HDU 5442 Favorite Donut

第一种方法，最小表示法
其实呢，你将每一个字母反转一下，将’a’变成’z’，就是最小表示法。
但是反转之后，我们如果用最小表示法，得到的是，在原串上位置最靠后的情况，与题意不服，所以我这里就强行将之往后硬判，最坏复杂度是当串所以的字符都相同的情况，退化成O(n2)。

//      whn6325689//      Mr.Phoebe//      http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<50#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define debug(a) cout << #a" = " << (a) << endl;#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,rtemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=20010;char str[MAXN],rev[MAXN];int n,ch[30];int init(){    int now='z';    for(int i=0; i<26; i++)        ch[i]=now--;}int getsmall(string s, bool flag){    int i,j,k,l;    int N=s.length();    s+=s;    int ans=0;    for(i=0,j=1; j<N;)    {        for(k=0; k<N&&s[i+k]==s[j+k]; k++);        if(k>=N)        {            if(flag)            {                ans=j;                j++;                continue;            }            break;        }        if(s[i+k]<s[j+k])        {            j+=k+1;        }        else        {            l=i+k;            i=j;            ans=i;            j=max(l,j)+1;        }    }    return ans;}bool compare(int ans0,int ans1){    int pos1=ans0,pos2=ans1;    for(int i=0; i<n; i++)    {        if(str[pos1]==str[pos2])        {            if(++pos1>=n) pos1-=n;            if(--pos2<0) pos2+=n;        }        else        {            return str[pos1]>str[pos2];        }    }    return ans0<=ans1;}string st,en;int main(){//    freopen("data.txt", "r", stdin);    init();    int T;    read(T);    while(T--)    {        st.clear();        en.clear();        read(n);        scanf("%s",str);        for(int i=0; i<n; i++)            st+=ch[str[i]-'a'];        en=st;        reverse(en.begin(),en.end());        int ans0=getsmall(st,false);        int tmp=getsmall(en,true);        int ans1=(n-tmp-1)%n;        if (compare(ans0,ans1))            printf("%d 0\n",ans0+1);        else            printf("%d 1\n",ans1+1);    }    return 0;}

后缀数组法，裸的后缀数组应用

//      xixihaha#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define debug(a) cout << #a" = " << (a) << endl;#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,rtemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=100010;char C[MAXN],rC[MAXN];int idx[MAXN],mp[MAXN];int rk[MAXN],sa[MAXN],height[MAXN],w[MAXN],wa[MAXN],ret[MAXN];void getSa (int len,int up){    int *k = rk;    int *id = height;    int *r = ret;    int  *cnt = wa;    for(int i=0;i<up;i++) cnt[i] = 0;    for(int i=0;i<len;i++) cnt[k[i] = w[i]]++;    for(int i=0;i<up;i++) cnt[i+1] += cnt[i];    for(int i = len - 1; i >= 0; i--) {        sa[--cnt[k[i]]] = i;    }    for(int d = 1,p = 0; p < len;){        for(int i = len - d; i < len; i++) id[p++] = i;        for(int i=0;i<len;i++)if(sa[i] >= d) id[p++] = sa[i] - d;        for(int i=0;i<len;i++) r[i] = k[id[i]];        for(int i=0;i<up;i++) cnt[i] = 0;        for(int i=0;i<len;i++) cnt[r[i]]++;        for(int i=0;i<up;i++) cnt[i+1] += cnt[i];        for(int i = len - 1; i >= 0; i--) {            sa[--cnt[r[i]]] = id[i];        }        swap(k,r);        p = 0;        k[sa[0]] = p++;        for(int i=0;i<len-1;i++) {            if(sa[i]+d < len && sa[i+1]+d <len &&r[sa[i]] == r[sa[i+1]]&& r[sa[i]+d] == r[sa[i+1]+d])                k[sa[i+1]] = p - 1;            else k[sa[i+1]] = p++;        }        if(p >= len) return ;        d *= 2,up = p, p = 0;    }}void getHeight(int len) {    for(int i=0;i<len;i++) rk[sa[i]] = i;    height[0] =  0;    for(int i = 0,p = 0; i < len - 1; i++) {        int j = sa[rk[i]-1];        while(i+p < len&& j+p < len&& w[i+p] == w[j+p]) {            p++;        }        height[rk[i]] = p;        p = max(0,p - 1);    }}int getSuffix(char s[]) {    int len = strlen(s),up = 0;    for(int i = 0; i < len; i++) {        w[i] = s[i];        up = max(up,w[i]);    }    w[len++] = 0;    getSa(len,up+1);    getHeight(len);    return len;}int solve(char s[],int n){    int len=getSuffix(s);    /*   for(int i=1;i<len;i++){     printf("%s\n",s+sa[i]);     }*/    int start=len-1;    while(sa[start]>n/2)start--;    int ans=idx[sa[start]];    while(height[start]>=n/2){        start--;        ans=min(ans,idx[sa[start]]);    }    return ans;}int cmp(int x,int y,int n){    for(int i=x,j=y,k=0;k<n;k++,i++,j++){        if(C[i]==rC[j])continue;        if(C[i]<rC[j])return -1;        if(C[i]>rC[j])return 1;    }    return 0;}int main(){    //    freopen("in","r",stdin);    int T;scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        scanf("%s",C);        for(int i=0;i<n;i++)rC[i]=C[i];        reverse(rC,rC+n);        for(int i=0;i<n;i++){            C[i+n]=C[i];            idx[i]=idx[i+n]=i;        }        C[n*2]='\0';        int ans=solve(C,n*2),dir=0;        for(int i=0;i<n;i++){            rC[i+n]=rC[i];            idx[i]=idx[n+i]=n-i-1;            mp[n-i-1]=i;        }        rC[n*2]='\0';        //        cout<<rC<<endl;        int res=solve(rC,n*2);        //        cout<<ans<<"!!"<<res<<endl;        //        cout<<string(C+ans,n)<<endl;        //        cout<<string(rC+mp[res],n)<<endl;        int flag=cmp(ans,mp[res],n);        if(flag==-1){            ans=res,dir=1;        }        else if(flag==0){            if(res<ans){                ans=res;                dir=1;            }        }        printf("%d %d\n",ans+1,dir);    }    return 0;}