hdoj 5023 线段树+位压缩

hdoj 5023

题意：给一个全是2的序列，进行若干操作，将某区间修改为任意数字或询问区间内的有哪些数字。

思路：线段树水题，用位压缩一下就好。

#include <cstdio>
#include <cstring>
#include <set>
using namespace std;
const int M = 1000020;
struct Tree{
int l, r, color;
bool flag;
}tree[M * 4];
void init(){
memset(tree, 0, sizeof tree);
}
void buildtree(int rt, int l, int r) {
tree[rt].l = l, tree[rt].r = r;
tree[rt].flag = false;
tree[rt].color = 2;
if(l == r){
return ;
}
int mid = (l + r) / 2;
buildtree(rt * 2, l, mid);
buildtree(rt * 2 + 1, mid + 1, r);
}
void pullup(int rt) {
tree[rt].color = (tree[rt * 2].color | tree[rt * 2 + 1].color);
}
void pushdown(int rt){
if(tree[rt].flag) {
tree[rt * 2].color = tree[rt].color;
tree[rt * 2 + 1].color = tree[rt].color;
tree[rt].flag = false;
tree[rt * 2].flag = true;
tree[rt * 2 + 1].flag = true;
}
}
void modify(int rt, int l, int r, int Color) {
if(tree[rt].l == l && tree[rt].r == r){
tree[rt].color = Color;
tree[rt].flag = true;
return ;
}
pushdown(rt);
int mid = (tree[rt].l + tree[rt].r) / 2;
if(l > mid) modify(rt * 2 + 1, l, r, Color);
else if(r <= mid) modify(rt * 2, l, r, Color);
else {
modify(rt * 2, l, mid, Color);
modify(rt * 2 + 1, mid + 1, r, Color);
}
pullup(rt);
}
int color = 0;
void query(int rt, int l, int r) {
if(tree[rt].l == l && tree[rt].r == r){
color |= tree[rt].color;
return ;
}
pushdown(rt);
int mid = (tree[rt].l + tree[rt].r) / 2;
if(r <= mid) query(rt * 2, l, r);
else if(l > mid) query(rt * 2 + 1, l, r);
else {
query(rt * 2, l, mid);
query(rt * 2 + 1, mid + 1, r);
}
}
int main() {
int n, m;
while(~scanf("%d %d", &n, &m)) {
if(n == m && n == 0) break;
init();
buildtree(1, 1, n);
for(int i = 0; i < m; i++) {
char str[10];
int a, b, c;
scanf("%s", str);
if(str[0] == 'Q') {
scanf("%d %d", &a, &b);
color = 0;
query(1, a, b);
bool first = true;
for(int i = 1; i <= 30; i++) {
if(color & (1 << (i - 1))) {
if(!first) printf(" %d", i);
else {
first = false;
printf("%d", i);
}
}
}
putchar('\n');
}
else {
scanf("%d %d %d", &a, &b, &c);
c--;
modify(1, a, b, (1 << c));
}
}
}
return 0;
}