HDU 6351 Beautiful Now(记忆化搜索+回溯)
Beautiful Now
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1208 Accepted Submission(s): 435
Problem Description
Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?
Input
The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.
Output
For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Sample Input
5
12 1
213 2
998244353 1
998244353 2
998244353 3
Sample Output
12 21
123 321
298944353 998544323
238944359 998544332
233944859 998544332
这题不想说什么了,比赛的时候用暴搜无脑TLE,剪枝就WA,心态爆炸。后来看了同学的代码,他用记忆化搜索过了,后来写了一份。菜是原罪。
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
const int maxn = 105;
int len;
char num[maxn],n[maxn];
map<pii, int> m;
int getmin(int cnt)
{
int x = atoi(num);
if (cnt == 0) return x;
if (m[pii(x, cnt)]) return m[pii(x, cnt)]; //记忆化
int res = x;
for (int i = 0; i < len; i++)
{
for (int j = 0; j < i; j++)
{
if (num[i] < num[j])
{
swap(num[i], num[j]);
if (num[0] != '0')//前导零
{
int tmp = getmin(cnt - 1);
res = min(tmp, res);
}
swap(num[i], num[j]);
}
}
}
return m[pii(x, cnt)] = res; //记忆化
}
int getmax(int cnt)
{
int x = atoi(num);
if (cnt == 0) return x;
if (m[pii(x, cnt)]) return m[pii(x, cnt)];
int res = x;
for (int i = 0; i < len; i++)
{
for (int j = 0; j < i; j++)
{
if (num[i] > num[j])
{
swap(num[i], num[j]);
int tmp = getmax(cnt - 1);
res = max(tmp, res);
swap(num[i], num[j]);
break;
}
}
}
return m[pii(x, cnt)] = res;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int k;
scanf("%s%d",&n,&k);
len = strlen(n);
k = min(k, 10);//九位数,不会交换太多次
strcpy(num,n);
m.clear();
int minn = getmin(k);
strcpy(num,n);
m.clear();
int maxx = getmax(k);
printf("%d %d\n", minn, maxx);
}
return 0;
}