线段树直接处理,lazy标记记得下放。每个区间的各种颜色可以用二进制保存,因为最多只有30种颜色,所以用int就行。
两个区间合并后的颜色用或运算。
AC代码:
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 4e5 + 5;
struct node{
int L, R;
int color, tag;
}t[maxn];
void Build(int l, int r, int cur){ //Build the tree
t[cur].L = l; t[cur].R = r;
t[cur].color = t[cur].tag = 1;
if(l == r) return;
int mid = (l + r) / 2;
Build(l, mid, cur << 1);
Build(mid + 1, r, (cur << 1) + 1);
}
void Update(int l, int r, int c, int cur){
//printf("%d %d\n",l,r);
int l1 = t[cur].L, r1 = t[cur].R;
if(l1 == l && r1 == r){
t[cur].color = 1 << (c - 1);
t[cur].tag = 1;
return;
}
if(t[cur].tag){ //Let the lazy tag down
t[(cur << 1) + 1].tag = t[cur << 1].tag = t[cur].tag;
t[(cur << 1) + 1].color = t[cur << 1].color = t[cur].color;
}
int mid = (l1 + r1) / 2;
if(r <= mid) Update(l, r, c, cur << 1);
else if(l >= mid + 1) Update(l, r, c, (cur << 1) + 1);
else {
Update(l, mid, c, cur << 1);
Update(mid + 1, r, c, (cur << 1) + 1);
}
t[cur].color = t[cur << 1].color | t[(cur << 1) + 1].color;
int x = t[cur].color;
t[cur].tag = (x & (x - 1)) ? 0 :1;
}
int Find(int l, int r, int cur){
int l1 = t[cur].L, r1 = t[cur].R;
if(l1 == l && r1 == r || t[cur].tag) return t[cur].color;
if(t[cur].tag){ //Let the lazy tag down
t[(cur << 1) + 1].tag = t[cur << 1].tag = t[cur].tag;
t[(cur << 1) + 1].color = t[cur << 1].color = t[cur].color;
}
int mid = (l1 + r1) / 2;
if(r <= mid) return Find(l, r, cur << 1);
else if(l >= mid + 1) return Find(l, r, (cur << 1) + 1);
else {
int x = Find(l, mid, cur << 1);
int y = Find(mid + 1, r, (cur << 1) + 1);
return x | y;
}
}
int main(){
int L, T, Q;
while(scanf("%d%d%d", &L, &T, &Q) == 3){
Build(1, L, 1);
getchar();
char ch;
int a, b, c;
for(int i = 0; i < Q; ++i){
scanf("%c",&ch);
if(ch == 'C') {
scanf("%d%d%d", &a, &b, &c);
if(a > b) swap(a, b);
Update(a, b, c, 1);
}
else {
scanf("%d%d", &a, &b);
if(a > b) swap(a, b);
int x = Find(a, b, 1);
int ans = 0;
while(x > 0){
if(x & 1) ++ans;
x >>= 1;
}
printf("%d\n", ans);
}
getchar();
}
}
return 0;
}
如有不当之处欢迎指出!