题目链接：

解题思路：比较巧妙，状态压缩----最多三十种颜色，每一位表示每个颜色状态，那么使用逻辑或运算即可避免颜色重复计算的问题，统计颜色的时候判断1的位数即可。

延迟标记的传递时候需要更改左右孩子的值以及其标记！

代码：

 

1 const int maxn = 1e5 + 5; 2  3 int tree[maxn * 4], tag[maxn * 4]; 4 int hc[32]; 5 int n, t, o; 6  7 void build(int l, int r, int k){ 8     tree[k] = hc[1]; tag[k] = 1; 9     if(l == r) return; 10     int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;11     build(l, mid, lc);12     build(mid + 1, r, rc);13 }14 void update(int ul, int ur, int x, int l, int r, int k){15     if(ul <= l && ur >= r){16         tree[k] = hc[x];17         tag[k] = x;18         return;19     }20     if(ul > r || ur < l) return;21     22     int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;23     if(tag[k] != 0){24         tree[k] = hc[tag[k]];25         tag[lc] = tag[k]; tree[lc] = hc[tag[k]];26         tag[rc] = tag[k]; tree[rc] = hc[tag[k]];27         tag[k] = 0;28     }29     30     update(ul, ur, x, l, mid, lc);31     update(ul, ur, x, mid + 1, r, rc);32     tree[k] = tree[lc] | tree[rc];33 }34 int query(int ql, int qr, int l, int r, int k){35     if(ql <= l && qr >= r) {36         if(tag[k]) tree[k] = hc[tag[k]];37         return tree[k];38     }39     if(ql > r || qr < l) return 0;40     41     int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1;42     if(tag[k] != 0){43         tree[k] = hc[tag[k]];44         tag[lc] = tag[k]; tree[lc] = hc[tag[k]];45         tag[rc] = tag[k]; tree[rc] = hc[tag[k]];46         tag[k] = 0;47     }48     int q1 = query(ql, qr, l, mid, lc);49     int q2 = query(ql, qr, mid + 1, r, rc);50     return q1 | q2;51 }52 int getAmo(int x){53     int ans = 0;54     while(x > 0){55         if(x & 1) ans++;56         x >>= 1;57     }58     return ans;59 }60 61 int main(){62     for(int i = 1, x = 1; i <= 30; i++){63         hc[i] = x;64         x <<= 1;    65     }66     scanf("%d %d %d", &n, &t, &o);67     build(1, n, 1);68     for(int i = 0; i < o; i++){69         char ch;70         scanf(" %c", &ch);71         if(ch == 'C'){72             int u, v, c;73             scanf("%d %d %d", &u, &v, &c);74             if(u > v) swap(u, v);75             update(u, v, c, 1, n, 1);76         }77         else{78             int u, v;79             scanf("%d %d", &u, &v);80             if(u > v) swap(u, v);81             int tmp = query(u, v, 1, n, 1);82             tmp = getAmo(tmp);83             printf("%d\n", tmp);84         }85     }86 }

 

 

 

题目：

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46657		Accepted: 14131

Description

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. 

There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board: 

1. "C A B C" Color the board from segment A to segment B with color C. 

2. "P A B" Output the number of different colors painted between segment A and segment B (including). 

In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your. 

Input

First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.

Output

Ouput results of the output operation in order, each line contains a number.

Sample Input

2 2 4C 1 1 2P 1 2C 2 2 2P 1 2

Sample Output

21