Codeforces Round #104 (Div. 1) E. Lucky Queries 线段树

链接:

http://codeforces.com/contest/145/problem/E

题意:

维护一个只有数字4和7的数列,以下有两种操作: 

1.将[l,r]区间4变为7,7变为4。 

2.询问全局最长不下降子序列的长度。 

输出所有2操作的答案

题解:

线段树维护。 

维护四个信息: 

1.区间4的个数。 

2.区间7的个数。 

3.区间不降子序列的长度。 

4.区间不升子序列的长度。

 

具体更新的看代码

代码:

31 int n, m;
32 string s;
33 struct Node {
34     int f, s, fs, sf, rev;
35 }Tree[MAXN << 2];
36 
37 void pushup(int rt) {
38     Tree[rt].f = Tree[rt << 1].f + Tree[rt << 1 | 1].f;
39     Tree[rt].s = Tree[rt << 1].s + Tree[rt << 1 | 1].s;
40     Tree[rt].fs = max(Tree[rt << 1].f + Tree[rt << 1 | 1].fs, Tree[rt << 1].fs + Tree[rt << 1 | 1].s);
41     Tree[rt].sf = max(Tree[rt << 1].s + Tree[rt << 1 | 1].sf, Tree[rt << 1].sf + Tree[rt << 1 | 1].f);
42 }
43 
44 void pushdown(int rt) {
45     if (Tree[rt].rev) {
46         swap(Tree[rt << 1].f, Tree[rt << 1].s);
47         swap(Tree[rt << 1 | 1].f, Tree[rt << 1 | 1].s);
48         swap(Tree[rt << 1].sf, Tree[rt << 1].fs);
49         swap(Tree[rt << 1 | 1].sf, Tree[rt << 1 | 1].fs);
50         Tree[rt << 1].rev ^= 1;
51         Tree[rt << 1 | 1].rev ^= 1;
52         Tree[rt].rev = 0;
53     }
54 }
55 
56 void build(int l, int r, int rt) {
57     if (l == r) {
58         if (s[l - 1] == '4') Tree[rt].f = 1;
59         else Tree[rt].s = 1;
60         Tree[rt].fs = Tree[rt].sf = 1;
61         return;
62     }
63     int m = (l + r) >> 1;
64     build(lson);
65     build(rson);
66     pushup(rt);
67 }
68 
69 void update(int L, int R, int l, int r, int rt) {
70     if (L <= l && r <= R) {
71         Tree[rt].rev ^= 1;
72         swap(Tree[rt].f, Tree[rt].s);
73         swap(Tree[rt].fs, Tree[rt].sf);
74         return;
75     }
76     pushdown(rt);
77     int m = (l + r) >> 1;
78     if (L <= m) update(L, R, lson);
79     if (R > m) update(L, R, rson);
80     pushup(rt);
81 }
82 
83 int main() {
84     ios::sync_with_stdio(false), cin.tie(0);
85     cin >> n >> m >> s;
86     build(1, n, 1);
87     while (m--) {
88         string s;
89         cin >> s;
90         if (s == "switch") {
91             int l, r;
92             cin >> l >> r;
93             update(l, r, 1, n, 1);
94         }
95         else cout << Tree[1].fs << endl;
96     }
97     return 0;
98 }

 

posted @ 2017-08-17 22:18  Flowersea  阅读(183)  评论(0编辑  收藏  举报