线段树入门pku2777 涂颜色问题

 /*pku 2777http://poj.org/problem?id=2777一板子L长，有两个操作一个着色l到r为c这种颜色，一个是查询l到r有多少种颜色就是插入和查询操作Count ColorTime Limit: 1000MS  Memory Limit: 65536KTotal Submissions: 20838  Accepted: 6034DescriptionChosen 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.InputFirst 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.OutputOuput results of the output operation in order, each line contains a number.Sample Input2 2 4C 1 1 2P 1 2C 2 2 2P 1 2Sample Output21SourcePOJ Monthly--2006.03.26,dodo*/#include <iostream>using namespace std;const int N= 100010;const int C= 35;const int nocolor = -1;const int mucolor = -2;struct node{    int l,r,c;};node tree[N*4];int color[C];void create(int k,int l,int r){    tree[k].l = l;    tree[k].r = r;    tree[k].c = nocolor;    if(l == r) return;    int m = (l+r)>>1;    create(k+k,l,m);    create(k+k+1,m+1,r);}void insert(int k,int l,int r,int c){    if(tree[k].c==c) return;    if(l==tree[k].l && r==tree[k].r) {        tree[k].c = c;        return;    }    // 要传递下去     if(mucolor != tree[k].c ) {        tree[k+k].c = tree[k].c;        tree[k+k+1].c = tree[k].c;        tree[k].c = mucolor;    }        int m = (tree[k].l+tree[k].r)>>1;    if(r<=m) { insert(k+k,l,r,c); return ;}    if(l>m) { insert(k+k+1,l,r,c); return;}    insert(k+k,l,m,c);    insert(k+k+1,m+1,r,c);}void query(int k,int l,int r){    if(tree[k].c == nocolor) return;    if(tree[k].c != mucolor ) {        color[tree[k].c] = 1;        return ;    }         int m = (tree[k].l + tree[k].r) >>1;    if(r<=m) { query(k+k,l,r); return; }    if(l>m) { query(k+k+1,l,r); return; }    query(k+k,l,m);    query(k+k+1,m+1,r);}int main(){    int L,T,O;    int i,l,r,c,t,sum;    char s[5];    scanf("%d%d%d",&L,&T,&O);        create(1,1,L);    tree[1].c = 1;    while(O--) {        scanf("%s",s);        if(s[0]=='C') {            scanf("%d%d%d",&l,&r,&c);            if(l<=r) insert(1,l,r,c);            else insert(1,r,l,c);        }else {            scanf("%d%d",&l,&r);                        for(i=1;i<=T;++i) color[i] = 0;            sum = 0;            if(l<=r) query(1,l,r);            else query(1,r,l);                        for(i=1;i<=T;++i) {                sum+=color[i];            }            printf("%d\n",sum);        }    }    return 0;}