HDU 4162 Shape Number

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4162

题意：

求给定字符的一阶差分链的最小表示。

题解：

先求一阶差分链，再求一阶差分链的最小表示法。

代码：

跑了670MS

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 3e5 + ;

char s1[maxn],s2[maxn];

int solve(char *s) {
    int i = , j = , k = ,len=strlen(s);
    while (i < len&&j < len&&k<len) {
        int t = s[(i + k) % len] - s[(j + k) % len];
        if (!t) k++;
        else {
            if (t > ) i = i + k + ;
            else j = j + k + ;
            if (i == j) j++;
            k = ;
        }
    }
    return i < j ? i : j;
}

int main() {
    while (scanf("%s", s1) == ) {
        int len = strlen(s1);
        for (int i = ; i < len; i++) {
            s2[i] = (s1[(i + ) % len] - s1[i] + ) %  + '';
        }
        s2[len] = '\0';
        //cout << s2 << endl;
        int pos = solve(s2);
        for (int i = ; i < len; i++) {
            printf("%c", s2[(pos + i) % len]);
        }
        printf("\n");
    }
    return ;
}

贴个后缀数组的解法：

跑了2527MS

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-;

const int maxn = 3e5 + ;

char s1[maxn],s2[maxn];

struct SuffixArray{
    char s[maxn];
    int sa[maxn],t[maxn],t2[maxn],c[maxn];
    int n,m;
    void init(int n,int m){
        this->n=n;
        this->m=m;
    }
    void build_sa(){
        int i,*x=t,*y=t2;
        for(i=;i<m;i++) c[i]=;
        for(i=;i<n;i++) c[x[i]=s[i]]++;
        for(i=;i<m;i++) c[i]+=c[i-];
        for(i=n-;i>=;i--) sa[--c[x[i]]]=i;
        for(int k=;k<=n;k<<=){
            int p=;
//            for(i=n-k;i<n;i++) y[p++]=i;
//            for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
            for(i=;i<n;i++) y[p++]=(sa[i]-k+n)%n;

            for(i=;i<m;i++) c[i]=;
            for(i=;i<n;i++) c[x[y[i]]]++;
            for(i=;i<m;i++) c[i]+=c[i-];
            for(i=n-;i>=;i--) sa[--c[x[y[i]]]]=y[i];

            swap(x,y);
            p=; x[sa[]]=;
            for(i=;i<n;i++){
                x[sa[i]]=y[sa[i-]]==y[sa[i]]&&y[sa[i-]+k]==y[sa[i]+k]?p-:p++;
            }
            if(p>=n) break;
            m=p;
        }
    }
}mysa;

int main() {
    while (scanf("%s", s1) == ) {
        int len = strlen(s1);
        mysa.init(len,);
        for (int i = ; i < len; i++) {
            s2[i] = (s1[(i + ) % len] - s1[i] + ) %  + '';
        }
        s2[len] = '\0';
        strcpy(mysa.s,s2);
        mysa.build_sa();

        for(int p=mysa.sa[];p<mysa.sa[]+mysa.n;p++){
            printf("%c",mysa.s[p%mysa.n]);
        }
        puts("");
    }
    return ;
}