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综合性很强的一道题目,结合manacher,后缀数组,哈希,RMQ,二分可解。
基本思路是通过manacher可以找到所有可能的回文串,哈希去重,后缀数组二分找数目。最后暴力求解。
需要注意kth需要为__int64。
1 /* 4426 */ 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 __int64 mod = 777777777LL; 44 const int seed = 50009; 45 const int INF = 0x3f3f3f3f; 46 const int maxn = 1e5+5; 47 // input 48 char s[maxn]; 49 int val[26]; 50 __int64 V[maxn]; 51 // manacher 52 char ss[maxn*2]; 53 int P[maxn*2]; 54 // sa 55 int a[maxn]; 56 int height[maxn], sa[maxn], rrank[maxn]; 57 int wa[maxn], wb[maxn], wc[maxn], wv[maxn]; 58 // RMQ 59 int dp[maxn][17]; 60 // Hash 61 unsigned __int64 H[maxn]; 62 // init 63 unsigned __int64 Base[maxn], Power[maxn]; 64 vpii pal; 65 set<unsigned __int64> st; 66 67 void init() { 68 Base[0] = Power[0] = 1; 69 rep(i, 1, maxn) { 70 Base[i] = 1LL * Base[i-1] * 26 % mod; 71 Power[i] = 1LL * Power[i-1] * seed; 72 } 73 } 74 75 void init_Hash(int n) { 76 H[0] = s[0] - ‘a‘ + 1; 77 rep(i, 1, n) 78 H[i] = H[i-1] * seed + s[i]-‘a‘+1; 79 } 80 81 unsigned __int64 getHash(int l, int r) { 82 if (l == 0) 83 return H[r]; 84 return H[r] - H[l-1] * Power[r-l+1]; 85 } 86 87 bool cmp(int *r, int a, int b, int l) { 88 return r[a]==r[b] && r[a+l]==r[b+l]; 89 } 90 91 void da(int *r, int *sa, int n, int m) { 92 int i, j, *x=wa, *y=wb, *t, p; 93 94 for (i=0; i<m; ++i) wc[i] = 0; 95 for (i=0; i<n; ++i) wc[x[i]=r[i]]++; 96 for (i=1; i<m; ++i) wc[i] += wc[i-1]; 97 for (i=n-1; i>=0; --i) sa[--wc[x[i]]] = i; 98 for (j=1,p=1; p<n; j*=2, m=p) { 99 for (p=0,i=n-j; i<n; ++i) y[p++] = i; 100 for (i=0; i<n; ++i) if (sa[i] >= j) y[p++] = sa[i]-j; 101 for (i=0; i<n; ++i) wv[i] = x[y[i]]; 102 for (i=0; i<m; ++i) wc[i] = 0; 103 for (i=0; i<n; ++i) wc[wv[i]]++; 104 for (i=1; i<m; ++i) wc[i] += wc[i-1]; 105 for (i=n-1; i>=0; --i) sa[--wc[wv[i]]] = y[i]; 106 for (t=x,x=y,y=t,p=1,x[sa[0]]=0, i=1; i<n; ++i) 107 x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++; 108 } 109 } 110 111 void calheight(int *r, int *sa, int n) { 112 int i, j, k = 0; 113 114 for (i=1; i<=n; ++i) rrank[sa[i]] = i; 115 for (i=0; i<n; height[rrank[i++]]=k) 116 for (k?k--:0, j=sa[rrank[i]-1]; r[j+k]==r[i+k]; ++k) ; 117 } 118 119 void init_RMQ(int n) { 120 int i, j; 121 122 for (i=1; i<=n; ++i) 123 dp[i][0] = height[i]; 124 dp[1][0] = INF; 125 for (j=1; (1<<j)<=n; ++j) 126 for (i=1; i+(1<<j)-1<=n; ++i) 127 dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]); 128 } 129 130 int RMQ(int l, int r) { 131 if (l > r) 132 swap(l, r); 133 134 ++l; 135 int k = 0; 136 137 while (1<<(k+1) <= r-l+1) 138 ++k; 139 140 return min(dp[l][k], dp[r-(1<<k)+1][k]); 141 } 142 143 void Manacher(char *s, int *P, int n) { 144 pal.clr(); 145 st.clr(); 146 147 int i, j, mx = 0, id = 0; 148 149 for (i=1; i<n; ++i) { 150 P[i] = mx>i ? min(P[2*id-i], mx-i) : 1; 151 while (s[i+P[i]] == s[i-P[i]]) 152 ++P[i]; 153 if (i+P[i] > mx) { 154 for (j=mx; j<i+P[i]; ++j) { 155 int l = 2*i-j, r = j; 156 l >>= 1; 157 r = (r & 1) ? (r>>1) : (r>>1)-1; 158 if (l > r) 159 continue; 160 161 unsigned __int64 hval = getHash(l, r); 162 if (st.find(hval) == st.end()) { 163 st.insert(hval); 164 pal.pb(mp(l, r)); 165 } 166 } 167 mx = i + P[i]; 168 id = i; 169 } 170 } 171 } 172 173 int getCnt(int fr, int to, int n) { 174 int len = to - fr + 1; 175 int rankfr = rrank[fr]; 176 int l, r, mid, tmp; 177 int L = rankfr, R = rankfr; 178 179 // find left most 180 l = 1, r = rankfr - 1; 181 while (l <= r) { 182 mid = (l + r) >> 1; 183 tmp = RMQ(mid, rankfr); 184 if (tmp >= len) { 185 L = mid; 186 r = mid - 1; 187 } else { 188 l = mid + 1; 189 } 190 } 191 192 // find right most 193 l = rankfr + 1, r = n; 194 while (l <= r) { 195 mid = (l + r) >> 1; 196 tmp = RMQ(rankfr, mid); 197 if (tmp >= len) { 198 R = mid; 199 l = mid + 1; 200 } else { 201 r = mid - 1; 202 } 203 } 204 205 return R - L + 1; 206 } 207 208 void init_Val(int n) { 209 V[0] = val[s[0]-‘a‘]; 210 rep(i, 1, n) 211 V[i] = (V[i-1] * 26 + val[s[i]-‘a‘]) % mod; 212 } 213 214 __int64 getVal(int l, int r) { 215 if (l == 0) 216 return V[r]; 217 return (V[r] - V[l-1] * Base[r-l+1]%mod + mod) % mod; 218 } 219 220 void printSa(int n) { 221 for (int i=1; i<=n; ++i) 222 printf("%d ", sa[i]); 223 putchar(‘\n‘); 224 } 225 226 void printHeight(int n) { 227 for (int i=1; i<=n; ++i) 228 printf("%d ", height[i]); 229 putchar(‘\n‘); 230 } 231 232 int main() { 233 ios::sync_with_stdio(false); 234 #ifndef ONLINE_JUDGE 235 freopen("data.in", "r", stdin); 236 freopen("data.out", "w", stdout); 237 #endif 238 239 int t; 240 int n, q; 241 __int64 kth; 242 int ans; 243 vpii vc; 244 245 init(); 246 scanf("%d", &t); 247 while (t--) { 248 scanf("%d %d", &n, &q); 249 scanf("%s", s); 250 251 // init sa 252 rep(i, 0, n) 253 a[i] = s[i]-‘a‘+1; 254 a[n] = 0; 255 da(a, sa, n+1, 30); 256 calheight(a, sa, n); 257 init_RMQ(n); 258 259 // init Hash 260 init_Hash(n); 261 262 // init Manacher 263 int l = 0; 264 ss[l++] = ‘@‘; 265 // ss[l++] = ‘#‘; 266 rep(i, 0, n) { 267 ss[l++] = s[i]; 268 ss[l++] = ‘#‘; 269 } 270 ss[l] = ‘\0‘; 271 Manacher(ss, P, l); 272 273 // find count of palindromic 274 int sz = SZ(pal); 275 rep(i, 0, sz) { 276 wc[i] = getCnt(pal[i].fir, pal[i].sec, n); 277 pal[i].sec = (pal[i].fir + pal[i].sec) >> 1; 278 } 279 280 while (q--) { 281 scanf("%I64d", &kth); 282 rep(i, 0, 26) 283 scanf("%d", &val[i]); 284 init_Val(n); 285 vc.clr(); 286 rep(i, 0, sz) { 287 int tmp = getVal(pal[i].fir, pal[i].sec); 288 vc.pb(mp(tmp, wc[i])); 289 } 290 sort(all(vc)); 291 ans = -1; 292 rep(i, 0, sz) { 293 if (kth <= vc[i].sec) { 294 ans = vc[i].fir; 295 break; 296 } else { 297 kth -= vc[i].sec; 298 } 299 } 300 printf("%d\n", ans); 301 } 302 putchar(‘\n‘); 303 } 304 305 #ifndef ONLINE_JUDGE 306 printf("time = %d.\n", (int)clock()); 307 #endif 308 309 return 0; 310 }
数据发生器。
1 from random import randint, shuffle 2 import shutil 3 import string 4 5 6 def GenDataIn(): 7 with open("data.in", "w") as fout: 8 t = 20 9 bound = 10**3 10 lc = list(string.lowercase) 11 uc = list(string.uppercase) 12 fout.write("%d\n" % (t)) 13 for tt in xrange(t): 14 n = randint(10**4, 10**5) 15 q = randint(30, 40) 16 fout.write("%d %d\n" % (n, q)) 17 line = "" 18 for j in xrange(n): 19 idx = randint(0, 25) 20 line += lc[idx] 21 fout.write("%s\n" % (line)) 22 for i in xrange(q): 23 kth = randint(1, 1005) 24 dataList = [kth] 25 for j in xrange(26): 26 val = randint(0, 25) 27 dataList.append(val) 28 fout.write(" ".join(map(str, dataList)) + "\n") 29 30 31 def MovDataIn(): 32 desFileName = "F:\eclipse_prj\workspace\hdoj\data.in" 33 shutil.copyfile("data.in", desFileName) 34 35 36 if __name__ == "__main__": 37 GenDataIn() 38 MovDataIn()
【HDOJ】4426 Palindromic Substring
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原文地址:http://www.cnblogs.com/bombe1013/p/5182512.html