标签:
后缀数组的应用。和男人八题那个后缀数组差不多。
1 /* 2890 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <algorithm> 12 #include <cstdio> 13 #include <cmath> 14 #include <ctime> 15 #include <cstring> 16 #include <climits> 17 #include <cctype> 18 #include <cassert> 19 #include <functional> 20 #include <iterator> 21 #include <iomanip> 22 using namespace std; 23 //#pragma comment(linker,"/STACK:102400000,1024000") 24 25 #define sti set<int> 26 #define stpii set<pair<int, int> > 27 #define mpii map<int,int> 28 #define vi vector<int> 29 #define pii pair<int,int> 30 #define vpii vector<pair<int,int> > 31 #define rep(i, a, n) for (int i=a;i<n;++i) 32 #define per(i, a, n) for (int i=n-1;i>=a;--i) 33 #define clr clear 34 #define pb push_back 35 #define mp make_pair 36 #define fir first 37 #define sec second 38 #define all(x) (x).begin(),(x).end() 39 #define SZ(x) ((int)(x).size()) 40 #define lson l, mid, rt<<1 41 #define rson mid+1, r, rt<<1|1 42 43 const int maxn = 50050; 44 int a[maxn], b[maxn], aa[maxn]; 45 int pos[maxn], pn; 46 int rrank[maxn], height[maxn], sa[maxn]; 47 int wa[maxn], wb[maxn], wc[maxn], wv[maxn]; 48 int n, k, at; 49 50 bool cmp(int *r, int a, int b, int l) { 51 return r[a]==r[b] && r[a+l]==r[b+l]; 52 } 53 54 void da(int *r, int *sa, int n, int m) { 55 int i, j, *x=wa, *y=wb, *t, p; 56 57 for (i=0; i<m; ++i) wc[i] = 0; 58 for (i=0; i<n; ++i) wc[x[i]=r[i]]++; 59 for (i=1; i<m; ++i) wc[i] += wc[i-1]; 60 for (i=n-1; i>=0; --i) sa[--wc[x[i]]] = i; 61 for (j=1,p=1; p<n; j*=2, m=p) { 62 for (p=0,i=n-j; i<n; ++i) y[p++] = i; 63 for (i=0; i<n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j; 64 for (i=0; i<n; ++i) wv[i] = x[y[i]]; 65 for (i=0; i<m; ++i) wc[i] = 0; 66 for (i=0; i<n; ++i) wc[wv[i]]++; 67 for (i=1; i<m; ++i) wc[i] += wc[i-1]; 68 for (i=n-1; i>=0; --i) sa[--wc[wv[i]]] = y[i]; 69 for (t=x, x=y, y=t, i=1,p=1,x[sa[0]]=0; i<n; ++i) 70 x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1:p++; 71 } 72 } 73 74 void calheight(int *r, int *sa, int n) { 75 int i, j, k=0; 76 77 for (i=1; i<=n; ++i) rrank[sa[i]] = i; 78 for (i=0; i<n; height[rrank[i++]]=k) 79 for (k?k--:0, j=sa[rrank[i]-1]; r[i+k]==r[j+k]; ++k); 80 } 81 82 bool judge(int bound) { 83 int i = 1; 84 int pn = 0; 85 86 while (i <= n) { 87 if (height[i] < bound) { 88 at = sa[i-1]; 89 sort(pos, pos+pn); 90 int kk = 1, p = pos[0]; 91 rep(j, 1, pn) { 92 if (pos[j]-p >= bound) { 93 p = pos[j]; 94 ++kk; 95 } 96 } 97 98 if (kk >= k) 99 return true; 100 pn = 0; 101 pos[pn++] = sa[i++]; 102 } else { 103 pos[pn++] = sa[i++]; 104 } 105 } 106 at = sa[n-1]; 107 sort(pos, pos+pn); 108 int kk = 1, p = pos[0]; 109 rep(j, 1, pn) { 110 if (pos[j]-p >= bound) { 111 p = pos[j]; 112 ++kk; 113 } 114 } 115 116 if (kk >= k) 117 return true; 118 119 return false; 120 } 121 122 void printSa(int n) { 123 for (int i=1; i<=n; ++i) 124 printf("%d ", sa[i]); 125 putchar(‘\n‘); 126 } 127 128 void printHeight(int n) { 129 for (int i=1; i<=n; ++i) 130 printf("%d ", height[i]); 131 putchar(‘\n‘); 132 } 133 134 void solve() { 135 sort(b, b+n); 136 int nn = unique(b, b+n) - b; 137 138 rep(i, 0, n) 139 aa[i] = lower_bound(b, b+nn, a[i]) - b + 1; 140 aa[n] = 0; 141 142 da(aa, sa, n+1, nn+4); 143 calheight(aa, sa, n); 144 145 #ifndef ONLINE_JUDGE 146 // printSa(n); 147 // printHeight(n); 148 #endif 149 150 int l = 1, r = n, mid; 151 int ans = 0, fr; 152 153 while (l <= r) { 154 mid = (l + r) >> 1; 155 if (judge(mid)) { 156 ans = mid; 157 fr = at; 158 l = mid + 1; 159 } else { 160 r = mid - 1; 161 } 162 } 163 164 printf("%d\n", ans); 165 rep(i, 0, ans) 166 printf("%d\n", b[aa[fr+i]-1]); 167 } 168 169 int main() { 170 ios::sync_with_stdio(false); 171 #ifndef ONLINE_JUDGE 172 freopen("data.in", "r", stdin); 173 freopen("data.out", "w", stdout); 174 #endif 175 176 int t; 177 178 scanf("%d", &t); 179 while (t--) { 180 scanf("%d %d", &n, &k); 181 rep(i, 0, n) { 182 scanf("%d", &a[i]); 183 b[i] = a[i]; 184 } 185 solve(); 186 if (t) 187 putchar(‘\n‘); 188 } 189 190 #ifndef ONLINE_JUDGE 191 printf("time = %d.\n", (int)clock()); 192 #endif 193 194 return 0; 195 }
【HDOJ】2890 Longest Repeated subsequence
标签:
原文地址:http://www.cnblogs.com/bombe1013/p/5178848.html