SRM 607 D2 L3：CombinationLockDiv2，DP

题目：http://community.topcoder.com/stat?c=problem_statement&pm=12969&rd=15840

参考：http://apps.topcoder.com/wiki/display/tc/SRM+607

有难度，dp状态不好想。

代码：

#include <algorithm>
#include <iostream>
#include <sstream>

#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>

using namespace std;


#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)

/*************** Program Begin **********************/

static const int INF = 1000000000;
static const int MAX_N = 50;
static const int MAX_OP = 450;
int dp[MAX_N + 1][MAX_OP + 1][2];

class CombinationLockDiv2 {
public:
	int N;
	vector<int> d;

	int rec(int p, int x, int up )
	{
		int & res = dp[p][x][up];
		if (p == N) {		// base case
			res = 0;
			return res;
		}
		if (res != -1) {
			return res;
		}
		res = INF;
		for (int i = 0; i <= 1; i++) {
			for (int y = 0; y <= MAX_OP; y++) {
				if (0 == i) {	// down
					if (d[p] - y % 10 != 0) {
						// invalid
						continue;
					}
				} else {	// up
					if ( (d[p] + y) % 10 != 0 ) {
						// invalid
						continue;
					}
				}

				if (i == up) {	// not necessary open new intervals
					res = min(res, max(y - x, 0) + rec(p + 1, y, i) );
				} else {	// must open y new intervals
					res = min(res, y + rec(p + 1, y, i) );
				}
			}
		}
		return res;
	}

	int minimumMoves(string s, string t)
	{
		this->N = s.size();
		d.resize(this->N);
		for (int i = 0; i < this->N; i++) {
			if (s[i] >= t[i]) {
				d[i] = s[i] - t[i];
			} else {
				d[i] = s[i] + 10 - t[i];
			}
		}
		memset(dp, -1, sizeof(dp));
		return rec(0,0,0);
	}
};


/************** Program End ************************/

SRM 607 D2 L3：CombinationLockDiv2，DP

原文：http://blog.csdn.net/xzz_hust/article/details/19006487