先贴个lrj的板子, kmp基本照这个写了, 其中f数组表示失配后应转移的新状态
int main() {
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;
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.
字符串循环节板子, 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]);
}
原文：https://www.cnblogs.com/uid001/p/10536194.html