Black And White
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3811 Accepted Submission(s): 1129
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]
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 2011 Multi-University Training Contest 8 - Host by HUST 裸裸的区间更新，注意点已在代码中写出
#include#include #include #include #include #include #include using namespace std; #define MAXN 100000 + 1000 struct node{ int lsum0,rsum0,msum0,lsum1,rsum1,msum1; int ck; int l,r; int mid(){ return (l+r)>>1; } }tree[MAXN<<2]; int a[MAXN]; void PushUp(int rt){ int ll = tree[rt<<1].r-tree[rt<<1].l+1;//左子树区间的长度 int rr = tree[rt<<1|1].r-tree[rt<<1|1].l+1;//右子树区间的长度 tree[rt].lsum1 = tree[rt<<1].lsum1; if(tree[rt<<1].lsum1 == ll){ tree[rt].lsum1 = tree[rt].lsum1+tree[rt<<1|1].lsum1; } tree[rt].rsum1 = tree[rt<<1|1].rsum1; if(tree[rt<<1|1].rsum1 == rr){ tree[rt].rsum1 = tree[rt].rsum1+tree[rt<<1].rsum1; } tree[rt].lsum0 = tree[rt<<1].lsum0; if(tree[rt<<1].lsum0 == ll){ tree[rt].lsum0+=tree[rt<<1|1].lsum0; } tree[rt].rsum0 = tree[rt<<1|1].rsum0; if(tree[rt<<1|1].rsum0 == rr){ tree[rt].rsum0+=tree[rt<<1].rsum0; } tree[rt].msum0=max(max(tree[rt<<1].msum0,tree[rt<<1|1].msum0),tree[rt<<1].rsum0+tree[rt<<1|1].lsum0); tree[rt].msum1 = max(max(tree[rt<<1].msum1,tree[rt<<1|1].msum1),tree[rt<<1].rsum1+tree[rt<<1|1].lsum1); } void PushDown(int rt){ if(tree[rt].ck == 1){ tree[rt<<1].ck ^= 1;//现在左子树父亲节点的所有01都需要交换，所以左子树也是需要交换，那么左子树的左右子树也是需要交换，如果原来左子树就要交换 tree[rt<<1|1].ck ^= 1;//，那么交换了两次就相当于不需要交换，这样异或之后就可以不需要交换 tree[rt].ck = 0; swap(tree[rt<<1].lsum0,tree[rt<<1].lsum1); swap(tree[rt<<1].rsum0,tree[rt<<1].rsum1); swap(tree[rt<<1].msum0,tree[rt<<1].msum1); swap(tree[rt<<1|1].lsum0,tree[rt<<1|1].lsum1); swap(tree[rt<<1|1].rsum0,tree[rt<<1|1].rsum1); swap(tree[rt<<1|1].msum0,tree[rt<<1|1].msum1); } } void build(int l,int r,int rt){ tree[rt].l = l; tree[rt].r = r; tree[rt].ck = 0;//开始的时候都不转换 if(l == r){ if(a[l] == 0){ tree[rt].lsum0=tree[rt].rsum0=tree[rt].msum0=1; tree[rt].lsum1=tree[rt].rsum1=tree[rt].msum1=0; } else{ tree[rt].lsum1=tree[rt].rsum1=tree[rt].msum1=1; tree[rt].lsum0=tree[rt].rsum0=tree[rt].msum0=0; } return ; } //int m = (l+r)>>1; int m = tree[rt].mid(); build(l,m,rt<<1); build(m+1,r,rt<<1|1); PushUp(rt); } void update(int l,int r,int rt){ if(l<=tree[rt].l && tree[rt].r<=r){ tree[rt].ck ^= 1; swap(tree[rt].lsum0,tree[rt].lsum1); swap(tree[rt].rsum0,tree[rt].rsum1); swap(tree[rt].msum0,tree[rt].msum1); return ; } PushDown(rt); int m = tree[rt].mid(); if(r <= m){ update(l,r,rt<<1); } else if(l > m){ update(l,r,rt<<1|1); } else{ update(l,m,rt<<1); update(m+1,r,rt<<1|1); } PushUp(rt); } int query(int l,int r,int rt){ if(l<=tree[rt].l && tree[rt].r<=r){ return tree[rt].msum1; } PushDown(rt); int m = tree[rt].mid(); if(r <= m){ return query(l,r,rt<<1); } else if(l > m){ return query(l,r,rt<<1|1); } else{ int lr = query(l,m,rt<<1); int rr = query(m+1,r,rt<<1|1); int a = tree[rt<<1].rsum1;//左子树最长的连续的1 if(a > tree[rt<<1].r-l+1){//后面的那个是最大范围 a = tree[rt<<1].r-l+1; } int b = tree[rt<<1|1].lsum1; if(b > r-tree[rt<<1|1].l+1){ b = r-tree[rt<<1|1].l+1; } return max(max(lr,rr),a+b); } } int main(){ int i; int n,m,op,aa,b; while(~scanf("%d",&n)){ for(i=1;i<=n;i++){ scanf("%d",&a[i]); } build(1,n,1); scanf("%d",&m); while(m--){ scanf("%d%d%d",&op,&aa,&b); if(op == 0){ int ans = query(aa,b,1); printf("%d\n",ans); } else{ update(aa,b,1); } } } return 0; }