POJ - 3185The Water Bowls (反转)

The Water Bowls Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5852 Accepted: 2296

Description

The cows have a line of 20 water bowls from which they drink. The bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which holds no water). They want all 20 water bowls to be right-side-up and thus use their wide snouts to flip bowls.  Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls).  Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up?

Input

Line 1: A single line with 20 space-separated integers

Output

Line 1: The minimum number of bowl flips necessary to flip all the bowls right-side-up (i.e., to 0). For the inputs given, it will always be possible to find some combination of flips that will manipulate the bowls to 20 0's.

Sample Input

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0

Sample Output

3

Hint

Explanation of the sample:  Flip bowls 4, 9, and 11 to make them all drinkable:  0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state]  0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4]  0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9]  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11] 对于我们自己增加的前置元素，要考虑它所有可能的值 然后就是用反转了 /*?????*/ #include <map> #include <set> #include <cmath> #include <ctime> #include <queue> #include <vector> #include <cctype> #include <cstdio> #include <string> #include <cstring> #include <sstream> #include <cstdlib> #include <typeinfo> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define pb push_back #define mp make_pair #define mem(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) #define lson rt << 1, l, mid #define rson rt << 1|1, mid + 1, r #define FIN freopen("input.txt", "r", stdin) #define FOUT freopen("output.txt", "w", stdout) typedef long long LL; typedef pair<int, int > PII; typedef pair<int,string> PIS; typedef unsigned long long uLL; template<typename T> void print(T* p, T* q, string Gap = " ", bool flag = false) { int d = p < q ? 1 : -1; while(p != q) { if(flag) cout << Gap[0] << *p << Gap[1]; else cout << *p; p += d; if(p != q && !flag) cout << Gap; } cout << endl; } template<typename T> void print(const T &a, string bes = "") { int len = bes.length(); if(len >= 2)cout << bes[0] << a << bes[1] << endl; else cout << a << endl; } void IO_Init() { ios::sync_with_stdio(false); } LL LLabs(LL a) { return a > 0 ? a : -a; } const double PI = 3.1415926535898; const double eps = 1e-10; const int MAXM = 1e4 + 5; const int MAXN = 1e5 + 5; const LL INF = 0x3f3f3f3f; /*?????*/ int dir[MAXN]; int f[MAXN]; int calc(int x) { memset(f, 0, sizeof(f)); int res = 0, sum = 0; dir[0] = x; for(int i = 0; i < 20; i ++) { if((dir[i] + sum) & 1) { res ++; f[i] = 1; } sum += f[i]; if(i - 3 + 1 >= 0) { sum -= f[i - 3 + 1]; } } if((dir[20] + sum) & 1){ return INF; } return res; } void solve() { printf("%d\n", min(calc(0), calc(1))); } int main() { //FIN; while(~scanf("%d", &dir[1])) { for(int i = 2; i <= 20; i ++) { scanf("%d", &dir[i]); } solve(); } return 0; }