HihoCoder 1264 神奇字符串（暴力）

题意:

两个字符串A和B,|A|,|B|≤100,每个单位时间可以交换字符串A的相邻的两个字符
我们说两个字符串是非常相似的,当且仅当它们的编辑距离≤1
问最少需要多少时间,让A和B变得非常相似,保证存在一种这样的方案

分析:

首先看到100,显然O(n3)的大暴力
直接枚举跟B编辑是1的字符串,添加1个字母,删除1个字母,替换1个字母
然后根据冒泡排序模拟就好了(也就是计算逆序对的个数)

代码:

////  Created by TaoSama on 2016-02-14//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#include <cassert>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int cc1[26], cc2[26];string s, t;int calc(string s, string t) {    if(s.size() != t.size()) return INF;    int ret = 0;    for(int i = 0; i < t.size(); ++i) {        char c = t[i];        string::size_type idx = s.find(c, i);        if(idx == string::npos) return INF;        for(int j = idx; j > i; --j) {            ++ret;            swap(s[j], s[j - 1]);        }    }    assert(s == t);    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    cin >> s >> t;    int ans = INF;    //add    for(int i = 0; i <= t.size(); ++i) {        for(char c = 'a'; c <= 'z'; ++c) {            string tmp = t;            tmp.insert(i, 1, c);            ans = min(ans, calc(s, tmp));        }    }    //del    for(int i = 0; i < t.size(); ++i) {        string tmp = t;        tmp.erase(i, 1);        ans = min(ans, calc(s, tmp));    }    //replace    for(int i = 0; i < t.size(); ++i) {        for(char c = 'a'; c <= 'z'; ++c) {            string tmp = t;            tmp[i] = c;            ans = min(ans, calc(s, tmp));        }    }    cout << ans << '\n';    return 0;}