B P5 第十三届北航程序设计竞赛预赛
https://buaacoding.cn/contest-ng/index.html#/188/problems
其实这题挺简单的。
注意到答案的大小最多是22
二分,check长度是mid的不同子串有多少个,然后,如果不是1 << mid个,那么肯定是不行的。
想到了这个,我居然用了字符串hash去判,傻逼了,注意到是01串,相当于二进制。
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#include <bits/stdc++.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; const int maxn = 5e6 + 20; typedef unsigned long long int ULL; char str[maxn]; int lenstr; int cnt[1 << 23], DFN; int po[maxn]; bool check(int mid) { DFN++; int val = 0; for (int i = 1; i <= mid; ++i) { val = val * 2 + str[i] - '0'; } cnt[val] = DFN; for (int i = mid + 1; i <= lenstr; ++i) { val -= po[mid - 1] * (str[i - mid] - '0'); val *= 2; val += str[i] - '0'; cnt[val] = DFN; } int en = (1 << mid) - 1; for (int i = 0; i <= en; ++i) { if (cnt[i] != DFN) { return true; } } return false; } void work() { scanf("%s", str + 1); lenstr = strlen(str + 1); int be = 1, en = min(22, lenstr); while (be <= en) { int mid = (be + en) >> 1; if (check(mid)) en = mid - 1; else be = mid + 1; } printf("%d\n", be); } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif po[0] = 1; for (int i = 1; i <= 23; ++i) po[i] = po[i - 1] * 2; int t; scanf("%d", &t); while (t--) work(); return 0; }
很奇怪为什么能卡我dc3,我能算出长度是i的不同子串的个数,
思路就是,每一个sa[i],有lenstr - sa[i] + 1个前缀,然后和前面的sa[i - 1] 会有重叠的地方,也就是height[i]个,然后减去即可。
区间减法打标记。但是,,TLE额。
why
可能多了点常数
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/* Author: stupid_one Result: TLE Submission_id: 506947 Created at: Sun Dec 17 2017 10:45:12 GMT+0800 (CST) Problem_id: 959 Time: 1000 Memory: 0 */ #include <bits/stdc++.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; const int maxn = 9e6 + 20; typedef unsigned long long int ULL; char str[maxn]; int lenstr; int cnt[maxn]; const int N = maxn; #define F(x) ((x)/3+((x)%3==1?0:tb)) #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) int r[maxn]; int wa[maxn],wb[maxn],wv[maxn],WS[maxn]; int sa[maxn]; int c0(int *r,int a,int b) { return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2]; } int c12(int k,int *r,int a,int b) { if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1); else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1]; } void sort(int *r,int *a,int *b,int n,int m) { int i; for(i=0; i<n; i++) wv[i]=r[a[i]]; for(i=0; i<m; i++) WS[i]=0; for(i=0; i<n; i++) WS[wv[i]]++; for(i=1; i<m; i++) WS[i]+=WS[i-1]; for(i=n-1; i>=0; i--) b[--WS[wv[i]]]=a[i]; return; } void dc3(int *r,int *sa,int n,int m) { //涵义与DA 相同 int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p; r[n]=r[n+1]=0; for(i=0; i<n; i++) if(i%3!=0) wa[tbc++]=i; sort(r+2,wa,wb,tbc,m); sort(r+1,wb,wa,tbc,m); sort(r,wa,wb,tbc,m); for(p=1,rn[F(wb[0])]=0,i=1; i<tbc; i++) rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++; if(p<tbc) dc3(rn,san,tbc,p); else for(i=0; i<tbc; i++) san[rn[i]]=i; for(i=0; i<tbc; i++) if(san[i]<tb) wb[ta++]=san[i]*3; if(n%3==1) wb[ta++]=n-1; sort(r,wb,wa,ta,m); for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i; for(i=0,j=0,p=0; i<ta && j<tbc; p++) sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++]; for(; i<ta; p++) sa[p]=wa[i++]; for(; j<tbc; p++) sa[p]=wb[j++]; return; } int RANK[maxn], height[maxn]; void calheight(int *r,int *sa,int n) { // 此处N为实际长度 int i,j,k=0; // height[]的合法范围为 1-N, 其中0是结尾加入的字符 for(i=1; i<=n; i++) RANK[sa[i]]=i; // 根据SA求RANK for(i=0; i<n; height[RANK[i++]] = k ) // 定义:h[i] = height[ rank[i] ] for(k?k--:0,j=sa[RANK[i]-1]; r[i+k]==r[j+k]; k++); //根据 h[i] >= h[i-1]-1 来优化计算height过程 } void work() { scanf("%s", str); lenstr = strlen(str); for (int i = 0; i < lenstr; ++i) r[i] = str[i]; r[lenstr] = 0; dc3(r, sa, lenstr + 1, 128); calheight(r, sa, lenstr); // printf("%d\n", sa[1]); for (int i = 1; i <= lenstr; ++i) { cnt[1]++; cnt[lenstr - sa[i] + 1]--; cnt[1]--; cnt[height[i] + 1]++; } for (int i = 1; i < 23; ++i) cnt[i] += cnt[i - 1]; for (int i = 1; i <= 22; ++i) { if (cnt[i] != (1 << (i))) { printf("%d\n", i); for (int k = 1; k < 23; ++k) cnt[k] = 0; return; } } } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif int t; scanf("%d", &t); while (t--) work(); return 0; }
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/* Author: stupid_one Result: TLE Submission_id: 506775 Created at: Sun Dec 17 2017 01:21:28 GMT+0800 (CST) Problem_id: 959 Time: 1000 Memory: 0 */ #include <bits/stdc++.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; const int maxn = 5e6 + 20; typedef unsigned long long int ULL; ULL po[maxn]; char str[maxn]; int lenstr; ULL hs[maxn]; const int seed = 131; const int mod = 10007; struct MP { int first[mod + 2], num; ULL eval[maxn]; int nxt[maxn]; void init() { memset(first, -1, sizeof first); num = 0; } bool addEdge(ULL val) { int u = val % mod; for (int i = first[u]; ~i; i = nxt[i]) { if (val == eval[i]) return false; } eval[num] = val; nxt[num] = first[u]; first[u] = num++; return true; } } h; bool check(int val) { int t = 1 << val; if (lenstr - val + 1 < t) return false; h.init(); int ans = 0; for (int i = val; i <= lenstr; ++i) { ans += h.addEdge(hs[i] - po[val] * hs[i - val]); if (ans == t) return true; if (ans + (lenstr - i) < t) return false; } return false; } int sa[maxn], x[maxn], y[maxn], book[maxn]; //book[]大小起码是lenstr,book[rank[]] bool cmp(int r[], int a, int b, int len) { //这个必须是int r[], return r[a] == r[b] && r[a + len] == r[b + len]; } void da(char str[], int sa[], int lenstr, int mx) { int *fir = x, *sec = y, *ToChange; for (int i = 0; i <= mx; ++i) book[i] = 0; //清0 for (int i = 1; i <= lenstr; ++i) { fir[i] = str[i]; //开始的rank数组,只保留相对大小即可,开始就是str[] book[str[i]]++; //统计不同字母的个数 } for (int i = 1; i <= mx; ++i) book[i] += book[i - 1]; //统计 <= 这个字母的有多少个元素 for (int i = lenstr; i >= 1; --i) sa[book[fir[i]]--] = i; // <=str[i]这个字母的有x个,那么,排第x的就应该是这个i的位置了。 //倒过来排序,是为了确保相同字符的时候,前面的就先在前面出现。 //p是第二个关键字0的个数 for (int j = 1, p = 1; p <= lenstr; j <<= 1, mx = p) { //字符串长度为j的比较 //现在求第二个关键字,然后合并(合并的时候按第一关键字优先合并) p = 0; for (int i = lenstr - j + 1; i <= lenstr; ++i) sec[++p] = i; //上面的位置,再跳j格就是越界了的,所以第二关键字是0,排在前面 for (int i = 1; i <= lenstr; ++i) if (sa[i] > j) //如果排名第i的起始位置在长度j之后 sec[++p] = sa[i] - j; //减去这个长度j,表明第sa[i] - j这个位置的第二个是从sa[i]处拿的,排名靠前也//正常,因为sa[i]排名是递增的 //sec[]保存的是下标,现在对第一个关键字排序 for (int i = 0; i <= mx; ++i) book[i] = 0; //清0 for (int i = 1; i <= lenstr; ++i) book[fir[sec[i]]]++; for (int i = 1; i <= mx; ++i) book[i] += book[i - 1]; for (int i = lenstr; i >= 1; --i) sa[book[fir[sec[i]]]--] = sec[i]; //因为sec[i]才是对应str[]的下标 //现在要把第二关键字的结果,合并到第一关键字那里。同时我需要用到第一关键//字保存的记录,所以用指针交换的方式达到快速交换数组中的值 ToChange = fir, fir = sec, sec = ToChange; fir[sa[1]] = 0; //固定的是0 因为sa[1]固定是lenstr那个0 p = 2; for (int i = 2; i <= lenstr; ++i) //fir是当前的rank值,sec是前一次的rank值 fir[sa[i]] = cmp(sec, sa[i - 1], sa[i], j) ? p - 1 : p++; } return ; } int height[maxn], RANK[maxn]; void calcHight(char str[], int sa[], int lenstr) { for (int i = 1; i <= lenstr; ++i) RANK[sa[i]] = i; //O(n)处理出rank[] int k = 0; for (int i = 1; i <= lenstr - 1; ++i) { //最后一位不用算,最后一位排名一定是1,然后sa[0]就尴尬了 k -= k > 0; int j = sa[RANK[i] - 1]; //排名在i前一位的那个串,相似度最高 while (str[j + k] == str[i + k]) ++k; height[RANK[i]] = k; } return ; } int cnt[maxn]; void work() { scanf("%s", str + 1); lenstr = strlen(str + 1); for (int i = 1; i <= lenstr; ++i) cnt[i] = 0; str[lenstr + 1] = '$'; str[lenstr + 2] = 0; da(str, sa, lenstr + 1, 128); calcHight(str, sa, lenstr + 1); for (int i = 2; i <= lenstr + 1; ++i) { cnt[1]++; cnt[lenstr - sa[i] + 2]--; cnt[1]--; cnt[height[i] + 1]++; } for (int i = 1; i <= lenstr; ++i) cnt[i] += cnt[i - 1]; for (int i = 1; i <= lenstr; ++i) { if (cnt[i] != (1 << i)) { printf("%d\n", i); return; } } // printf("\n"); // for (int i = 1; str[i]; ++i) hs[i] = hs[i - 1] * seed + str[i]; // for (int i = 1; i <= 22; ++i) { // if (!check(i)) { // printf("%d\n", i); // return ; // } // } } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif po[0] = 1; for (int i = 1; i <= maxn - 20; ++i) po[i] = po[i - 1] * seed; int t; scanf("%d", &t); while (t--) work(); return 0; }
既然选择了远方,就要风雨兼程~
posted on 2017-12-18 10:02 stupid_one 阅读(360) 评论(0) 编辑 收藏 举报