[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4592

[算法]

对于操作1 , 我们首先查询区间[l0 , r0]中有多少个1 , 然后二分求出最大的x(x <= r1)使得[l1 , x]中0的个数 <= [l0 , r0]中1的个数

对于操作2 , 对于一段区间[l , r] , 我们将它分成[l , mid]和[mid + 1 , r]两个子区间 , 那么 , 最长连续的0的个数有三种情况 :

1. 在[l , mid]中 2. 在[mid + 1 , r]中 3. mid向前延伸最多的0的个数 + (mid + 1)向后延伸最多的0的个数

线段树维护即可

时间复杂度 : O(NlogN ^ 2)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200010
typedef long long ll;
typedef long double ld;
const int inf = 1e9; int n , m; template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
T f = ; x = ;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
x *= f;
} struct Segment_Tree
{
struct Node
{
int l , r;
int lm0 , rm0 , value , cnt;
int tag;
} a[MAXN << ];
inline void build(int index , int l , int r)
{
a[index].l = l , a[index].r = r;
a[index].cnt = r - l + ;
a[index].tag = -;
if (l == r) return;
int mid = (l + r) >> ;
build(index << , l , mid);
build(index << | , mid + , r);
}
inline void update(int x)
{
int l = a[x].l , r = a[x].r;
int mid = (l + r) >> ;
a[x].cnt = a[x << ].cnt + a[x << | ].cnt;
if (a[x << ].lm0 == mid - l + ) a[x].lm0 = mid - l + + a[x << | ].lm0;
else a[x].lm0 = a[x << ].lm0;
if (a[x << | ].rm0 == r - mid) a[x].rm0 = r - mid + a[x << ].rm0;
else a[x].rm0 = a[x << | ].rm0;
a[x].value = max(a[x << ].value , a[x << | ].value);
chkmax(a[x].value , a[x << ].rm0 + a[x << | ].lm0);
}
inline void pushdown(int index)
{
int l = a[index].l , r = a[index].r;
int mid = (l + r) >> ;
if (l == r) return;
if (a[index].tag == )
{
a[index << ].cnt = a[index << | ].cnt = ;
a[index << ].value = mid - l + ;
a[index << | ].value = r - mid;
a[index << ].lm0 = a[index << ].rm0 = mid - l + ;
a[index << | ].lm0 = a[index << | ].rm0 = r - mid;
a[index << ].tag = a[index << | ].tag = a[index].tag;
a[index].tag = -;
}
if (a[index].tag == )
{
a[index << ].cnt = mid - l + ;
a[index << | ].cnt = r - mid;
a[index << ].value = a[index << | ].value = ;
a[index << ].lm0 = a[index << ].rm0 = ;
a[index << | ].lm0 = a[index << | ].rm0 = ;
a[index << ].tag = a[index << | ].tag = a[index].tag;
a[index].tag = -;
}
}
inline void modify(int index , int l , int r)
{
if (a[index].l == l && a[index].r == r)
{
a[index].cnt = ;
a[index].lm0 = a[index].rm0 = r - l + ;
a[index].value = r - l + ;
a[index].tag = ;
return;
}
pushdown(index);
int mid = (a[index].l + a[index].r) >> ;
if (mid >= r) modify(index << , l , r);
else if (mid + <= l) modify(index << | , l , r);
else
{
modify(index << , l , mid);
modify(index << | , mid + , r);
}
update(index);
}
inline int queryA(int index , int l , int r)
{
if (a[index].l == l && a[index].r == r)
return a[index].cnt;
pushdown(index);
int mid = (a[index].l + a[index].r) >> ;
if (mid >= r) return queryA(index << , l , r);
else if (mid + <= l) return queryA(index << | , l , r);
else return queryA(index << , l , mid) + queryA(index << | , mid + , r);
}
inline int queryB(int index , int l , int r)
{
if (a[index].l == l && a[index].r == r)
return a[index].value;
pushdown(index);
int mid = (a[index].l + a[index].r) >> ;
if (mid >= r) return queryB(index << , l , r);
else if (mid + <= l) return queryB(index << | , l , r);
{
int ret = max(queryB(index << , l , mid) , queryB(index << | , mid + , r));
chkmax(ret , min(mid - l + , a[index << ].rm0) + min(r - mid , a[index << | ].lm0));
return ret;
}
}
inline void change(int index , int l , int r)
{
if (l > r) return;
if (a[index].l == l && a[index].r == r)
{
a[index].cnt = r - l + ;
a[index].lm0 = a[index].rm0 = a[index].value = ;
a[index].tag = ;
return;
}
pushdown(index);
int mid = (a[index].l + a[index].r) >> ;
if (mid >= r) change(index << , l , r);
else if (mid + <= l) change(index << | , l , r);
else
{
change(index << , l , mid);
change(index << | , mid + , r);
}
update(index);
}
} SGT; int main()
{ read(n); read(m);
SGT.build( , , n);
for (int i = ; i <= m; i++)
{
int type;
read(type);
if (type == )
{
int x , y;
read(x); read(y);
SGT.modify( , x , y);
} else if (type == )
{
int l0 , r0 , l1 , r1;
read(l0); read(r0); read(l1); read(r1);
int cnt = SGT.queryA( , l0 , r0);
SGT.modify( , l0 , r0);
int l = l1 , r = r1 , loc = ;
while (l <= r)
{
int mid = (l + r) >> ;
if ((mid - l1 + ) - SGT.queryA( , l1 , mid) <= cnt)
{
loc = mid;
l = mid + ;
} else r = mid - ;
}
SGT.change( , l1 , loc);
} else
{
int l , r;
read(l); read(r);
printf("%d\n" , SGT.queryB( , l , r));
}
} return ;
}

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