SPOJ SUBLEX

来源：互联网发布：大数据分析视频教程编辑：程序博客网时间：2024/05/18 12:36

求字典序第k大字串

SAM的节点和转移所构成的是一个DAG，并且SAM从起点开始的每一条路径都是模式串的一个字串，根据这两个性质在SAM上dfs即可

#include<bits/stdc++.h>using namespace std;const int maxn = 91234*2 ,mlen = 26;char tem[maxn];struct Sam{    int len[maxn],fa[maxn],nex[maxn][mlen];    int siz[maxn];    int _cnt,root,omg;    int newNode(int L = 0){        len[_cnt] = L;        memset(nex[_cnt],fa[_cnt] = -1,sizeof(nex[_cnt]));        return _cnt++;    }    void init(){        _cnt = 0;        root = omg = newNode();    }    void extend(int x){        int ox = newNode(len[omg]+1);        while(omg != -1 && nex[omg][x] == -1){            nex[omg][x] = ox;            omg = fa[omg];        }        if(omg == -1) fa[ox] = root;        else{            int omgx = nex[omg][x];            if(len[omgx] == len[omg]+1) fa[ox] = omgx;            else{                int mgx = newNode(len[omg]+1);                for(int i=0;i<mlen;i++)                    nex[mgx][i] = nex[omgx][i];                fa[mgx] = fa[omgx];                fa[omgx] = fa[ox] = mgx;                while(omg != -1 && nex[omg][x] == omgx)                    nex[omg][x] = mgx,omg = fa[omg];            }        }        omg = ox;    }    void out(){        for(int i=0;i<_cnt;i++){            printf("id = %d fa = %d len = %d siz = %d: ",i,fa[i],len[i],siz[i]);            for(int j=0;j<5;j++){                printf("(%c,%2d) ",j + 'a',nex[i][j]);            }            puts("");        }    }    void build(char *arr){        init();        for(int i=0;arr[i];i++){            extend(arr[i] - 'a');        }    }    void initDfs(int st = -1){        if(st == -1){            st = root;            memset(siz,-1,sizeof(siz));        }        if(siz[st] != -1) return;        siz[st] = 1;        for(int i=0;i<mlen;i++){            if(nex[st][i] != -1){                initDfs(nex[st][i]);                siz[st] += siz[nex[st][i]];            }        }    }    void dfs(int x,int st=-1,int len = 0){        if(st == -1) st = root;        if(x <= 1){            tem[len] = 0;            printf("%s\n",tem);            return;        }        x--;        for(int i=0;i<mlen;i++){            int p = nex[st][i];            if(p != -1){                if(x > siz[p]) x -= siz[p];                else{                    tem[len] = 'a' + i;                    dfs(x,p,len+1);                    return;                }            }        }    }}SAM;char arr[maxn];int main(){    scanf("%s",arr);    SAM.build(arr);    SAM.initDfs();    int T;    scanf("%d",&T);    int k;    while(T-- && ~scanf("%d",&k)){        SAM.dfs(k+1);    }    return 0;}

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