poj 2777 Count Color (线段树)

题目链接：http://poj.org/problem?id=2777

颜色数量很少，所以状压进 (int)，线段树维护即可

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;

const int maxn = 100010;

int n, T, m;

struct Node{
	int tag;
	int color;
}t[maxn << 2];

void pushup(int i){
	t[i].color = (t[i << 1].color | t[i << 1 | 1].color);
}

void pushdown(int i, int l, int r){
	if(t[i].tag != -1){
		t[i << 1].tag = t[i << 1 | 1].tag = t[i].tag;
		t[i << 1].color = t[i].color;
		t[i << 1 | 1].color = t[i].color;
		t[i].tag = -1;
	}
}

void build(int i, int l, int r){
	t[i].tag = -1;
	if(l == r){
		t[i].color = 1;
		t[i].tag = -1;
		return;
	}
	int mid = (l + r) >> 1;
	build(i << 1, l, mid);
	build(i << 1 | 1, mid + 1, r);
	pushup(i); 
}

void modify(int i, int c, int l, int r, int x, int y){
	if(x <= l && r <= y){
		t[i].color = 0;
		t[i].tag = c;
		t[i].color = (1 << c);
		return;
	}
	
	pushdown(i, l, r);
	
	int mid = (l + r) >> 1;
	if(x <= mid) modify(i << 1, c, l, mid, x, y);
	if(y > mid) modify(i << 1 | 1, c, mid + 1, r, x, y);
	
	pushup(i);
}

int query(int i, int l, int r, int x, int y){
	if(x <= l && r <= y){
		return t[i].color;
	}
	
	pushdown(i, l, r);
	
	int color = 0;
	int mid = (l + r) >> 1;
	if(x <= mid) color |= query(i << 1, l, mid, x, y);
	if(y > mid) color |= query(i << 1 | 1, mid + 1, r, x, y);
	return color;
}

int count(int x){
	int res = 0;
	while(x){
		if(x & 1) ++res;
		x >>= 1;
	}
	return res;
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
		scanf("%d%d%d", &n, &T, &m); --T;
		
		build(1, 1, n);
		
		char op[10]; int a, b, c;
		for(int i = 1 ; i <= m ; ++i){
			scanf("%s", op);
			if(op[0] == 'C'){
				scanf("%d%d%d", &a, &b, &c);
				if(a > b) swap(a, b);
				--c;
				modify(1, c, 1, n, a, b);
			} else{
				scanf("%d%d", &a, &b);
				if(a > b) swap(a, b);
				int C = query(1, 1, n, a, b);
				printf("%d
", count(query(1, 1, n, a, b)));
			}
		}

	return 0;
}