\[Count Color\]
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 50865
Accepted: 15346
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:
- "C A B C" Color the board from segment A to segment B with color C.
- "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 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
Source
Submit
题解
题目大意：我们需要设计一个程序,支持两种操作
- 修改:把区间\([l,r]\)内的颜色全部换成\(Q\)
- 询问:区间\([l,r]\)内有多少种不同的颜色
- 其中颜色数T\(<=30\)
本来博主蒟蒻看错题了,以为是SDOI2011染色那样询问区间内有多少个颜色段,然后迅速码完交了一发,结果WA,然后还以为是自己打错了,一直在调,后来发现是自己理解错题意了,重新打了一次,看到了颜色\(T<=30\),这对int类型状态压缩是完全没有问题的,然后就是乱打了,我们把颜色i表示为\(1<<(i-1)\)代表第i中颜色,这样在统计颜色数的时候把每一位拆开,看着一位是否是1
Code
#include<iostream>
#include<cstdio>
#include<cstring>
#define in(i) (i=read())
#define ll(i) (i<<1)
#define rr(i) (i<<1|1)
#define mid (l+r>>1)
using namespace std;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') {ans=(ans<<1)+(ans<<3)+i-'0'; i=getchar();}
return ans*f;
}
int n,m,T,ans;
int sum[400010],lazy[400010];
inline void pushup(int node) {
sum[node]=sum[ll(node)]|sum[rr(node)];
}
inline void pushdown(int node) {
lazy[ll(node)]=lazy[node];
lazy[rr(node)]=lazy[node];
sum[ll(node)]=sum[rr(node)]=lazy[node];
lazy[node]=0;
}
void build(int node,int l,int r) {
if(l==r) {
sum[node]=1;
return;
}
build(ll(node),l,mid);
build(rr(node),mid+1,r);
pushup(node);
}
void update(int node,int l,int r,int left,int right,int k) {
if(l>right || r<left) return;
if(left<=l && r<=right) {
lazy[node]=1<<k-1;
sum[node]=1<<k-1;
return;
}
if(lazy[node]) pushdown(node);
update(ll(node),l,mid,left,right,k);
update(rr(node),mid+1,r,left,right,k);
pushup(node);
}
inline int work(int x) {
int ans=0;
while(x) {
if(x&1) ans++;
x>>=1;
}
return ans;
}
void check(int node,int l,int r,int left,int right) {
if(l>right || r<left) return;
if(left<=l && r<=right) {
ans|=sum[node];
return;
}
if(lazy[node]) pushdown(node);
check(ll(node),l,mid,left,right);
check(rr(node),mid+1,r,left,right);
}
int main()
{
while(scanf("%d%d%d",&n,&T,&m)!=EOF) {
memset(lazy,0,sizeof(lazy));
int x,y,k; build(1,1,n);
for(int i=1;i<=m;i++) {
char op[10]; scanf("%s",op);
if(op[0]=='C') {
in(x); in(y); in(k);
if(x>y) swap(x,y);
update(1,1,n,x,y,k);
}
else {
ans=0; in(x); in(y);
if(x>y) swap(x,y);
check(1,1,n,x,y); ans=work(ans);
printf("%d\n",ans);
}
}
}
return 0;
}