poj2777 Count Color

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 37779 Accepted: 11355

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

Sample Output

21

这道题和just a hook差不多，在那道题的基础上加深了一些，每一次询问都要把数组a[]清空，然后往数组力添加代表不同颜色的数字，如果数组里已经有了该颜色就不用添加了。还学到了"<<"运算符，当它和"+"一起使用的时候，要加括号，因为其优先级小于"+"，用位运算的速度确实比直接乘和除快啊。1<<i代表2的i次,i<<1代表i*2.

#include<stdio.h>#include<string.h>char s[10];int a[100],t;struct node{int l,r,num;}b[4*200006];void build(int l,int r,int i){int mid;b[i].l=l;b[i].r=r;b[i].num=1;if(l==r)return;mid=(l+r)/2;build(l,mid,i<<1);build(mid+1,r,(i<<1)+1);}void update(int l,int r,int num,int i){int mid;if(b[i].num==num)return;if(b[i].l==l && b[i].r==r){b[i].num=num;return;}if(b[i].num!=-1){b[i<<1].num=b[(i<<1)+1].num=b[i].num;b[i].num=-1;}mid=(b[i].l+b[i].r)/2;if(r<=mid)update(l,r,num,i<<1);else if(l>mid) update(l,r,num,(i<<1)+1);else if(r>mid && l<=mid) {update(l,mid,num,i<<1);update(mid+1,r,num,(i<<1)+1);}}void question(int l,int r,int i){int mid,j,flag;if(b[i].num!=-1){flag=1;for(j=1;j<=t;j++){if(a[j]==b[i].num){flag=0;break;}}if(flag==1){a[++t]=b[i].num;}return;}mid=(b[i].l+b[i].r)/2;if(r<=mid) question(l,r,i<<1);else if(l>mid) question(l,r,(i<<1)+1);else {question(l,mid,i<<1);question(mid+1,r,(i<<1)+1);}}int main(){int n,m,i,j,c,d,e,T,temp,num,h;while(scanf("%d%d%d",&n,&m,&T)!=EOF){build(1,n,1);for(i=1;i<=T;i++){scanf("%s",s);if(s[0]=='C'){scanf("%d%d%d",&c,&d,&e);if(c>d){temp=c;c=d;d=temp;}update(c,d,e,1);}else if(s[0]=='P'){scanf("%d%d",&c,&d);if(c>d){temp=c;c=d;d=temp;}memset(a,0,sizeof(a));t=0;question(c,d,1);printf("%d\n",t);}}}return 0;}