bzoj 3704 昊昊的机油之GRST - 贪心

题目传送门

　　传送门

题目大意

　　给定一个数组$a$和数组$b$，每次操作可以选择$a$的一个子区间将其中的数在模4意义下加1，问把$a$变成$b$的最少操作次数。

　　首先求$b - a$，再差分，令这个数组为$c$。

　　那么操作的次数是这个数组$c$中正数的和。

　　现在可以做的是选择一个地方+4，它后面的某个地方-4。

　　考虑什么时候可以使得当前的答案减少。

• 在$-3$处$+4$，在$3$处$-4$，答案减少2
• 在$-3$处$+4$，在$2$处$-4$，答案减少1
• 在$-2$处$+4$，在$3$处$-4$，答案减少1

　　首先我们将尽量靠后的$-3$和$3$匹配，然后将对它操作。

　　这个用一个栈就能做。

　　剩下就贪心一下就好了。注意当$-3$和$2$操作时会多出$-2$。

Code

 /**
  * bzoj
  * Problem#3704
  * Accepted
  * Time: 1400ms
  * Memory: 128316k
  */
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <queue>
 #include <stack>
 using namespace std;
 typedef bool boolean;
 
 typedef class Input {
     protected:
         const static int limit = ;
         FILE* file; 
 
         int ss, st;
         char buf[limit];
     public:
 
         Input():file(NULL)  {   };
         Input(FILE* file):file(file) {  }
 
         void open(FILE *file) {
             this->file = file;
         }
 
         void open(const char* filename) {
             file = fopen(filename, "r");
         }
 
         char pick() {
             if (ss == st)
                 st = fread(buf, , limit, file), ss = ;//, cerr << "str: " << buf << "ed " << st << endl;
             return buf[ss++];
         }
 }Input;
 
 #define digit(_x) ((_x) >= '0' && (_x) <= '9')
 
 Input& operator >> (Input& in, unsigned& u) {
     char x;
     while (~(x = in.pick()) && !digit(x));
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     return in;
 }
 
 Input& operator >> (Input& in, unsigned long long& u) {
     char x;
     while (~(x = in.pick()) && !digit(x));
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     return in;
 }
 
 Input& operator >> (Input& in, int& u) {
     char x;
     while (~(x = in.pick()) && !digit(x) && x != '-');
     int aflag = ((x == '-') ? (x = in.pick(), -) : ());
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     u *= aflag;
     return in;
 }
 
 Input& operator >> (Input& in, long long& u) {
     char x;
     while (~(x = in.pick()) && !digit(x) && x != '-');
     int aflag = ((x == '-') ? (x = in.pick(), -) : ());
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     u *= aflag;
     return in;
 }
 
 Input& operator >> (Input& in, char* str) {
     for (char x; ~(x = in.pick()) && x != '\n' && x != ' '; *(str++) = x);
     return in;
 }
 
 Input in (stdin);
 
 template <typename T>
 void pfill(T* pst, const T* ped, T val) {
     for ( ; pst != ped; *(pst++) = val);
 }
 
 int n;
 int *a, *b, *c;
 boolean *used;
 
 inline void init() {
     in >> n;
     a = new int[(n + )];
     b = new int[(n + )];
     c = new int[(n + )];
     for (int i = ; i <= n; i++) {
         in >> a[i];
     }
     for (int i = ; i <= n; i++) {
         in >> b[i];
         c[i] = (b[i] - a[i] + ) % ;
     }
 }
 
 stack<int> sta;
 inline void solve() {
     int ans = ;
     for (int i = n; i > ; i--)
         c[i] = c[i] - c[i - ];
     used = new boolean[(n + )];
     pfill(used, used + n + , false);
     for (int i = ; i <= n; i++)
         if (c[i] > )
             ans += c[i];
     for (int i = , y; i <= n; i++)
         if (c[i] == -)
             sta.push(i);
         else if (c[i] ==  && !sta.empty()) {
             y = sta.top();
             sta.pop();
             used[y] = used[i] = true;
             ans -= ;
         }
 
     int have_2 = , have_3 = ;
     for (int i = ; i <= n; i++) {
         if (used[i])
             continue;
         switch (c[i]) {
             case :
                 if (have_3)
                     have_3--, ans--;
                 else if (have_2)
                     have_2--, ans--;
                 break;
             case :
                 if (have_3)
                     have_3--, have_2++, ans--;
                 break;
             case -:
                 have_2++;
                 break;
             case -:
                 have_3++;
                 break;
         }
     }
     printf("%d\n", ans);
 }
 
 int main() {
     init();
     solve();
     return ;
 }