codeforces 906E Reverses

本文介绍了一种通过合并两个字符串并将其分解为最少数量的偶数长度回文串的方法来解决串翻转问题。该方法首先将两个串合并,接着采用特定算法找到最小翻转次数及方案。

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Description

给你两个串s和t,其中t是由s中选择若干个不相交的区间翻转得到的,现在要求求出最少的翻转次数以及给出方案。
1|s|=|t|500000

Solution

先将s和t串合并成一个新的串a,其中若i为奇数,那么a[i]=s[(i+1)/2]否则a[i]=t[i/2]
然后问题就变成了将a分解成最少的长度为偶数的回文串,跟将字符串分解成最少的回文串个数类似,算法详情见论文:https://arxiv.org/pdf/1403.2431v2.pdf

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
    char ch;
    while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
    if (ch=='-'){
        int s=0;
        while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
        return -s;
    }
    int s=ch-'0';
    while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
    return s;
}

const int N = 1e6+5;

int n;
char s[N];
struct element{
    int v,pos;
    element(const int v_=0,const int pos_=0){v=v_;pos=pos_;}
    friend bool operator < (element a,element b){return a.v<b.v;}
}PL[N],GPL[N];
struct triple{
    int st,delt,tot;
    triple(const int st_=0,const int delt_=0,const int tot_=0){st=st_;delt=delt_;tot=tot_;}
}G[N],G1[N],G2[N];
int k;
char a[N],b[N];

element min(element a,element b){return a<b?a:b;}

int main(){
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    scanf("%s",a+1);
    scanf("%s",b+1);
    int l=strlen(a+1);
    n=l*2;
    fo(i,1,n)if (i&1)s[i]=a[(i+1)/2];else s[i]=b[i/2];
    k=0;
    fo(i,0,n)GPL[i]=element(1e9,0);
    fo(w,1,n){
        int k1=0;
        fo(i,1,k)
        if (G[i].st>1&&s[G[i].st-1]==s[w])G1[++k1]=triple(G[i].st-1,G[i].delt,G[i].tot);
        int lst=-w;
        int k2=0;
        fo(i,1,k1){
            if (G1[i].st-lst!=G1[i].delt){
                G2[++k2]=triple(G1[i].st,G1[i].st-lst,1);
                if (G1[i].tot>1)G2[++k2]=triple(G1[i].st+G1[i].delt,G1[i].delt,G1[i].tot-1);
            }
            else G2[++k2]=G1[i];
            lst=G1[i].st+G1[i].delt*(G1[i].tot-1);
        }
        if (s[w-1]==s[w]){
            G2[++k2]=triple(w-1,w-1-lst,1);
            lst=w-1;
        }
        G2[++k2]=triple(w,w-lst,1);
        lst=w;

        G[k=1]=G2[1];
        fo(i,2,k2){
            if (G2[i].delt==G[k].delt)G[k].tot+=G2[i].tot;
            else G[++k]=G2[i];
        }

        PL[w]=element(1e9,0);
        if (w%2==0){
            if (s[w]==s[w-1])PL[w]=min(PL[w],element(PL[w-2].v,w-2));
        }
        fo(i,1,k){
            int r=G[i].st+(G[i].tot-1)*G[i].delt;
            element v=element(1e9,0);
            if (r&1)v=element(PL[r-1].v+1,r-1);
            if (G[i].tot>1)v=min(v,GPL[G[i].st-G[i].delt]);
            if (G[i].delt<=G[i].st)GPL[G[i].st-G[i].delt]=v;
            if (w%2==0)PL[w]=min(PL[w],v);
        }
    }
    if (PL[n].v>n)printf("-1\n");
    else{
        printf("%d\n",PL[n].v);
        for(int i=n;i;i=PL[i].pos)
        if (i-PL[i].pos>2)printf("%d %d\n",PL[i].pos/2+1,i/2);
    }
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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