POJ 2777 Count Color（区间更新 + 状压）

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题目链接：http://poj.org/problem?id=2777

题意：给定长度L的区间，初始全部为1。

两种操作
1. 更新：C A B C 将区间[A，B]全部赋值为C。
2. 查询：P A B 查询区间[A，B]的颜色种数。

思路：基本为裸的区间更新。
因为最多只有30种颜色，所以在存储某个区间的颜色类型和种数时，可以利用一个int数来存储（利用每一位的二进制，1为有该颜色，0位无该颜色）。

代码：

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <algorithm>#include <math.h>#include <vector>using namespace std;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const int N = 1e5 + 5;const int INF = 1e9 + 10;const double EPS = 1e-6;int board[N << 2];int lazy[N << 2];void pushup(int rt) {    board[rt] = board[rt << 1] | board[rt << 1 | 1];}void pushdown(int rt) {    if (lazy[rt] != -1) {        lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];        board[rt << 1] = lazy[rt];        board[rt << 1 | 1] =  lazy[rt];        lazy[rt] = -1;    }}void build(int l, int r, int rt) {    lazy[rt] = -1;    if (l == r) {        board[rt] = 1;        return ;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    pushup(rt);}void update(int c, int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) {        board[rt] = c;        lazy[rt] = c;        return ;    }    pushdown(rt);    int m = (l + r) >> 1;    if (L <= m)        update(c, L, R, lson);    if (R > m)        update(c, L, R, rson);    pushup(rt);}int query(int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) {        return board[rt];    }    pushdown(rt);    int m = (l + r) >> 1;    int ql = 0, qr = 0, res;    if (L <= m)        ql = query(L, R, lson);    if (R > m)        qr = query(L, R, rson);    res = ql | qr;    pushup(rt);    return res;}int main() {    int L, T, O;    while (scanf("%d%d%d", &L, &T, &O) != EOF) {        build(1, L, 1);        for (int i_q = 1; i_q <= O; i_q++) {            char str[2];            int l, r;            scanf("%s%d%d", str, &l, &r);            if (l > r)                swap(l, r);            if (str[0] == 'C') {                int c;                scanf("%d", &c);                c = (1 << (c - 1));                update(c, l, r, 1, L, 1);            }            else {                int res = query(l, r, 1, L, 1);                int co = 0;                while (res) {                    if (res & 1)                        co++;                    res >>= 1;                }                printf("%d\n", co);            }        }        cout << endl;    }    return 0;}

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