uva 11039 - Building designing（贪心）

题目链接：uva 11039 - Building designing

题目大意：给出n个数，均为非0数，要求选出尽量多的数组成序列，序列要求正负交替，绝对值递增。

解题思路：读入时将正数与负数分别储存在两个数组，负数可以直接储存绝对值。然后将两个数组按照绝对大小分别排序，然后在两个数组中交替选出尽量小的数，满足绝对值大于前面一个选出的数。

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

const int N = 500005;

int n, l, r, left
, right
;

void init() {
l = r = 0;
memset(right, 0, sizeof(right));
memset(left, 0, sizeof(left));

int c;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &c);
if (c < 0) left[l++] = -c;
else right[r++] = c;
}

sort(left, left + l);
sort(right, right + r);
}

int solve() {
if (l && r) {
int p, q, begin, ans = 1, tmp;
if (left[0] < right[0]) {
p = 1, q = 0, begin = 0;
tmp = left[0];
}
else {
p = 0, q = 1, begin = 1;
tmp = right[0];
}

for (int i = begin; p < l || q < r; i++) {
bool flag = true;

if (i % 2) {
// if (p >= l) break;

while (p < l) {
if (tmp <= left[p]) {
ans++;
tmp = left[p++];
flag = false;
break;
} else p++;
}
} else {
// if (q >= r) break;

while (q < r) {
if (tmp <= right[q]) {
ans++;
tmp = right[q++];
flag = false;
break;
} else q++;
}
}
if (flag) break;
}
return ans;
} else return 1;
}

int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init();
printf("%d\n", solve() );
}
return 0;
}