題目鏈接： http://poj.org/problem?id=2777
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
Output
Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
Source
題意：
給一個(gè)固定長(zhǎng)度為L(zhǎng)的畫(huà)板
有兩個(gè)操作：
C A B C：區(qū)間A--B內(nèi)涂上顏色C。
P A B：查詢區(qū)間AB內(nèi)顏色種類數(shù)。
PS:此題和 HDU：5023 是類似的！
附題解： http://blog.csdn.net/u012860063/article/details/39434665
代碼例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define LL int
const int maxn = 110017;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
//把當(dāng)前結(jié)點(diǎn)的信息更新到父結(jié)點(diǎn)
sum[rt] = sum[rt<<1] | sum[rt<<1|1];//總共的顏色
}
void PushDown(int rt,int m)
{
if(add[rt])
{
add[rt<<1] = add[rt];
add[rt<<1|1] = add[rt];
sum[rt<<1] = add[rt];
sum[rt<<1|1] = add[rt];
add[rt] = 0;//將標(biāo)記向兒子節(jié)點(diǎn)移動(dòng)后，父節(jié)點(diǎn)的延遲標(biāo)記去掉
//傳遞后，當(dāng)前節(jié)點(diǎn)標(biāo)記域清空
}
}
void build(int l,int r,int rt)
{
add[rt] = 0;//初始化為全部結(jié)點(diǎn)未被標(biāo)記
if (l == r)
{
sum[rt] = 1;//初始顏色為1
return ;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if (L <= l && r <= R)
{
add[rt] =1<<(c-1);//位運(yùn)算左移表示有某種顏色
sum[rt] =1<<(c-1);
return ;
}
PushDown(rt , r - l + 1);//----延遲標(biāo)志域向下傳遞
int mid = (l + r) >> 1;
if (L <= mid)
update(L , R , c , lson);//更新左兒子
if (mid < R)
update(L , R , c , rson);//更新右兒子
PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
if (L <= l && r <= R)
{
return sum[rt];
}
//要取rt子節(jié)點(diǎn)的值時(shí)，也要先把rt的延遲標(biāo)記向下移動(dòng)
PushDown(rt , r - l + 1);
int mid = (l + r) >> 1;
LL ret = 0;
if (L <= mid)
ret |= query(L , R , lson);
if (mid < R)
ret |= query(L , R , rson);
return ret;
}
int main()
{
int L, T, O;
int a, b, c;
while(~scanf("%d%d%d",&L,&T,&O))
{
build(1, L, 1);//建樹(shù)
while(O--)//Q為詢問(wèn)次數(shù)
{
char op[2];
scanf("%s",op);
if(op[0] == 'P')
{
scanf("%d%d",&a,&b);
if(a > b)
{
int t = a;
a = b;
b = t;
}
LL tt=query(a, b, 1, L, 1);
int ans = 0;
while(tt)
{
if(tt&1)
{
ans++;
}
tt>>=1;
}
printf("%d\n",ans);
}
else
{
scanf("%d%d%d",&a,&b,&c);
if(a > b)
{
int t = a;
a = b;
b = t;
}
update(a, b, c, 1, L, 1);
}
}
}
return 0;
}
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