长度不小于k的公共子串的个数

长度不小于k的公共子串的个数

长度不小于k的公共子串的个数,论文里有题解,卡了一上午,因为sum没开long long!!!

没开long long毁一生again~~~

以后应该早看POJ里的Discuss啊QAQ

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 200003; int t1[N], t2[N], c[N];
void st(int *x, int *y, int *sa, int n, int m) {
int i;
for(i = 0; i < m; ++i) c[i] = 0;
for(i = 0; i < n; ++i) ++c[x[y[i]]];
for(i = 1; i < m; ++i) c[i] += c[i - 1];
for(i = n - 1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
}
void mkhz(int *a, int *sa, int n, int m) {
int *t, *x = t1, *y = t2, i, j, p;
for(i = 0; i < n; ++i) x[i] = a[i], y[i] = i;
st(x, y, sa, n, m);
for(p = 1, j = 1; p < n; j <<= 1, m = p) {
for(p = 0, i = n - j; i < n; ++i) y[p++] = i;
for(i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
st(x, y, sa, n, m);
for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + j] == y[sa[i - 1] + j] ? p - 1 : p++;
}
}
void mkh(int *r, int *sa, int *rank, int *h, int n) {
int i, j, k = 0;
for(i = 1; i <= n; ++i) rank[sa[i]] = i;
for(i = 1; i <= n; h[rank[i++]] = k)
for(k ? --k : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
} char s[N];
int a[N], sa[N], rank[N], h[N], n, k, sta[N], top, tmp, mark[N];
long long sum[N];
int main() {
while (scanf("%d", &k), k) {
scanf("%s", s + 1);
tmp = strlen(s + 1);
s[++tmp] = 1;
scanf("%s", s + tmp + 1);
n = strlen(s + 1);
for(int i = 1; i <= n; ++i) a[i] = s[i];
mkhz(a, sa, n + 1, 130);
mkh(a, sa, rank, h, n);
long long ans = 0;
for(int i = 1; i <= n; ++i) {
if (h[i] < k) {
top = sum[0] = sum[1] = 0;
} else {
for(int j = top; sta[j] > h[i] - k + 1 && j; --j) {
sum[mark[j]] += h[i] - k + 1 - sta[j];
sta[j] = h[i] - k + 1;
}
sta[++top] = h[i] - k + 1;
if (sa[i - 1] < tmp) mark[top] = 0;
if (sa[i - 1] > tmp) mark[top] = 1;
sum[mark[top]] += h[i] - k + 1;
if (sa[i] < tmp) ans += sum[1];
if (sa[i] > tmp) ans += sum[0];
}
}
printf("%lld\n", ans);
}
return 0;
}

没开long long,毁我青春,耗我钱财,颓我精神==

05-11 11:30