hdu 4046 Panda

http://acm.hdu.edu.cn/showproblem.php?pid=4046
　　简单线段树【单点修改，查询区间】，合并两个目标区间，查找区间中目标子串的个数，子串可以有重叠的部分！
AC代码：
View Code
  1 #include <cstring>
  2 #include <cstdio>
  3 #include <cassert>
  4 #include <algorithm>
  5 
  6 using namespace std;
  7 
  8 const int maxn = 50001;
  9 int cnt[maxn << 2];
 10 
 11 #define lson l, m, rt << 1
 12 #define rson m + 1, r, rt << 1 | 1
 13 
 14 char str[maxn];
 15 
 16 void up(int rt, int l, int m, int r) {
 17     cnt[rt] = cnt[rt << 1] + cnt[rt << 1 | 1];
 18     if (m - l) {
 19         if (str[m - 1] == 'w' && str[m] == 'b' && str[m + 1] == 'w') {
 20             cnt[rt]++;
 21         }
 22         if (r - m > 1) {
 23             if (str[m] == 'w' && str[m + 1] == 'b' && str[m + 2] == 'w') {
 24                 cnt[rt]++;
 25             }
 26         }
 27     }
 28 }
 29 
 30 void build(int l, int r, int rt) {
 31     if (l == r) {
 32         cnt[rt] = 0;
 33 
 34         return ;
 35     }
 36     int m = (l + r) >> 1;
 37 
 38     build(lson);
 39     build(rson);
 40     up(rt, l, m, r);
 41 }
 42 
 43 void update(int pos, char ch, int l, int r, int rt) {
 44     if (l == r) {
 45         str[pos] = ch;
 46 
 47         return ;
 48     }
 49     int m = (l + r) >> 1;
 50 
 51     if (pos <= m) update(pos, ch, lson);
 52     else update(pos, ch, rson);
 53     up(rt, l, m, r);
 54 }
 55 
 56 struct Answer {
 57     int l;
 58     int r;
 59     int cnt;
 60 };
 61 
 62 Answer query(int L, int R, int l, int r, int rt) {
 63     Answer ret;
 64 
 65     if (L <= l && r <= R) {
 66         ret.l = l;
 67         ret.r = r;
 68         ret.cnt = cnt[rt];
 69 
 70         return ret;
 71     }
 72     int m = (l + r) >> 1;
 73 
 74     if (L <= m) {
 75         ret = query(L, R, lson);
 76         if (m < R) {
 77             Answer tmp = query(L, R, rson);
 78             ret.cnt += tmp.cnt;
 79 
 80             int t = ret.r;
 81 
 82             if (ret.r - ret.l) {
 83 
 84                 if (str[t - 1] == 'w' && str[t] == 'b' && str[t + 1] == 'w') {
 85                     ret.cnt++;
 86                 }
 87                 if (tmp.r - tmp.l) {
 88                     if (str[t] == 'w' && str[t + 1] == 'b' && str[t + 2] == 'w') {
 89                         ret.cnt++;
 90                     }
 91                 }
 92             } else {
 93                 if (tmp.r - tmp.l) {
 94                     if (str[t] == 'w' && str[t + 1] == 'b' && str[t + 2] == 'w') {
 95                         ret.cnt++;
 96                     }
 97                 }
 98             }
 99             ret.r = tmp.r;
100         }
101     } else if (m < R) {
102         ret = query(L, R, rson);
103     }
104 
105     return ret;
106 }
107 
108 void deal() {
109     int n,m;
110 
111     scanf("%d%d", &n, &m);
112     scanf("%s", str);
113 
114     build(0, n - 1, 1);
115     while (m--) {
116         int op, l, r;
117         char ch[3];
118 
119         scanf("%d", &op);
120         if (op) {
121             scanf("%d %s", &l, ch);
122             update(l, ch[0], 0, n - 1, 1);
123         } else {
124             scanf("%d%d", &l, &r);
125             printf("%d\n", query(l, r, 0, n - 1, 1).cnt);
126         }
127     }
128 }
129 
130 int main() {
131     int T;
132 
133     scanf("%d", &T);
134     for (int i = 1; i <= T; i++) {
135         printf("Case %d:\n", i);
136         deal();
137     }
138 
139     return 0;
140 }
——written by Lyon
posted @ 2012-09-21 12:34 LyonLys 阅读(158) 评论(0) 收藏举报
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