Educational Codeforces Round 17 C. Two strings-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_45714303/article/details/125376446

本文介绍如何使用C++实现动态规划算法解决字符串编辑问题，通过pl和pr数组存储子问题最优解，寻找从s到t的最少操作次数。主要讨论了check函数的应用和mid点的选择策略，最后给出了完整的solve函数和main函数实现。

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#include<bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define eps  1e-2
#define gcd __gcd
#define pb push_back
#define PI acos(-1.0)
#define lowbit(x) (x)&(-x)
#define bug printf("!!!!!\n");
#define mem(x,y) memset(x,y,sizeof(x))

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef unsigned long long uLL;
#define int long long 

const int maxn = 1e5+2;
const int INF  = 1<<30;
const int mod  = 1e9+7;

string s,t;
int pl[maxn],pr[maxn];

bool check(int mid,int x){
    int L=x,R=mid,val1=(L==0?-1:pl[L-1]),val2=((R==t.size()-1)?1e9:pr[R+1]);
    // cout<<L<<" "<<R<<" ";
    // cout<<val1<<" "<<val2<<endl;
    return val1<val2;
}

void solve(){
    memset(pl,0x3f,sizeof pl);
    memset(pr,-0x3f,sizeof pr);
    cin>>s>>t;
    int pos1=0,pos2=0,len1=s.size(),len2=t.size();
    while(pos1<len1&&pos2<len2){
        if(s[pos1]==t[pos2]){
            pl[pos2]=pos1;
            ++pos1;++pos2;
        }else{
            ++pos1;
        }
    }
    // for(int i=0;i<len2;i++) cout<<pl[i]<<" ";cout<<endl;
    if(pos2==len2){
        cout<<t<<endl;
        return;
    }
    // check(7,2);
    pos1=s.size()-1;pos2=t.size()-1;
    while(pos1>=0&&pos2>=0){
        if(s[pos1]==t[pos2]){
            pr[pos2]=pos1;
            --pos1;--pos2;
        }else{
            --pos1;
        }
    }
    // for(int i=0;i<len2;i++) cout<<pr[i]<<" ";cout<<endl;
    int ans=1e9,pos;
    // check(1,0);
    for(int i=0;i<t.size();i++){
        int l=i,r=t.size()-1;
        while(l<r){
            int mid=(l+r)>>1;
            if(check(mid,i)) r=mid;
            else l=mid+1;
        }
        if(check(l,i)){
            // cout<<i<<" "<<l<<endl;
            if(ans>l-i+1){
                ans=l-i+1;
                pos=i;
            }
        }
    }
    if(ans==len2){
        cout<<"-"<<endl;
        return;
    }
    // cout<<ans<<" "<<pos<<endl;
    string res;
    for(int i=0;i<pos;i++) res+=t[i];
    for(int i=pos+ans;i<t.size();i++) res+=t[i];
    cout<<res<<endl;
    return;
}

int32_t main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//    ios::sync_with_stdio(false);
    int t = 1;
    //scanf("%d",&t);
    while(t--){
    //    printf("Case %d: ",cas++);
        solve();
    }
    return 0;
}