poj 2777 Count Color
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 26845 | Accepted: 8023 |
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
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=100003;
int flag[33],color[MAXN];
typedef struct
{
int left,right,col;
}line;
line tree[4*MAXN];
int sum;
void Create(int l,int r,int root) //建树
{
tree[root].left=l;
tree[root].right=r;
tree[root].col=1;
if(l==r) return;
int mid=(l+r)>>1;
Create(l,mid,root<<1);
Create(mid+1,r,(root<<1)+1);
}
void Updata(int l,int r,int colo,int root) //更新染色板
{
if(r<tree[root].left||l>tree[root].right) return;
if(l<=tree[root].left&&tree[root].right<=r)
{
tree[root].col=colo;
return;
}
if(tree[root].col==colo) return;
if(tree[root].left==tree[root].right) return;
if(tree[root].col>=0) //如果这一段的颜色超过1种,则向下更新之前的颜色,并将本段颜色赋值-1
{
tree[root<<1].col=tree[root].col;
tree[(root<<1)+1].col=tree[root].col;
tree[root].col=-1;
}
int mid=(tree[root].left+tree[root].right)>>1;
if(l>mid) Updata(l,r,colo,(root<<1)+1);
else if(r<=mid) Updata(l,r,colo,root<<1);
else
{
Updata(l,mid,colo,root<<1);
Updata(mid+1,r,colo,(root<<1)+1);
}
}
void solve(int l,int r,int root) //询问
{
if(r<tree[root].left||l>tree[root].right) return;
if(tree[root].col!=-1) //如果父节点有单一的颜色,说明子节点也有单一颜色,故直接更新
{
//printf("root = %d tree[root].col = %d\n",root,tree[root].col);
flag[tree[root].col]=1;//统计哪些颜色出现过
return;
}
if(tree[root].left==tree[root].right) return;
int mid=(tree[root].left+tree[root].right)>>1;
if(l>mid) solve(l,r,(root<<1)+1);
else if(r<=mid) solve(l,r,root<<1);
else
{
solve(l,mid,root<<1);
solve(mid+1,r,(root<<1)+1);
}
}
int main()
{
//freopen("input","r",stdin);
int n,t,o;
while(scanf("%d %d %d",&n,&t,&o)!=EOF)
{
memset(color,0,sizeof(color));
Create(1,n,1);
int tl,tr,tc;
char str[2];
for(int i=0;i<o;i++)
{
scanf("%s",str);
if(str[0]=='C')
{
scanf("%d %d %d",&tl,&tr,&tc);
if(tl>tr)
{
int temp=tl;
tl=tr;
tr=temp;
}
Updata(tl,tr,tc,1);
}
else if(str[0]=='P')
{
memset(flag,0,sizeof(flag));
scanf("%d %d",&tl,&tr);
if(tl>tr)
{
int temp=tl;
tl=tr;
tr=temp;
}
sum=0;
solve(tl,tr,1);
for(int i=1;i<=t;i++)
if(flag[i]) sum++;
printf("%d\n",sum);
}
}
}
return 0;
}