hdu5023 A Corrupt Mayor's Performance Art 线段树区间更新

每次操作可以对区间赋值，询问某一区间的值的集合

简单的区间更新

维护 sum[o] 表示该区间值的集合，用二进制表示

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <map>
using namespace std;
#define INF 1e9

#define rep(i,x,y) for(int i=x;i<=y;i++)
#define mset(x) memset(x,0,sizeof(x))
typedef __int64 ll;

const int maxnode = 1000005*4;
char cmd;
int n, m, x1, x2, c, ans;

struct IntervalTree{
	
	int setv[maxnode];
	ll sumv[maxnode];

	void init(){
		memset(setv, -1, sizeof(setv));
		mset(sumv);
	}

	void pushup(int o){
		sumv[o] = sumv[2*o+1] | sumv[2*o];
	}

	void pushdown(int o){
		int lc = 2*o+1, rc = o*2;
		if(setv[o]>=0){
			setv[lc] = setv[rc] = setv[o];
			sumv[lc] = sumv[rc] = sumv[o];
			setv[o] = -1;
		} 
	}

	void update(int o, int L, int R){
		int lc=2*o, rc=2*o+1;
		if(x1<=L && x2>=R){
			setv[o] = c;
			sumv[o] = (1<<c);
			return;
		}
		int M = L + (R-L)/2;
		pushdown(o);
		if(x1<=M)	update(lc, L, M);
		if(x2>M)	update(rc, M+1, R);
		pushup(o);
	}

	void query(int o, int L, int R){
		int lc=2*o, rc=2*o+1;
		if(x1<=L && x2>=R){
			ans |= sumv[o];
			return ;
		}
		pushdown(o);
		int M = L + (R-L)/2;
		if(x1<=M)	query(lc, L, M);
		if(x2>M)	query(rc, M+1, R);
	}

}solve;

int main(){
//	freopen("a.txt","r",stdin);
//	freopen(".out","w",stdout);
	while(cin>>n>>m){
		if(n==0 && m==0)
			break;
		solve.init();

		x1=1; x2=n; c=2;
		solve.update(1,1,n);

		while(m--){
			cin>>cmd;
			if(cmd=='P'){
				scanf("%d%d%d", &x1, &x2, &c);
				solve.update(1, 1, n);
			}
			else{
				scanf("%d%d", &x1, &x2);
				ans=0;
				solve.query(1, 1, n);
				int ii=0;
				rep(i,1,30){
					if((ans>>i)&1){
						if(ii) putchar(' ');
						printf("%d", i);
						ii++;
					}
				}
				puts("");
			}
		}

	}
	return 0;
}