﻿ poj2777Count Color（线段树）-C 语言_软件编程-脚本宝典

# poj2777Count Color（线段树）

## poj2777Count Color（线段树）

Count ColorTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 40404 Accepted: 12188DescriptionChosen Problem Solving and Program design as an optional
Count Color
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 40404 Accepted: 12188

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

```#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

struct node
{
int l,r,s;
} t[MAXN<<2];
int L,O,T;
int sum[33];//因为颜色数不多，可以用一个数组来装；1表示该区间有该颜色，0反之

void build(int l, int r, int i)
{
t[i].l = l;
t[i].r = r;
t[i].s = 1;
if(l == r) return ;
int mid = (l+r)>>1;
build(l, mid, i<<1);
build(mid+1, r, i<<1|1);
}

void update(int l, int r, int m, int i)
{
if(t[i].s == m) return ;
if(t[i].l == l && t[i].r == r)
{
t[i].s = m;
return ;
}
if(t[i].s != -1)
{
t[i<<1].s = t[i<<1|1].s = t[i].s;
t[i].s = -1;
}
int mid = (t[i].l+t[i].r)>>1;
if(l > mid) update(l, r, m, i<<1|1);
else if(r <= mid) update(l, r, m, i<<1);
else
{
update(l, mid, m, i<<1);
update(mid+1, r, m, i<<1|1);
}
}

void query(int l, int r, int i)
{
if(t[i].s != -1)//如果纯色把该颜色标记
{
sum[t[i].s] = 1;
return ;
}
else//否则继续查询子节点
{
int mid = (t[i].l+t[i].r)>>1;
if(l > mid) query(l, r, i<<1|1);
else if(r <= mid) query(l, r, i<<1);
else
{
query(l, mid, i<<1);
query(mid+1, r, i<<1|1);
}
}
}

int main()
{
char ch;
int a,b,c;
while(scanf("%d%d%d",&L,&T,&O)==3)
{
build(1, L, 1);
while(O--)
{
getchar();
scanf("%c",&ch);
if(ch == 'C')
{
scanf("%d%d%d",&a,&b,&c);
update(a, b, c, 1);
}
int ans = 0;
if(ch == 'P')
{
CL(sum, 0);//每次查询之前清零
scanf("%d%d",&a,&b);
query(a, b, 1);
for(int i=1; i<=T; i++)//统计出现过的颜色数
if(sum[i]) ans++;
printf("%d\n",ans);
}
}
}
return 0;
}```

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