Count Color POJ 2777

题目链接:点我

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    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

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题意:

给你一块木板,木板可以分成L份,你有两种操作:1. 将A到B的所有木块涂成颜色C, 2. 查询A到B共有多少种不同的颜色.

思路:

线段树的区间查询以及更新,利用lazy的方法延迟更新.

代码:

#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<iostream>using namespace std;struct ss{    int l, r, color;}a[400000];bool sum[40];void build(int cur, int l, int r){    a[cur].l = l;    a[cur].r = r;    a[cur].color = 1;//初始化颜色    if(l == r)        return ;    int mid = (l + r) >> 1;    build(cur << 1, l, mid);    build(cur << 1|1, mid + 1, r);}void update(int l, int r, int m, int cur){    if(a[cur].color == m) return ;    if(a[cur].l == l && a[cur].r == r){        a[cur].color = m;//如果区间完全符合,直接更新        return ;    }    if(a[cur].color != -1){//标记向下更新        a[cur << 1].color = a[cur << 1|1].color = a[cur].color;        a[cur].color = -1;    }    int mid = (a[cur].l + a[cur].r) >> 1;    if (l > mid) update( l, r, m ,cur << 1|1);    else if (r <= mid) update( l, r, m ,cur << 1);    else {        update( l, mid, m, cur << 1);        update( mid + 1, r, m, cur << 1|1);    }}void quiry(int l, int r, int cur){    if(a[cur].color != -1){        sum[a[cur].color] = true;        return;    }    int mid = (a[cur].l + a[cur].r) >> 1;    if(l > mid) quiry( l, r, cur << 1|1);    else if(r <= mid) quiry( l, r, cur << 1);    else {        quiry( l, mid, cur << 1);        quiry( mid + 1, r, cur << 1|1);    }}int main(){    int n,t,m;    while(scanf("%d %d %d", &n, &t, &m) !=EOF){          build( 1, 1, n);          while(m--){            char ch;            scanf(" %c", &ch);            if (ch == 'C'){                int x,y,z;                scanf("%d %d %d", &x, &y, &z);                update( x, y, z, 1);            }else{                 memset(sum,false,sizeof(sum));//用数组记录有多少种不同的颜色                int x, y;                scanf("%d %d", &x, &y);                quiry( x, y, 1);                int ans = 0;                for(int i = 1; i <= 30; ++ i)                    if (sum[i])                        ++ ans;                printf("%d\n", ans);            }          }    }    return 0;}