題目鏈接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=2806
(luogu) https://www.luogu.org/problemnew/show/P4022

題解:對“作文庫”中的串建廣義SAM。(感覺加個#拼在一起直接SAM也行啊，只是常數大了點，但是大家都寫的廣義SAM我也就跟著寫廣義SAM了233333)

詢問時二分\(L\), 變成求最少幾個位置不匹配。然后DP方程是\(dp[i]=\min(dp[i-1]+1,\min_{i-match[i]\le j\le i-L} dp[j])\). 其中\(match[j]\)表示以\(j\)結尾最長能匹配的子串長度，用后綴自動機求出。單調隊列優化即可。

應該a[i]-=96我寫成a[i]-=48; dp完了之后不更新ans; 二分分反……我是有多無腦

注意一個問題是，由于這里有\(mid\)的限制，只有\(\le i-mid\)的位置才能被放進單調隊列, 不可以一上來就把\(0\)放進去!

代碼

#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std;const int N = 1.1e6; const int S = 2; int son[(N<<1)+3][S+1]; int fa[(N<<1)+3]; int len[(N<<1)+3]; char a[N+3]; char b[N+3]; int match[N+3]; int dp[N+3]; int dq[N+3]; int n,m,siz,rtn,lstpos;void initSAM() {siz = rtn = lstpos = 1; }void insertchar(char ch) {int p = lstpos,np; siz++; np = lstpos = siz; len[np] = len[p]+1;for(; p && son[p][ch]==0; p=fa[p]) {son[p][ch] = np;}if(!p) {fa[np] = rtn;}else{int q = son[p][ch];if(len[q]==len[p]+1) {fa[np] = q;}else{siz++; int nq = siz; len[nq] = len[p]+1;memcpy(son[nq],son[q],sizeof(son[q]));fa[nq] = fa[q]; fa[np] = fa[q] = nq;for(; p && son[p][ch]==q; p=fa[p]) {son[p][ch] = nq;}}} }int main() {initSAM();scanf("%d%d",&n,&m);for(int i=1; i<=m; i++){lstpos = rtn;scanf("%s",a+1); int lena = strlen(a+1);for(int j=1; j<=lena; j++) {insertchar(a[j]-48);}}for(int i=1; i<=n; i++){scanf("%s",a+1); int lena = strlen(a+1);for(int j=1; j<=lena; j++) {a[j]-=48;}int u = rtn,cur = 0;for(int j=1; j<=lena; j++){while(u && son[u][a[j]]==0) {u = fa[u]; cur = len[u];}if(son[u][a[j]]) {u = son[u][a[j]]; cur++;}else {u = rtn; cur = 0;}match[j] = cur;}int left = 0,right = lena,mid,ans;while(left<right){mid = (left+right+1)>>1;dp[0] = 0; int head = 1,tail = 0;if(mid<=1) {tail++; dq[tail] = 0;}for(int j=1; j<=lena; j++){dp[j] = dp[j-1]+1;while(head<=tail && dq[head]<j-match[j]) {dq[head] = 0; head++;}if(head<=tail){dp[j] = min(dp[j],dp[dq[head]]);}if(j-mid+1>=0){while(head<=tail && dp[dq[tail]]>=dp[j-mid+1]) {dq[tail] = 0; tail--;}tail++; dq[tail] = j-mid+1;}}ans = dp[lena];if(ans*10<=lena) {left = mid;}else {right = mid-1;}}printf("%d\n",left);}return 0; }

總結

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