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  • 移除相邻(string操作+implement)

    https://codeforces.com/contest/1321/problem/C

    题意:给出一个字符串,消除规则:某个字符比左边相邻或右边相邻字符大1,则该字符可以删除,比如ba,b可以删除。

    问最多可以删除多少个字符。

    解法:贪心从字符z开始删除,判断能否从左边找和右边找比该字符小1的字符,找到则删除该字符,并继续从头开始枚举该字符,比如cbbbbcaaca。

    //#include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <stdio.h>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <string.h>
    #include <vector>
    typedef long long ll ;
    #define int ll
    #define mod 1000000007
    #define gcd __gcd
    #define rep(i , j , n) for(int i = j ; i <= n ; i++)
    #define red(i , n , j)  for(int i = n ; i >= j ; i--)
    #define ME(x , y) memset(x , y , sizeof(x))
    //ll lcm(ll a , ll b){return a*b/gcd(a,b);}
    ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
    //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
    //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
    #define SC scanf
    #define INF  0x3f3f3f3f
    #define PI acos(-1)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    using namespace std;
    const int N = 1e6+100;
    const int maxn = 1e2+9;
    int flag ;
    char c = 'z';
    string a ;
    
    bool search(int i){
        if(i - 1 >= 0 && a[i - 1] == c - 1){
            a.erase(a.begin() + i);
            return true;
        }
        if(i + 1 < a.size() && a[i + 1] == c - 1){
            a.erase(a.begin() + i);
            return true;
        }
        return false;
    }
    
    void solve(){
        int n ;
        cin >> n ;
        cin >> a ;
        while(c > 'a'){
            flag = 0 ;
            rep(i , 0 , a.size()){
                if(a[i] == c){
                    if(search(i)){
                        flag = 1 ;
                    }
                }
            }
            c--;
            if(flag) c++;
        }
        cout << n - a.size() << endl;
    }
    
    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        //int t ;
        //cin >> t ;
        //while(t--){
            solve();
        //}
    }
    
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  • 原文地址:https://www.cnblogs.com/nonames/p/12394364.html
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