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线段树(区间操作) POJ 3325 Help with Intervals

2016-03-24 20:05 387 查看
题目传送门

题意:四种集合的操作,对应区间的01,问最后存在集合存在的区间。

分析:U T [l, r]填充1; I T [0, l), (r, N]填充0; D T [l, r]填充0; C T[0, l), (r, N]填充0并且[l, r]xor; S T [l, r]xor

线段树结点两个属性,cover[o]: 该区间是否填充(1, 0, -1),_xor[o]: 该区间是否异或反转(1, 0)。最后(和[的区别可以原数*2判奇偶得

#include <cstdio>
#include <cstring>
#include <algorithm>

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 2 * 65535 + 5;
bool vis
;
int cover[N<<2], _xor[N<<2];

void Xor(int o) {
if (cover[o] != -1) {
cover[o] ^= 1;
} else {
_xor[o] ^= 1;
}
}
void push_down(int o) {
if (cover[o] != -1) {
cover[o<<1] = cover[o<<1|1] = cover[o];
_xor[o<<1] = _xor[o<<1|1] = 0;
cover[o] = -1;
}
if (_xor[o]) {
Xor (o<<1);
Xor (o<<1|1);
_xor[o] = 0;
}
}
void updata(int ql, int qr, char op, int l, int r, int o) {
if (ql <= l && r <= qr) {
if (op == 'U') {
cover[o] = 1;
_xor[o] = 0;
} else if (op == 'D') {
cover[o] = _xor[o] = 0;
} else if (op == 'C' || op == 'S') {
Xor (o);
}
return ;
}
push_down (o);
int mid = l + r >> 1;
if (ql <= mid) {
updata (ql, qr, op, lson);
} else if (op == 'I' || op == 'C') {
cover[o<<1] = _xor[o<<1] = 0;
}
if (qr > mid) {
updata (ql, qr, op, rson);
} else if (op == 'I' || op == 'C') {
cover[o<<1|1] = _xor[o<<1|1] = 0;
}
}
void query(int l, int r, int o) {
if (cover[o] == 1) {
for (int i=l; i<=r; ++i) {
vis[i] = true;
}
return ;
} else if (cover[o] == 0) {
return ;
}
if (l == r) {
return ;
}
push_down (o);
int mid = l + r >> 1;
query (lson);
query (rson);
}

int main() {
char op, l, r; int a, b;
int n = N - 5;
cover[1] = cover[1] = 0;
while (~scanf ("%c %c%d,%d%c\n", &op, &l, &a, &b, &r)) {
a <<= 1; b <<= 1;
if (l == '(') {
a++;
}
if (r == ')') {
b--;
}
if (a > b) {
if (op == 'C' || op == 'I') {
cover[1] = _xor[1] = 0;
}
} else {
updata (a, b, op, 0, n, 1);
}
}
memset (vis, false, sizeof (vis));
query (0, n, 1);
bool flag = false;
int s = -1, e = 0;
for (int i=0; i<=n; ++i) {
if (vis[i]) {
if (s == -1) {
s = i;
}
e = i;
} else {
if (s != -1) {
if (flag) putchar (' ');
flag = true;
printf ("%c%d,%d%c", s&1 ? '(' : '[', s>>1, (e+1)>>1, e&1 ? ')' : ']');
s = -1;
}
}
}
if (flag) {
puts ("");
} else {
puts ("empty set");
}

return 0;
}


  
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