POJ-2777 Count Color [线段树+bitset]

题目

题目在这里

思路

一共不超过30种颜色，可以bitset数组替换线段树中的int数组。套线段树模版就行了。

代码

#include <cstdio>
#include <algorithm>
#include <bitset>

using namespace std;

const int maxn = 1e5+5;
int L, T, O;
bitset<32>tree[4*maxn], lz[4*maxn];
void buildtree(int k, int l, int r) {
    if(l == r) {
        tree[k] = 2;
        return ;
    }
    int m = (l+r) / 2;
    int left = k*2, right = k*2+1;
    buildtree(left, l, m);
    buildtree(right, m+1, r);
    tree[k] = tree[left]|tree[right];
}
void pushdown(int k) {
    if(lz[k] != 0) {
        lz[k*2] = lz[k];
        lz[k*2+1] = lz[k];
        tree[k*2] = lz[k];
        tree[k*2+1] = lz[k];
        lz[k].reset();
    }
}
void update(int k, int l, int r, int L, int R, int c) {
    if(r <= R && l >= L) {
        tree[k].reset();
        tree[k].set(c);
        lz[k].reset();
        lz[k].set(c);
        return ;
    }
    pushdown(k);
    int m = (l+r) >> 1;
    if(m >= L)
        update(k*2, l, m, L, R, c);
    if(m < R)
        update(k*2+1, m+1, r, L, R, c);
    tree[k] = tree[k*2] | tree[k*2+1];
    return ;
}
bitset<32> query(int k, int l, int r, int L, int R) {
    if(l >= L && r <= R)
        return tree[k];
    if(r < L || l > R)
        return 0;
    pushdown(k);
    bitset<32> s(0);
    int m = (l+r) >> 1;
    if(m >= L) s |= query(k*2, l, m, L, R);
    if(m < R) s |= query(k*2+1, m+1, r, L, R);
    return s;
}
int main(void)
{
    /* freopen("in.txt", "r", stdin); */
    char ch;
    int a, b, c;
    scanf("%d%d%d", &L, &T, &O);
    buildtree(1, 1, L);
    for(int i = 0; i < O; ++i) {
        scanf(" %c", &ch);
        if(ch == 'C') {
            scanf("%d%d%d", &a, &b, &c);
            update(1, 1, L, a, b, c);
        } else {
            scanf("%d%d", &a, &b);
            if(a > b)
                swap(a, b);
            printf("%d\n", query(1, 1, L, a, b).count());
        }
    }
    return 0;
}