【POJ 2777】 Count Color（线段树区间更新与查询）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 40949 | Accepted: 12366 |
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.
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
感觉有些神经衰弱了。。。写发水水的线段树放松放松。。。
一直在搞图论 是该换换弦 磨得快锈了 太折磨人了……（弱也不知道说了些啥
一块木板 长L(1~L) 有T种颜色的油漆标号1~T 默认木板初始是1号颜色
进行O次操作 操作有两种
C a b c 表示木板a~b段涂c种油漆（若之前涂过其它颜色 则覆盖掉）
P a b 表示询问木板a~b段如今涂了几种油漆
两个数组 一个存树 一个存涂了哪几种油漆
存树的表示a~b涂的某种颜色 然后搞搞就出来了……好乏
代码例如以下：
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;
int tr[400400];
bool col[33];
int L,T,O;
void Color(int root,int l,int r,int a,int b,int c)
{
if(a == l && b == r)
{
tr[root] = c;
return;
}
int mid = (l+r)>>1;
if(tr[root])
tr[root<<1] = tr[root<<1|1] = tr[root];
tr[root] = 0;
if(mid >= b) Color(root<<1,l,mid,a,b,c);
else if(mid+1 <= a) Color(root<<1|1,mid+1,r,a,b,c);
else
{
Color(root<<1,l,mid,a,mid,c);
Color(root<<1|1,mid+1,r,mid+1,b,c);
}
}
void Search(int root,int l,int r,int a,int b)
{
//printf("root:%d l:%d r:%d co:%d
",root,l,r,tr[root]);
if(tr[root])
{
col[tr[root]] = 1;
return;
}
int mid = (l+r)>>1;
if(mid >= b) Search(root<<1,l,mid,a,b);
else if(mid+1 <= a) Search(root<<1|1,mid+1,r,a,b);
else
{
Search(root<<1,l,mid,a,mid);
Search(root<<1|1,mid+1,r,mid+1,b);
}
}
int main()
{
//fread();
//fwrite();
char opt[3];
int a,b,c;
while(~scanf("%d%d%d",&L,&T,&O))
{
memset(tr,0,sizeof(tr));
tr[1] = 1;
while(O--)
{
scanf("%s",opt);
scanf("%d%d",&a,&b);
if(a > b) swap(a,b);
if(opt[0] == 'C')
{
scanf("%d",&c);
Color(1,1,L,a,b,c);
}
else
{
memset(col,0,sizeof(col));
Search(1,1,L,a,b);
int ans = 0;
for(int i = 1; i <= T; ++i)
ans += col[i];
printf("%d
",ans);
}
}
}
return 0;
}