\[Count Color\]

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:

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

Sample Output

2
1

Source

POJ Monthly--2006.03.26,dodo

Submit

題解

題目大意：我們需要設計一個程序,支持兩種操作

• 修改:把區間\([l,r]\)內的顏色全部換成\(Q\)
• 詢問:區間\([l,r]\)內有多少種不同的顏色
• 其中顏色數T\(<=30\)

本來博主蒟蒻看錯題了,以為是SDOI2011染色那樣詢問區間內有多少個顏色段,然後迅速碼完交了一發,結果WA,然後還以為是自己打錯了,一直在調,後來發現是自己理解錯題意了,重新打了一次,看到了顏色\(T<=30\),這對int類型狀態壓縮是完全沒有問題的,然後就是亂打了,我們把顏色i表示為\(1<<(i-1)\)代表第i中顏色,這樣在統計顏色數的時候把每一位拆開,看著一位是否是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;
}

博主蒟蒻,隨意轉載.但必須附上原文鏈接

http://www.cnblogs.com/real-l/

[線段樹] Count Color