codeforces 906E Reverses

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
#include
#include
#include
#include
#include
#include
#include

#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.vstruct 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 aint 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;
}