SPOJ

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题目：求字符串字典序第k小的子串
思路：统计每个状态的子串的个数，按字典序寻找
代码：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int MAXN = 200050, SIZE = 26;struct SAM {    int len[MAXN], link[MAXN], next[MAXN][SIZE];    int total, last;    inline int newNode(int L) {        len[++total] = L; link[total] = 0;        for(int i = 0; i < SIZE; ++i) next[total][i] = 0;        return total;    }    inline void Add(int c) {        int i, p = last, cur = newNode(len[last] + 1);        for(; p && !next[p][c]; p = link[p]) next[p][c] = cur;        if(!p) link[cur] = 1;//令其指向初始状态        else {            int q = next[p][c];            if(len[q] == len[p] + 1) link[cur] = q;            else {//>                int clone = newNode(len[p] + 1);                for(i = 0; i < SIZE; ++i) next[clone][i] = next[q][i];                link[clone] = link[q];                link[q] = link[cur] = clone;for(; p && next[p][c] == q; p = link[p]) next[p][c] = clone;            }        }        last = cur;    }    void Init () {//根节点是1        total = 0;        last = newNode(0);    }}sam;char str[MAXN];int num[MAXN],idx[MAXN];int dp[MAXN];void solve(int k){    int now=1,L=0;    while(k){        for(int j=0;j<SIZE;j++){            if(sam.next[now][j]==0)                continue;            int x=sam.next[now][j];            if(k>dp[x])                k-=dp[x];            else{                str[L++]='a'+j;                now=x;                k--;                break;            }        }    }    str[L]='\0';    puts(str);}int main(){    int Q,k;    scanf("%s",str);        sam.Init();        for(int i=0;str[i]!='\0';i++)            sam.Add(str[i]-'a');        for(int i=1;i<=sam.total;i++)            num[i]=0;        for(int i=1;i<=sam.total;i++)            num[sam.len[i]]++;        for(int i=1;i<=sam.total;i++)            num[i]+=num[i-1];        for(int i=1;i<=sam.total;i++)            idx[num[sam.len[i]]--]=i;        for(int i=sam.total;i>=1;i--){            int now=idx[i];            dp[now]=1;            for(int j=0;j<SIZE;j++)                dp[now]+=dp[sam.next[now][j]];        }        scanf("%d",&Q);        while(Q--){            scanf("%d",&k);            solve(k);        }    return 0;}
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