CF240F (26颗线段树计数)

题目链接：Topcoder----洛谷

题目大意：

　　给定一个长为

思路：

　　用26颗线段树分别统计在每个位置上是否有对应的字母

　　每次操作：

　　　　1.有出现次数为奇数的字母：

　　　　　　大于1，不可能回文，无法操作。

　　　　　　等于1，放到中间

　　　　2.全是偶数次数：

　　　　　　分别放两端

  1 # include<bits/stdc++.h>
  2 using namespace std;
  3 #define endl "\n"
  4 # define int long long
  5 # define ls u<<1
  6 # define rs u<<1|1
  7 const int N = 1e5 + 10;
  8 char a[N], p;
  9 int n, m;
 10 struct segtree {
 11     int sum[4 * N], lazy[4 * N];
 12     void pushup(int u) {
 13         sum[u] = (sum[ls] + sum[rs]);
 14     }
 15 
 16     void build(int tr, int u, int l, int r) {
 17         lazy[u] = -1;
 18         if (l == r) {
 19             sum[u] = ((a[l] - 'a' ) == tr);
 20             return;
 21         }
 22         int mid = l + r >> 1;
 23         build(tr, ls, l, mid);
 24         build(tr, rs, mid + 1, r);
 25         pushup(u);
 26     }
 27 
 28     void pushdown(int u, int l, int r) {
 29         int mid = l + r >> 1;
 30         if (lazy[u] != -1) {
 31             sum[ls] = (mid - l + 1) * lazy[u];
 32             sum[rs] = (r - mid) * lazy[u];
 33 
 34             lazy[ls] = lazy[u];
 35             lazy[rs] = lazy[u];
 36             lazy[u] = -1;
 37         }
 38 
 39     }
 40 
 41     void modify(int u, int l, int r, int L, int R, int c) {
 42         if (l > r || l > R || r < L) return;
 43         if (L <= l && r <= R) {
 44             sum[u] = (r - l + 1) * c;
 45             lazy[u] = c;
 46             return;
 47         }
 48         int mid = l + r >> 1;
 49         pushdown(u, l, r);
 50         if (L <= mid) modify(ls, l, mid, L, R, c);
 51         if (mid + 1 <= R) modify(rs, mid + 1, r, L, R, c);
 52         pushup(u);
 53     }
 54 
 55     int query(int u, int l, int r, int L, int R) {
 56         if (l > r || l > R || r < L) return 0;
 57         if (l >= L && r <= R) {
 58             return sum[u];
 59         }
 60         pushdown(u, l, r);
 61         int mid = l + r >> 1;
 62         int ans = 0;
 63         if (L <= mid) ans += query(ls, l, mid, L, R);
 64         if (R > mid) ans += query(rs, mid + 1, r, L, R);
 65         return ans;
 66     }
 67 } tr[26];
 68 
 69 signed main() {
 70     ios::sync_with_stdio(false);
 71     cin.tie(0);
 72     cout.tie(0);
 73     freopen("input.txt", "r", stdin);
 74     freopen("output.txt", "w", stdout);
 75     cin >> n >> m;
 76     string s;
 77     cin >> s;
 78     for (int i = 1; i <= n; ++i) a[i] = s[i - 1];
 79     for (int i = 0; i < 26; ++i) tr[i].build(i, 1, 1, n);
 80     for (int t = 1; t <= m; ++t) {
 81         int l, r;
 82         cin >> l >> r;
 83         int odd = 0;
 84         int tmp[26] = {0}, key;
 85         for (int i = 0; i < 26; ++i) tmp[i] = tr[i].query(1, 1, n, l, r);//记录在区间[l,r]中每个字母出现的次数
 86         for (int i = 0; i < 26; ++i) if (tmp[i] & 1) odd++, key = i;
 87         if (odd > 1) continue;
 88         for (int i = 0; i < 26; ++i) tr[i].modify(1, 1, n, l, r, 0);
 89         if (odd) {
 90             --tmp[key];
 91             tr[key].modify(1, 1, n, (l + r) / 2, (l + r) / 2, 1);//奇数置中
 92         }
 93         int nl = l, nr = r;
 94         for (int i = 0; i < 26; ++i) {//偶数分两边放
 95             if (tmp[i]) {
 96                 tr[i].modify(1, 1, n, nl, nl + tmp[i] / 2 - 1, 1);
 97                 nl += tmp[i] / 2;
 98                 tr[i].modify(1, 1, n, nr - tmp[i] / 2 + 1, nr, 1);
 99                 nr -= tmp[i] / 2;
100             }
101         }
102     }
103     for (int i = 1; i <= n; ++i) {
104         for (int j = 0; j < 26; ++j) {
105             if (tr[j].query(1, 1, n, i, i)) {
106                 cout << (char)(j + 'a');
107             }
108         }
109     }
110     return 0;
111 }