題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3397 這道題是一個(gè)我認(rèn)為非常非常麻煩的題，操作非常多，而且要打兩個(gè)懶惰標(biāo)記，并且懶惰標(biāo)記有優(yōu)先級(jí)之分，當(dāng)一個(gè)節(jié)點(diǎn)收到一個(gè)覆蓋標(biāo)記的時(shí)候，取反標(biāo)記就應(yīng)該立即消失，這是一個(gè)麻煩，另外一個(gè)麻煩就是一個(gè)結(jié)點(diǎn)需要維持七個(gè)變量以滿足本題的操作和所求。區(qū)間合并 的基本思想不用贅述，下面有一個(gè)成形的代碼（貌似有bug），不過想法已經(jīng)體現(xiàn)出來。
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1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 #define lson l, m, rt<<1
6 #define rson m + 1, r, rt<<1|1
7 const int maxn = 111111;
8 int l1s[maxn<<2], r1s[maxn<<2], m1s[maxn<<2], l0s[maxn<<2], r0s[maxn<<2], m0s[maxn<<2], sum[maxn<<2];
9 int cover[maxn<<2], col[maxn<<2];
10 void PushDown(int rt, int m){
11 if (cover[rt] != -1){
12 cover[rt<<1] = cover[rt<<1|1] = cover[rt];
13 col[rt] = col[rt<<1] = col[rt<<1|1] = 0;
14 l1s[rt<<1] = r1s[rt<<1] = m1s[rt<<1] = sum[rt<<1] = cover[rt] ? m - (m >> 1) : 0;
15 l1s[rt<<1|1] = r1s[rt<<1|1] = m1s[rt<<1|1] = sum[rt<<1|1] = cover[rt] ? (m >> 1) : 0;
16 l0s[rt<<1] = r0s[rt<<1] = m0s[rt<<1] = cover[rt] ? 0 : m - (m >> 1);
17 l0s[rt<<1|1] = r0s[rt<<1|1] = m0s[rt<<1|1] = cover[rt] ? 0 : (m >> 1);
18 cover[rt] = -1;
19 }
20 else if (col[rt]){
21 col[rt<<1] += col[rt];
22 col[rt<<1|1] += col[rt];
23 col[rt] %= 2;
24 if (col[rt]){
25 swap(l1s[rt<<1], l0s[rt<<1]);
26 swap(l1s[rt<<1|1], l0s[rt<<1|1]);
27 swap(r1s[rt<<1], r0s[rt<<1]);
28 swap(r1s[rt<<1|1], r0s[rt<<1|1]);
29 swap(m1s[rt<<1], m0s[rt<<1]);
30 swap(m1s[rt<<1|1], m0s[rt<<1|1]);
31 sum[rt<<1] = (m - (m>>1)) - sum[rt<<1];
32 sum[rt<<1|1] = (m>>1) - sum[rt>>1|1];
33 }
34 col[rt] = 0;
35 }
36 }
37 void PushUp(int rt, int m){
38 l1s[rt] = l1s[rt<<1]; l0s[rt] = l0s[rt<<1];
39 r1s[rt] = r1s[rt<<1|1]; r0s[rt] = r0s[rt<<1|1];
40 if (l1s[rt] == m - (m >> 1)) l1s[rt] += l1s[rt<<1|1];
41 if (r1s[rt] == (m >> 1)) r1s[rt] += r1s[rt<<1];
42 if (l0s[rt] == m - (m >> 1)) l0s[rt] += l0s[rt<<1|1];
43 if (r0s[rt] == (m >> 1)) r0s[rt] += r0s[rt<<1];
44 m1s[rt] = max(l1s[rt<<1|1] + r1s[rt<<1], max(m1s[rt<<1], m1s[rt<<1|1]));
45 m0s[rt] = max(l0s[rt<<1|1] + r0s[rt<<1], max(m0s[rt<<1], m0s[rt<<1|1]));
46 sum[rt] = sum[rt<<1] + sum[rt<<1|1];
47 }
48 void build(int l, int r, int rt){
49 cover[rt] = -1; col[rt] = 0;
50 if (l == r){
51 scanf("%d", &sum[rt]);
52 l1s[rt] = r1s[rt] = m1s[rt] = cover[rt] = sum[rt];
53 l0s[rt] = r0s[rt] = m0s[rt] = 1 - sum[rt];
54 return;
55 }
56 int m = (l + r) >> 1;
57 build(lson);
58 build(rson);
59 PushUp(rt, r - l + 1);
60 }
61 void update(int ll, int rr, int c, int l, int r, int rt){
62 if (ll <= l && rr >= r){
63 if (c == 2){
64 col[rt] += 1;
65 sum[rt] = (r - l + 1) - sum[rt];
66 swap(m1s[rt], m0s[rt]);
67 swap(l1s[rt], l0s[rt]);
68 swap(r1s[rt], r0s[rt]);
69 }
70 else{
71 l1s[rt] = r1s[rt] = m1s[rt] = sum[rt] = c ? (r - l + 1) : 0;
72 l0s[rt] = r0s[rt] = m0s[rt] = c ? 0 : r - l + 1;
73 col[rt] = 0;
74 cover[rt] = c;
75 }
76 }
77 PushDown(rt, r - l + 1);
78 int m = (l + r) >> 1;
79 if (ll <= m) update(ll, rr, c, lson);
80 if (rr > m) update(ll, rr, c, rson);
81 PushUp(rt, r - l + 1);
82 }
83 int query_sum(int ll, int rr, int l, int r, int rt){
84 if (ll <= l && rr >= r) return sum[rt];
85 PushDown(rt, r - l + 1);
86 int m = (l + r) >> 1;
87 int ret = 0;
88 if (ll <= m) ret += query_sum(ll, rr, lson);
89 if (rr > m) ret += query_sum(ll, rr, rson);
90 return ret;
91 }
92 int query_maxlen(int ll, int rr, int l, int r, int rt){
93 if (ll == l && rr == r) return m1s[rt];
94 PushDown(rt, r - l + 1);
95 int m = (l + 1) >> 1;
96 if (rr <= m) return query_maxlen(ll, rr, lson);
97 else if (ll >= m) return query_maxlen(ll, rr, rson);
98 else{
99 int a = query_maxlen(ll, m, lson);
100 int b = query_maxlen(m+1, rr, rson);
101 int c = 0;
102 c += max(r1s[rt<<1], m - ll + 1);
103 c += max(l1s[rt<<1|1], rr - m);
104 return max(a, max(b, c));
105 }
106 }
107 int main(){
108 int t;
109 scanf("%d", &t);
110 while(t--){
111 int op, a, b;
112 int n, m;
113 scanf("%d%d", &n, &m);
114 build(1, n, 1);
115 while(m--){
116 scanf("%d%d%d", &op, &a, &b);
117 a += 1; b += 1;
118 if (op <= 2)
119 update(a, b, op, 1, n, 1);
120 else if (op == 3){
121 int ans = query_sum(a, b, 1, n, 1);
122 printf("%d\n", ans);
123 }
124 else{
125 int ans = query_maxlen(a, b, 1, n, 1);
126 printf("%d\n", ans);
127 }
128 }
129 }
130 return 0;
131
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