POJ 3255 Help with Intervals （线段树区间更新）

题意: 给定一个一开始是空集 给出一些集合操作 求最后得到的集合 (按照右端点升序排列输出)

分析: 把每个操作转化成对线段的操作 可以将其中的一些操作分解成其他操作

          这里给出一种机智的实现办法 分解那种可以自己写写也是比较容易的

代码:

//
// Created by TaoSama on 2015-09-18
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 65535 * 2 + 5, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n;
int cov[N << 2], Xor[N << 2];

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

void fxor(int rt) {
if(~cov[rt]) cov[rt] ^= 1;
else Xor[rt] ^= 1;
}

void push_down(int rt) {
if(~cov[rt]) {
Xor[rt << 1] = Xor[rt << 1 | 1] = 0;
cov[rt << 1] = cov[rt << 1 | 1] = cov[rt];
cov[rt] = -1;
}
if(Xor[rt]) {
fxor(rt << 1);
fxor(rt << 1 | 1);
Xor[rt] = 0;
}
}

//T-> [l, r] all about segments
//U: S∪T => [l,r]->1
//I: S∩T => (-∞,l)∪(r,+∞)->0
//S: S⊕T => [l,r] 1/0 reverse
//D: S-T => [l,r]->0
//C: T-S => (-∞,l)∪(r,+∞)->0, [l,r] 1/0 reverse

void update(int L, int R, char op, int l, int r, int rt) {
if(L <= l && r <= R) {
if(op == 'U') {
cov[rt] = 1;
Xor[rt] = 0;
}
if(op == 'D') {
cov[rt] = 0;
Xor[rt] = 0;
}
if(op == 'S' || op == 'C') fxor(rt);
return;
}
int m = l + r >> 1;
push_down(rt);
if(L <= m) update(L, R, op, lson);
else if(op == 'I' || op == 'C') cov[rt << 1] = Xor[rt << 1] = 0;
if(R > m) update(L, R, op, rson);
else if(op == 'I' || op == 'C') cov[rt << 1 | 1] = Xor[rt << 1 | 1] = 0;
}

bool vis
;
void query(int l, int r, int rt) {
if(cov[rt] == 1) {
fill(vis + l, vis + r + 1, 1);
return;
}
if(cov[rt] == 0 || l == r) return;
int m = l + r >> 1;
push_down(rt);
query(lson);
query(rson);
}

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

cov[1] = Xor[1] = 0;
char op, a, b; int l, r;
while(~scanf("%c %c%d,%d%c ", &op, &a, &l, &r, &b)) {
l <<= 1; r <<= 1;
if(a == '(') ++l;
if(b == ')') --r;
if(l > r) {
if(op == 'C' || op == 'I')
cov[1] = Xor[1] = 0;
continue;
}
update(l, r, op, 0, N, 1);
}

memset(vis, false, sizeof vis);
query(0, N, 1);
bool ok = false;
int s = -1, e;
for(int i = 0; i < N; ++i) {
if(vis[i]) {
if(s == -1) s = i;
e = i;
} else {
if(s == -1) continue;
if(ok) putchar(' ');
ok = true;
printf("%c%d,%d%c", "[("[s & 1], s >> 1, e + 1 >> 1, "])"[e & 1]);
s = -1;
}
}
if(!ok) printf("empty set");
puts("");
return 0;
}