kmp板子如下, 失配数组不优化的话, $f_i$就表示子串[0...i]前后缀最大匹配长度
int main() { scanf("%s%s", t, p); int n = strlen(t), m = strlen(p); f[0]=f[1]=0; int j = 0; REP(i,1,m-1) { while (j&&p[i]!=p[j]) j=f[j]; if (p[i]==p[j]) ++j; f[i+1] = j; } j = 0; REP(i,0,n-1) { while (j&&p[j]!=t[i]) j=f[j]; if (p[j]==t[i]) ++j; if (j==m) printf("%d", i-m+1); }}
练习1: hdu5763
大意: 给定字符串T, 模板串P, 可以将T中与P匹配的子串替换为'*', 求多少种替换方案.
一个板子题, kmp求出可以替换的位置, 然后dp就好了
const int N = 1e6+10;char t[N], p[N];int f[N], ff[N], *dp=ff+1;void add(int &a, int b) {a+=b;if (a>=P) a-=P;}void work() { scanf("%s%s", t, p); int n = strlen(t), m = strlen(p); f[0]=f[1]=0; REP(i,1,m-1) { int j = f[i]; while (j&&p[i]!=p[j]) j=f[j]; if (p[i]==p[j]) ++j; f[i+1] = j; } int j = 0; dp[-1] = 1; REP(i,0,n-1) { while (j&&p[j]!=t[i]) j=f[j]; if (p[j]==t[i]) ++j; dp[i] = 0; if (j==m) add(dp[i],dp[i-m]); add(dp[i],dp[i-1]); } printf("%d\n",dp[n-1]);}int main() { int t; scanf("%d", &t); REP(i,1,t) printf("Case #%d: ",i),work();}
练习2 CF825F
大意: 给定字符串$s$, 可以将连续$c1$个相同的子串$s1$压缩为|c1|+|s1|, 求压缩若干次后$s$最短长度.
要用到一个字符串循环节的结论, 假设一个长$n$的字符串$s$, 失配函数为$f$, 则循环节为n-f[n]或n
const int N = 8e3+10;int n;char s[N];int f[N], g[N][N], ff[N], *dp=ff+1;void getFail(char *s) { int m = strlen(s); f[0]=f[1]=0; REP(i,1,m-1) { int j=f[i]; while (j&&s[i]!=s[j]) j=f[j]; if (s[i]==s[j]) ++j; f[i+1] = j; }}int calc(int x) {int r=0;while (x) ++r,x/=10;return r;}int main() { scanf("%s", s); n = strlen(s); REP(i,0,n-1) { getFail(s+i); REP(j,i,n-1) { int len = j-i+1; if (len%(len-f[len])==0) g[i][j]=len-f[len]+calc(len/(len-f[len])); else g[i][j]=len+1; } } dp[-1] = 0; REP(i,0,n-1) { dp[i] = INF; REP(j,0,i) dp[i] = min(dp[i], dp[j-1]+g[j][i]); } printf("%d\n", dp[n-1]);}
练习3 CF494B
这题想了好长时间, 计数还是不熟练啊
$dp[i]=j+1+\sum\limits_{k\le j-1} sum[k]$
dp[i]是以i为右端点的方案, sum[i]是dp[i]的前缀和, j为上次匹配位置
int n, m;char t[N], p[N];int dp[N], s1[N], s2[N], f[N], g[N];void add(int &a, int b) {a+=b;if (a>=P)a-=P;}int main() { scanf("%s%s", t, p); n = strlen(t), m = strlen(p); f[0]=f[1]=0; REP(i,1,m-1) { int j=f[i]; while (j&&p[j]!=p[i]) j=f[j]; if (p[j]==p[i]) ++j; f[i+1]=j; } int j=0; REP(i,0,n-1) { while (j&&t[i]!=p[j]) j=f[j]; if (t[i]==p[j]) ++j; if (j==m) g[i]=1; } REP(i,0,n-1) { if (g[i]) j=i-m+1,dp[i]=s2[j-1],add(dp[i],j+1); else if (i) dp[i] = dp[i-1]; if (i) s1[i] = s1[i-1]; add(s1[i],dp[i]); if (i) s2[i] = s2[i-1]; add(s2[i],s1[i]); } printf("%d\n",s1[n-1]);}
练习4 CF 1015 Bracket Substring
大意:求长度为2n, 给定括号串$s$, 求所有以$s$为子串的合法括号序列种类数.
kmp经典套路了, 设$dp[i][j][k]$表示到$i$位时, 已经匹配$j$位, 左右括号差为$k$的方案数
对于放'(', 若s[j]为'('则匹配位数转移到j+1, 否则沿失配边走. 对于')'的情况同理.
要注意特殊处理已经转移完毕的情况. 这里可以将失配函数优化一下, 复杂度是$O(\omega^2 n^2m), \omega$为字符种类数, 不优化的话是$O(\omega n^2m^2)$实际上也能跑过.
const int N = 210;int n, m;char s[N];int dp[N][N][N], f[N];void add(int &a, int b) {a+=b;if (a>=P)a-=P;}int main() { scanf("%d%s", &n, s); m = strlen(s); REP(i,1,m-1) { int j = f[i]; while (j&&s[i]!=s[j]) j=f[j]; if (s[i]==s[j]) ++j; f[i+1]=j; } REP(i,1,m-1) if (s[i]==s[f[i]]) f[i]=f[f[i]]; dp[0][0][0]=1; REP(i,1,2*n) REP(j,0,m) REP(k,0,n) if (dp[i-1][j][k]) { int nxt = j; while (nxt&&s[nxt]!='(') nxt=f[nxt]; if (s[nxt]=='(') ++nxt; if (j==m) nxt=m; add(dp[i][nxt][k+1],dp[i-1][j][k]); if (!k) continue; nxt = j; while (nxt&&s[nxt]!=')') nxt=f[nxt]; if (s[nxt]==')') ++nxt; if (j==m) nxt=m; add(dp[i][nxt][k-1],dp[i-1][j][k]); } printf("%d\n", dp[2*n][m][0]);}
练习5 HDU3689 Infinite monkey theorem
跟上题类似, 借助kmp进行转移
const int N = 1e3+10;int n, m;double q[N], dp[N][22];char s[N];int f[N];void work() { REP(i,'a','z') q[i]=0; REP(i,1,n) { char c; double x; scanf(" %c%lf", &c, &x); q[c] = x; } scanf("%s", s); n = strlen(s); REP(i,1,n-1) { int j=f[i]; while (j&&s[j]!=s[i]) j=f[j]; if (s[j]==s[i]) ++j; f[i+1]=j; } REP(i,1,n-1) if (s[i]==s[f[i]]) f[i]=f[f[i]]; memset(dp,0,sizeof dp); dp[0][0] = 1; REP(i,1,m) REP(j,0,n) { REP(k,'a','z') { int nxt = j; while (nxt&&s[nxt]!=k) nxt=f[nxt]; if (s[nxt]==k) ++nxt; if (j==n) nxt=n; dp[i][nxt] += q[k]*dp[i-1][j]; } } printf("%.2lf%%\n", 100*dp[m][n]);}int main() { for (; scanf("%d%d", &n, &m), n||m; ) work();}
练习6 CF 432D Prefixes and Suffixes
大意: 给定字符串$s$, 求$s$有多少个前缀等于后缀, 并且输出它们的出现次数
kmp经典题了, 失配函数如果不优化的话, 那么$f_i$就表示子串[0,i-1]中最大的使前缀等于后缀的前缀长度.
也就是说$s$满足要求的最大前缀长度是$f[n]$, 其余前缀一定在$f[n]$内, 这样就可以迭代求出满足要求的前缀位置.
再用$dp$求出每个前缀的出现次数即可.
const int N = 1e6+10;int n;char s[N];int f[N], dp[N], a[N];int main() { scanf("%s", s); n = strlen(s); REP(i,1,n-1) { int j = f[i]; while (j&&s[i]!=s[j]) j=f[j]; if (s[i]==s[j]) ++j; f[i+1]=j; } for (int nxt=n; nxt; nxt=f[nxt]) a[++*a]=nxt; PER(i,1,n) dp[f[i]]+=++dp[i]; printf("%d\n", *a); PER(i,1,*a) printf("%d %d\n", a[i], dp[a[i]]);}
练习7 CF808G Anthem of Berland
先预处理了一下第i位是否能匹配到第j位, 最后再dp一遍
const int N = 1e6+10;int n, m;char s[N], t[N];vector> g;int dp[N], mx[N], f[N], ff[N];int main() { scanf("%s%s", s+1, t); n = strlen(s+1), m = strlen(t); if (n