POJ 2777 Count Color ( 线段树&&位运算

Count Color

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

题意

有两个操作
1 把一个区间变成一个颜色
2 查询区间中的颜色总量

AC代码

#include <cstdio>#include <algorithm>using namespace std;#define LL long longconst int MAXN = 1e5+10;struct node{    int l, r;    int lazy;    int color;}tree[MAXN*4];void build(int l,int r,int st){    tree[st].l = l; tree[st].r = r;    tree[st].lazy = 0; tree[st].color = 0;    if(l == r) return ;    int mid = (l+r) >> 1;    build(l,mid,st<<1); build(mid+1,r,st<<1|1);}void update(int l, int r,int st,int val){    if(tree[st].l==l && tree[st].r==r) {        tree[st].lazy = 1;        tree[st].color = 1<<(val-1);  //我们知道颜色的种类数一共不超过30,所以TAT        return;    }    if(tree[st].lazy) {        tree[st].lazy = 0;        tree[st<<1].lazy = tree[st<<1|1].lazy = 1;        tree[st<<1].color = tree[st<<1|1].color = tree[st].color;    }    int mid = (tree[st].l+tree[st].r) >> 1;    if(r <= mid) update(l,r,st<<1,val);    else if(l > mid) update(l,r,st<<1|1,val);    else {        update(l,mid,st<<1,val); update(mid+1,r,st<<1|1,val);    }    tree[st].color = tree[st<<1].color | tree[st<<1|1].color;    if(tree[st<<1].lazy && tree[st<<1|1].lazy && tree[st<<1].color==tree[st<<1|1].color)        tree[st].lazy = 1;}int query(int l,int r,int st){    if(l==tree[st].l && r==tree[st].r) return tree[st].color;    if(tree[st].lazy) return tree[st].color;     //剪下枝    int mid = (tree[st].l+tree[st].r) >> 1;    if(r <= mid) return query(l,r,st<<1);    else if(l > mid) return query(l,r,st<<1|1);    else {        return query(l,mid,st<<1)|query(mid+1,r,st<<1|1);    }}int main(){    int n, t, q;    while(~scanf("%d%d%d",&n,&t,&q)) {        build(1,n,1);        tree[1].lazy = 1;  tree[1].color = 1;        while(q--) {            char ch[10];            scanf("%s",ch);            int x, y, z;            if(ch[0] == 'C') {                scanf("%d%d%d",&x,&y,&z);                if(x > y) swap(x,y);                update(x,y,1,z);            }            else {                scanf("%d%d",&x,&y);                if(x > y) swap(x,y);                LL k = query(x,y,1);                int num = 0;                for(int i = 0; i < t; i++) {                    if(k&(1<<i)) num++;                }                printf("%d\n",num);            }        }    }return 0;}