/*
build的时候初始化初始颜色。
区间为[l,r]，闭区间。
inseart(a, b, val, 1);将[a,b]染色为val
find(a, b, 1);询问[a,b]内的颜色，散列存储在color[]中，由ans统计。
*/
#include <stdio.h>
#include <string.h>
#define M 300010
struct POS {
int left, right, col;
} tree[M];
bool color[31];
void swap(int &a, int &b) {
int t = a;
a = b;
b = t;
}
void build(int l, int r, int now) {
tree[now].left = l;
tree[now].right = r;
tree[now].col = 1;
int mid = (l + r) / 2;
if (r > l) {
build(mid + 1, r, now * 2 + 1);
build(l, mid, now * 2);
}
}
void inseart(int a, int b, int val, int now) {
if (tree[now].left >= a && tree[now].right <= b) {
tree[now].col = val;
return;
}
if (tree[now].col > 0) {
tree[now * 2].col = tree[now].col;
tree[now * 2 + 1].col = tree[now].col;
}
tree[now].col = -1;
int mid = (tree[now].left + tree[now].right) / 2;
if (b <= mid)
inseart(a, b, val, now * 2);
else if (a > mid)
inseart(a, b, val, now * 2 + 1);
else {
inseart(a, mid, val, now * 2);
inseart(mid + 1, b, val, now * 2 + 1);
}
}
void find(int a, int b, int now) {
if (tree[now].col > 0) {
color[tree[now].col] = true;
return;
}
if (tree[now].left < tree[now].right) {
int mid = (tree[now].left + tree[now].right) / 2;
if (b <= mid)
find(a, b, now * 2);
else if (a > mid)
find(a, b, now * 2 + 1);
else {
find(a, mid, now * 2);
find(mid + 1, b, now * 2 + 1);
}
}
}
int main() {
int i, l, t, o, a, b, val;
char c[2];
while (scanf("%d %d %d", &l, &t, &o) == 3) {
build(1, l, 1);
while (o--) {
scanf(" %s", &c);
if (c[0] == 'C') {
scanf(" %d %d %d", &a, &b, &val);
if (a > b)
swap(a, b);
inseart(a, b, val, 1);
} else {
scanf(" %d %d", &a, &b);
if (a > b)
swap(a, b);
memset(color, false, sizeof (color));
find(a, b, 1);
int ans = 0;
for (i = 1; i <= t; i++) {
if (color[i])
ans++;
}
printf("%d\n", ans);
}
}
}
return 0;
}
/*测试样例
开始默认颜色为1，C为染色操作，P为查询操作。
Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
*/
Description
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
be larger than B) as an operation defined previously.
Output
Sample Input
2 2 4 C 1 1 2 P 1 2 C 2 2 2 P 1 2
Sample Output
2 1