2015长春网络赛 1006（后缀数组或者最小表示法）

  1 #pragma warning(disable:4996)
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <math.h>
  5 #include <vector>
  6 #include <queue>
  7 #include <stack>
  8 #include <map>
  9 #include <set>
 10 #include <algorithm>
 11 #include <iostream>
 12 #include <functional>
 13 #include <string>
 14 const int INF = 1<<30;
 15 const double eps = 1e-5;
 16 typedef long long LL;
 17 /*
 18 
 19 */
 20 const int N = 40000 + 10;
 21 int rank[N], r[N], height[N], c[N], bucket[N], sa[N], tmp[N];
 22 //[0,n)字符串的长度，[0,m)字符集的大小
 23 void buildSa(int n, int m)
 24 {
 25     int *x = rank, *y = tmp, *t, k, p;
 26     for (int i = 0; i<m; ++i) c[i] = 0;
 27     for (int i = 0; i<n; ++i) c[x[i] = r[i]]++;
 28     for (int i = 1; i<m; ++i) c[i] += c[i - 1];
 29     for (int i = n - 1; i >= 0; --i) sa[--c[x[i]]] = i;
 30 
 31     for (k = 1; k <= n; k <<= 1)
 32     {
 33         p = 0;
 34         for (int i = n - k; i<n; ++i) y[p++] = i;
 35         for (int i = 0; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;
 36 
 37         for (int i = 0; i<n; ++i) bucket[i] = x[y[i]];//按排名收集第一关键字的大小, y[i]是下标，所以把y[i]看做是下标就好理解多了
 38         for (int i = 0; i<m; ++i) c[i] = 0;
 39         for (int i = 0; i<n; ++i) c[bucket[i]]++;
 40         for (int i = 1; i<m; ++i) c[i] += c[i - 1];
 41         for (int i = n - 1; i >= 0; --i) sa[--c[bucket[i]]] = y[i];
 42 
 43         t = x, x = y, y = t;
 44         p = 1; x[sa[0]] = 0;
 45         for (int i = 1; i<n; ++i)
 46             x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++;
 47         if (p >= n) break;
 48         m = p;
 49     }
 50 }
 51 
 52 //h[i] = height[rank[i]]   h[i] >= h[i-1] - 1;
 53 //[0,n]
 54 void makeHeight(int n)
 55 {
 56     int i, j, k = 0;
 57     for (i = 0; i <= n; ++i) rank[sa[i]] = i;
 58     //第n个字符的排名是0，sa[rank[i]-1]越界，所以不用去算
 59     for (i = 0; i<n; height[rank[i++]] = k)
 60     for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);
 61 }
 62 
 63 char str[N];
 64 std::string ans1,ans2;
 65 int idx1,idx2;
 66 int main()
 67 {
 68    int t,n,m;
 69    scanf("%d",&t);
 70    while(t--)
 71    {
 72         ans1 = ans2 = "";
 73         scanf("%d",&n);
 74         scanf("%s",str);
 75         for(int i=0;i<n;++i)
 76             r[i] = str[i] - ‘a‘ + 1;
 77         m = n;
 78         for(int i=0;i<n;++i)
 79             r[m++] = str[i] - ‘a‘ + 1;
 80         r[m++] = 0;
 81         buildSa(m,27);
 82         idx1 = sa[m-1];
 83         //printf("%d\n",sa[m-1]);
 84         m = n;
 85         for(int i=idx1; m--;++i)
 86             ans1 += r[i] + ‘a‘ - 1;
 87         m = 0;
 88         for(int i=n-1;i>=0; --i)
 89             r[m++] = str[i] -‘a‘;
 90         for(int i=n-1;i>=0; --i)
 91             r[m++] = str[i] -‘a‘;
 92         r[m++] = 26;
 93         buildSa(m,27);
 94         int ma = rank[0];
 95         idx2 = 0;
 96         for(int i=1;i<n;++i)
 97         {
 98             if(rank[i] >= ma)
 99             {
100                 ma = rank[i];
101                 idx2 = i;
102             }
103         }
104         m = n;
105         for(int i=idx2;m--;++i)
106             ans2 += r[i] + ‘a‘;
107         idx1 ++;
108         idx2 = n - idx2;
109         if(ans1 > ans2)
110         {
111             printf("%d %d\n",idx1,0);
112         }
113         else if(ans1 < ans2)
114         {
115             printf("%d %d\n",idx2,1);
116         }
117         else
118         {
119             if(idx1<=idx2)
120                 printf("%d %d\n",idx1,0);
121             else
122                 printf("%d %d\n",idx2,1);
123         }
124    }
125     return 0;
126 }