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\n", count(query(1, 1, n, a, b)));
}
}
return 0;
}