题目大意:给定一个序列,然后有M次操作;
- 0 l r:表示询问l,r中最大连续1的个数
- 1 l r:表示将l,r区间上的数取反
解题思路:线段树的一种题型,区间合并,由于有一个取反的操作,所以对于每一个节点要维护6个值,包含连续0,1最长序列的长度,左边和右边的最长连续长度。须要注意的是,假设询问的区间最大值是从R[lson] + L[rson]来到,要推断是否比长度大于r - l + 1。一開始没注意,所以WA了,上网搜了下别人的题解,发现非常多人在query里面没有pushup更新当前节点信息,这样肯定是不行的,有pushdown,肯定要在更新当前节点信息的。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int N, M, a[maxn];
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], filp[maxn << 2];
int L[maxn << 2][2], R[maxn << 2][2], S[maxn << 2][2];
void maintain (int u) {
filp[u] ^= 1;
swap(L[u][0], L[u][1]);
swap(R[u][0], R[u][1]);
swap(S[u][0], S[u][1]);
}
void pushup(int u) {
for (int i = 0; i < 2; i++) {
S[u][i] = max(max(S[lson(u)][i], S[rson(u)][i]), R[lson(u)][i] + L[rson(u)][i]);
L[u][i] = L[lson(u)][i] + (L[lson(u)][i] == rc[lson(u)] - lc[lson(u)] + 1 ? L[rson(u)][i] : 0);
R[u][i] = R[rson(u)][i] + (R[rson(u)][i] == rc[rson(u)] - lc[rson(u)] + 1 ? R[lson(u)][i] : 0);
}
}
void pushdown (int u) {
if (filp[u]) {
maintain(lson(u));
maintain(rson(u));
filp[u] = 0;
}
}
void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;
filp[u] = 0;
if (l == r) {
int d = a[l];
L[u][d] = R[u][d] = S[u][d] = 1;
L[u][d^1] = R[u][d^1] = S[u][d^1] = 0;
return ;
}
int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid+1, r);
pushup(u);
}
void modify (int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u);
return;
}
pushdown(u);
int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
modify(lson(u), l, r);
if (r > mid)
modify(rson(u), l, r);
pushup(u);
}
int query (int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return S[u][1];
pushdown(u);
int mid = (lc[u] + rc[u]) / 2, ret;
if (r <= mid)
ret = query(lson(u), l, r);
else if (l > mid)
ret = query(rson(u), l, r);
else {
int ll = query(lson(u), l, r);
int rr = query(rson(u), l, r);
int a = min(L[rson(u)][1], r - mid);
int b = min(R[lson(u)][1], mid - l + 1);
ret = max( max(ll, rr), a + b);
}
pushup(u);
return ret;
}
int main () {
int x, l, r;
while (scanf("%d", &N) == 1) {
for (int i = 1; i <= N; i++)
scanf("%d", &a[i]);
build (1, 1, N);
scanf("%d", &M);
while (M--) {
scanf("%d%d%d", &x, &l, &r);
if (x)
modify(1, l, r);
else
printf("%d
", query(1, l, r));
}
}
return 0;
}