暴力 hihoCoder 1251 Today Is a Rainy Day (15北京C)

题目传送门

题意：一串数字变成另一串数字，可以单个数字转变，或者一类数字转变，问最少操作次数

分析：首先一类转换比单个转换优，先枚举找出最优的映射方案，然后将零碎的操作加上。6位6进制表示map映射，比如123456，表示map[1] = 1, map[2] = 2, map[3] = 3...

先预处理出每种方案的的操作数，用BFS+dp写，然后暴力枚举取最小值。

#include <bits/stdc++.h>
using namespace std;

const int N = 110 + 5;
const int M = 5e4 + 5;
const int INF = 0x3f3f3f3f;
char ori[N], fin[N];
int dp[M];
int cnt[6];
int g[6][6];

int idx(int *c)	{
	int ret = 0;
	for (int i=0; i<6; ++i)	{
		ret = ret * 6 + c[i];
	}
	return ret;
}

void ridx(int x, int *c)	{
	for (int i=5; i>=0; --i)	{
		c[i] = x % 6;
		x /= 6;
	}
}

void BFS(void)	{
	memset (dp, INF, sizeof (dp));
	int c[6];
	for (int i=0; i<6; ++i)	c[i] = i;	//nomal, not change
	int s = idx (c);	dp[s] = 0;
	queue<int> que;	que.push (s);
	while (!que.empty ())	{
		int u = que.front ();	que.pop ();
		ridx (u, c);
		for (int i=0; i<6; ++i)	{
			for (int j=0; j<6; ++j)	{
				int t[6];
				memcpy (t, c, sizeof (t));
				for (int k=0; k<6; ++k)	{
					if (t[k] == i)	t[k] = j;
				}
				int v = idx (t);
				if (dp[v] > dp[u] + 1)	{
					dp[v] = dp[u] + 1;
					que.push (v);
				}
			}
		}
	}
}

int main(void)	{
	BFS ();
	while (scanf ("%s%s", ori, fin) == 2)	{
		int len = strlen (ori);
		memset (cnt, 0, sizeof (cnt));
		memset (g, 0, sizeof (g));
		for (int i=0; i<len; ++i)	{
			int d1 = fin[i] - '1', d2 = ori[i] - '1';
			cnt[d1]++;	g[d1][d2]++;
		}
		int ans = INF;
		int c[6];
		for (int i=0; i<M; ++i)	{
			ridx (i, c);
			int tmp = dp[i];
			for (int j=0; j<6; ++j)	{
				tmp += cnt[j] - g[j][c[j]];
			}
			ans = min (ans, tmp);
		}
		printf ("%d
", ans);
	}

	return 0;
}
/*
22345611
12345611
2234562221
1234561221
2234562211
1234561111
22345622112
12345611111
654321654321654321654321
123456123456123456123456
*/