【51nod1523非回文】dfs

ljq真的不能再吃了

于 2019-07-15 14:19:50 发布

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分类专栏： ACM dfs

本文链接：https://blog.youkuaiyun.com/heucodesong/article/details/95963392

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该博客讨论了51nod1523题目的解决方案，这是一个关于找到比给定非回文字符串字典序大且最小的非回文字符串的问题。作者首先尝试了贪心的方法，但发现该方法不正确。然后通过从第一个位置开始的深度优先搜索（DFS）策略，确保新加入的字符不会与左侧已有的字符形成回文，即避免偶数长度和奇数长度的回文子串。文章提到DFS的时间复杂度难以精确计算，并提供了最终通过19组数据的代码。

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51nod1523

题意

一个字符串是非回文的，当且仅当，他只由前p个小写字母构成，而且他不包含长度大于等于2的回文子串。

给出长度为n的非回文串s。请找出字典序比s大的，而且字典序要最小的长度为n的非回文。

一开始想了一个做法贪心的去改一个字符用manacher看是不是回文改成功就一定行
时间复杂度 1000261000
25组数据过了19组
但是其实是fake的因为你想
4 4
abcd
我这个做法会输出dbcd

但答案其实是 abda

我们考虑一下怎么样才能非回文呢
我们从 1位置开始dfs
那么左边的所有串一定是非回文
新加入的串不和左边所有串形成回文
就是不能形成偶数回文和奇数回文
也就是 arr[i]!=arr[i-1] 和 arr[i]!=arr[i-2]
dfs时间复杂度是玄学~~

AC代码

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 1025;
char str[MAX_N];
int arr[MAX_N],ans[MAX_N],n,m;
bool dfs(int pos,bool flag)
{
    if(pos==n+1) return true;
    int sta;
    if(flag) sta = 0;
    else
    {
        sta = arr[pos];
        if(pos==n) sta++;
    }
    for(;sta<m;++sta)
    {
        if((pos>1&&ans[pos-1]==sta) || (pos>2&&ans[pos-2]==sta)) continue;
        if(sta>arr[pos]) flag = 1;
        ans[pos] = sta;
        if(dfs(pos+1,flag)) return true;
    }
    return false;
}

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    for(int i = 1;i<=n;++i)
        arr[i] = str[i] - 'a';
    if(dfs(1,0))
    {
        for(int i = 1;i<=n;++i)
            printf("%c",ans[i]+'a');
        printf("\n");
    }
    else printf("NO\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

过19组数据的代码

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 1025;
const int maxn=1025;
char Ma[maxn*2+5];
int Mp[maxn*2+5];
void Manacher(char s[],int len)
{
    int l=0;
    Ma[l++]='$';
    Ma[l++]='#';
    for(int i=0; i<len; i++)
    {
        Ma[l++]=s[i];
        Ma[l++]='#';
    }
    Ma[l]=0;
    int mx=0,id=0;
    for(int i=0; i<l; i++)
    {
        Mp[i] = mx>i?min(Mp[2*id-i],mx-i):1;
        while(Ma[i+Mp[i]]==Ma[i-Mp[i]])
            Mp[i]++;
        if(i+Mp[i]>mx)
        {
            mx = i +Mp[i];
            id = i;
        }
    }
}
char s[MAX_N], str[MAX_N];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    scanf("%s",str);
    string ans_ = "";
    for(int i = 0;i<n;++i)
        s[i] = str[i];
    bool flag = false;
    for(int i = n-1;i>=0;--i)
    {
        for(int j = str[i]-'a'+1;j<m;j++)
        {
            s[i] = 'a'+j;
            Manacher(s,n);
            int ans=0;
            for(int i=0; i<2*n+2; i++)
                ans = max(ans,Mp[i]-1);
            if(ans==1)
            {
                flag = true;
                ans_ = s;
                break;
            }
        }
        if(flag) break;
        s[i] = str[i];
    }

    if(!flag) printf("NO\n");
    else printf("%s\n",ans_.c_str());
    return 0;
}