arr1[p].down^=1;
SWAP(p);
return;
}
int mid = l+(r-l)/2;
down(p,l,r);
if(x<=mid) modify(p*2,l,mid,x,y);
if(y>mid) modify(p*2+1,mid+1,r,x,y);
up(p,l,r);
}
node query(int p,int l,int r,int x,int y){
if(x<=l&&r<=y){
return arr1[p];
}
down(p,l,r);
int mid = l+(r-l)/2;
up(p,l,r);
if(x<=mid&&y>mid){
node nod1 = query(p*2,l,mid,x,y);
node nod2 = query(p*2+1,mid+1,r,x,y);
node nod;
if(nod1.len_l==(mid-l+1)){
nod.len_l = nod1.len_l+nod2.len_l;
}
else {
nod.len_l = nod1.len_l;
}
if(nod2.len_r==r-mid){
nod.len_r = nod1.len_r+nod2.len_r;
}
else {
nod.len_r = nod2.len_r;
}
int len_len = nod1.len_r+nod2.len_l;
nod.len=max(max(max(nod.len_r,nod.len_l),max(nod1.len,nod2.len)),len_len);
return nod;
}
else if(x<=mid){
return query(p*2,l,mid,x,y);
}
else if(x>mid){
return query(p*2+1,mid+1,r,x,y);
}
}
int main(){
int n,m;
while(~scanf("%d",&n)){
memset(arr1,0,sizeof(arr1));
memset(arr0,0,sizeof(arr0));
build(1,1,n);
scanf("%d",&m);
while(m--){
int d,a,b;
scanf("%d%d%d",&d,&a,&b);
if(d==1){
modify(1,1,n,a,b);
}
else if(d==0) {
printf("%d\n",query(1,1,n,a,b).len);
}
}
}
return 0;
}