题目链接

The Water Bowls POJ - 3185

思路

$$ 反转问题， 第一个反转与否决定全部 $$

时间复杂度

$$ O(N) $$

代码

#include <cstdio>
#include <algorithm>
using namespace std;

int a[25];
int b[25];

void f(int x) {
    b[x - 1] ^= 1;
    b[x] ^= 1;
    b[x + 1] ^= 1;
}

int gao(int x) {
    int cnt = 0;
    for (int i = 1; i <= 20; i++) {
        b[i] = a[i];
    }
    if (x == 1) {
        f(1);
        cnt++;
    }
    for (int i = 2; i <= 20; i++) {
        if (b[i - 1] == 1) {
            f(i);
            cnt++;
        }
    }
    if (b[20] == 1) {
        return 20;
    } else {
        return cnt;
    }

}

int main() {
    for (int i = 1; i <= 20; i++) {
        scanf("%d", &a[i]);// don't forget &
    }
    printf("%d", min(gao(0), gao(1)));
    return 0;
}