http://acm.hdu.edu.cn/showproblem.php?pid=3397
Sequence operation
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5801 Accepted Submission(s): 1713
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
表示再也不想看到线段树了>_<
题意：
0 a b:将区间[a,b]全部置为0;
1 a b:将区间[a,b]全部置为1;
2 a b:对区间[a,b]进行异或操作;
3 a b:询问区间[a,b]内有多少个1;
4 a b:询问区间[a,b]内最长连续的1有多长。
分析：
置0/1很简单成段更新lazy一下就行了，询问有多少个1就是个求和，维护区间内和值即可。询问最长连续的1是个区间合并，我们要维护3个值，区间左顶点开始的最长连续1,右顶点开始的最长联系1,区间内的最长连续1，这样维护父节点的时候要注意他的左右孩子是否连续。
异或操作就比较复杂，我们不仅要维护1的信息，还要维护0的信息。如果该区间内的值相同（setv!=-1）,那么setv^=1并且交换0/1信息的值，否则XOR^=1并且交换0/1信息的值。pushdown有些麻烦，如果setv[k]!=-1(k是当前节点,lc是左孩子,rc是右孩子),那么他两个孩子之前的异或操作都会被覆盖掉，所以下传setv[k],并将两个孩子的XOR标记清空;如果XOR[k]==1，那么将两个孩子自身的0/1信息互换并将XOR下传给左右孩子。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn 100007
using namespace std;
int XOR[maxn<<2],setv[maxn<<2],lm1[maxn<<2],rm1[maxn<<2],lm0[maxn<<2],rm0[maxn<<2],sum[maxn<<2],msum1[maxn<<2],msum0[maxn<<2];
inline void SWAP(int k,int l,int r)
{
swap(lm1[k],lm0[k]);swap(rm1[k],rm0[k]);swap(msum1[k],msum0[k]);sum[k]=r-l-sum[k];
}
inline void pushup(int k,int l,int r)
{
int lc=k*2+1,rc=k*2+2,m=l+r>>1;
sum[k]=sum[lc]+sum[rc];
if (lm1[lc]==m-l) lm1[k]=lm1[lc]+lm1[rc]; else lm1[k]=lm1[lc];
if (rm1[rc]==r-m) rm1[k]=rm1[rc]+rm1[lc]; else rm1[k]=rm1[rc];
msum1[k]=max(rm1[lc]+lm1[rc],max(msum1[lc],msum1[rc]));
if (lm0[lc]==m-l) lm0[k]=lm0[lc]+lm0[rc]; else lm0[k]=lm0[lc];
if (rm0[rc]==r-m) rm0[k]=rm0[rc]+rm0[lc]; else rm0[k]=rm0[rc];
msum0[k]=max(rm0[lc]+lm0[rc],max(msum0[lc],msum0[rc]));
}
inline void pushdown(int k,int l,int r)
{
int lc=k*2+1,rc=k*2+2,m=l+r>>1;
if (setv[k]!=-1)
{
setv[lc]=setv[rc]=setv[k];
XOR[lc]=XOR[rc]=0;
lm1[lc]=rm1[lc]=msum1[lc]=sum[lc]=setv[k]?m-l:0;
lm0[lc]=rm0[lc]=msum0[lc]=setv[k]?0:m-l;
lm1[rc]=rm1[rc]=msum1[rc]=sum[rc]=setv[k]?r-m:0;
lm0[rc]=rm0[rc]=msum0[rc]=setv[k]?0:r-m;
setv[k]=-1;
}
if (XOR[k])
{
if (setv[lc]!=-1) setv[lc]^=1; else XOR[lc]^=1;
if (setv[rc]!=-1) setv[rc]^=1; else XOR[rc]^=1;
SWAP(lc,l,m);SWAP(rc,m,r);
XOR[k]=0;
}
}
void update(int a,int b,int v,int k,int l,int r)
{
if (b<=l || r<=a) return ;
if (a<=l && r<=b)
{
setv[k]=v;
lm1[k]=rm1[k]=msum1[k]=sum[k]=v?r-l:0;
lm0[k]=rm0[k]=msum0[k]=v?0:r-l;
XOR[k]=0;
}
else
{
pushdown(k,l,r);
int m=l+r>>1;
update(a,b,v,k*2+1,l,m);
update(a,b,v,k*2+2,m,r);
pushup(k,l,r);
}
}
void change(int a,int b,int k,int l,int r)
{
if (b<=l || r<=a) return ;
if (a<=l && r<=b)
{
if (setv[k]!=-1)
setv[k]^=1;
else
XOR[k]^=1;
SWAP(k,l,r);
}
else
{
pushdown(k,l,r);
int m=l+r>>1;
change(a,b,k*2+1,l,m);
change(a,b,k*2+2,m,r);
pushup(k,l,r);
}
}
int query_sum(int a,int b,int k,int l,int r)
{
if (b<=l || r<=a) return 0;
if (a<=l && r<=b) return sum[k];
if (r-l!=1) pushdown(k,l,r);
int m=l+r>>1,v1=0,v2=0;
v1=query_sum(a,b,k*2+1,l,m);
v2=query_sum(a,b,k*2+2,m,r);
return v1+v2;
}
int query_lc1(itn a,int b,int k,int l,int r)
{
if (b<=l || r<=a) return 0;
if (a<=l && r<=b) return msum1[k];
if (r-l!=1) pushdown(k,l,r);
int m=l+r>>1,v1=0,v2=0,v3=0;
v1=query_lc1(a,b,k*2+1,l,m);
v2=query_lc1(a,b,k*2+2,m,r);
v3=min(b,m+lm1[k*2+2])-max(a,m-rm1[k*2+1]);
return max(v3,max(v1,v2));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE
int T_T,n,m,x,a,b,op;
scanf("%d",&T_T);
while (T_T--)
{
scanf("%d %d",&n,&m);
for (int i=0;i<n;i++)
{
scanf("%d",&x);
update(i,i+1,x,0,0,n);
}
while (m--)
{
scanf("%d %d %d",&op,&a,&b);
if (op==0)
{
update(a,b+1,0,0,0,n);
continue;
}
if (op==1)
{
update(a,b+1,1,0,0,n);
continue;
}
if (op==2)
{
change(a,b+1,0,0,n);
continue;
}
if (op==3)
{
printf("%d\n",query_sum(a,b+1,0,0,n));
continue;
}
if (op==4)
{
printf("%d\n",query_lc1(a,b+1,0,0,n));
continue;
}
}
}
return 0;
}