状态压缩DP SRM 667 Div1 OrderOfOperations 250-优快云博客

探讨了通过状态压缩动态规划解决程序指令最优执行顺序的问题，以实现最小化的总执行时间。

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Problem Statement

Cat Noku has just finished writing his first computer program. Noku's computer has m memory cells. The cells have addresses 0 through m-1. Noku's program consists of n instructions. The instructions have mutually independent effects and therefore they may be executed in any order. The instructions must be executed sequentially (i.e., one after another) and each instruction must be executed exactly once.

You are given a description of the n instructions as a vector <string> with n elements. Each instruction is a string of m characters. For each i, character i of an instruction is '1' if this instruction accesses memory cell i, or '0' if it does not.

Noku's computer uses caching, which influences the time needed to execute an instruction. More precisely, executing an instruction takes k^2 units of time, where k is the number of new memory cells this instruction accesses. (I.e., k is the number of memory cells that are accessed by this instruction but have not been accessed by any previously executed instruction. Note that k may be zero, in which case the current instruction is indeed executed in 0 units of time.)

Noku's instructions can be executed in many different orders. Clearly, different orders may lead to a different total time of execution. Find and return the shortest amount of time in which it is possible to execute all instructions.

Definition

Class:	OrderOfOperations
Method:	minTime
Parameters:	vector <string>
Returns:	int
Method signature:	int minTime(vector <string> s)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256
Stack limit (MB):	256

Constraints

n will be between 1 and 50, inclusive.

m will be between 1 and 20, inclusive.

s will have exactly n elements.

Each element of s will have exactly m characters.

Each character of s[i] will be either '0' or '1' for all valid i.

Examples

{
 "111",
 "001",
 "010"
}

Returns: 3

Cat Noku has 3 instructions. The first instruction ("111") accesses all three memory cells. The second instruction ("001") accesses only memory cell 2. The third instruction ("010") accesses only memory cell 1. If Noku executes these three instructions in the given order, it will take 3^2 + 0^2 + 0^2 = 9 units of time. However, if he executes them in the order "second, third, first", it will take only 1^2 + 1^2 + 1^2 = 3 units of time. This is one optimal solution. Another optimal solution is to execute the instructions in the order "third, second, first".

{
 "11101",
 "00111",
 "10101",
 "00000",
 "11000"
}

Returns: 9

{
  "11111111111111111111"
}

Returns: 400

A single instruction that accesses all 20 memory cells.

{
  "1000",
  "1100",
  "1110"
}

Returns: 3

{
  "111",
  "111",
  "110",
  "100"
}

Returns: 3

题意：给n个01串，设计一种顺序，使得每次新出现的1的个数的平方和最小

分析：比赛时不知道是div1的题，以为暴力贪心可以过，结果被hack掉了。题解说没有充分的证明使用贪心是很有风险的，正解是用状态压缩DP。

收获：爆零还能涨分，TC真奇怪。

官方题解

int dp[(1<<20)+10];
int a[55];

class OrderOfOperations {
public:
   int minTime( vector <string> s ) {
        int n = s.size (), m = s[0].length ();
        memset (a, 0, sizeof (a));
        int tot = 0;
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                if (s[i][j] == '1') a[i] |= (1<<j);
            }
            tot |= a[i];
        }
        memset (dp, INF, sizeof (dp));
        dp[0] = 0;
        for (int i=0; i<(1<<m); ++i)    {
            for (int j=0; j<n; ++j) {
                int x = i | a[j];                       //从i状态转移到x的状态
                int y = x - i;                          //表示新出现的1
                int k = __builtin_popcount (y);         //内置函数，快速得到二进制下1的个数
                dp[x] = min (dp[x], dp[i] + k * k);     //类似Bellman_Ford
            }
        }

        return dp[tot];
   }
};

转载于:https://www.cnblogs.com/Running-Time/p/4803452.html