2016 Multi-University Training Contest 4 1006 Substring（后缀数组）

题意

给出一个字符X和一个字符串，求字符串的字串中，包含字符的不同字串有多少个。

思路

很明显的后缀数组，但是要做一些变化，因为必须至少包含一个字符X，所以我们就记录一下对于每个sa[i]，离他最近的X的位置为Loc[sa[i]],这个位置到sa[i]的距离和height[i]的最大值就是不可取的数量，然后用n减去sa[i]和上面那个值就是这个sa的可选数量，累加就是答案。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef int main_int;
#define int ll
const int N = 1e5+10;

int wa
, wb
, ws
, wv
;
int ran
, height
;

bool cmp(int r[], int a, int b, int l) {
return r[a] == r[b] && r[a+l] == r[b+l];
}

void da(int r[], int sa[], int n, int m) {
int i, j, p, *x = wa, *y = wb;
for (i = 0; i < m; ++i) ws[i] = 0;
for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
for (i = 1; i < m; ++i) ws[i] += ws[i-1];
for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
for (j = 1, p = 1; p < n; j *= 2, 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;
for (i = 0; i < n; ++i) wv[i] = x[y[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) sa[--ws[wv[i]]] = y[i];
for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
}
}

void calheight(int r[], int sa[], int n) {
int i, j, k = 0;
for (i = 1; i <= n; ++i) ran[sa[i]] = i;
for (i = 0; i < n; height[ran[i++]] = k)
for (k?k--:0, j = sa[ran[i]-1]; r[i+k] == r[j+k]; k++);
}
int loc
;
char str
;
int
9a33
sa
, n, r
;
char ch[10];
main_int main() {
int T,kas =1;
//freopen("1006.in","r",stdin);
//freopen("1006","w",stdout);
scanf("%I64d", &T);
while(T--){
scanf("%s", ch);
scanf("%s", str);
n = strlen(str);
for(int i = 0; i < n; i++) {
r[i] = (int)str[i];
}
r
= 0;
loc
= n;
for(int i = n-1 ; i >= 0 ; i --){
if(str[i] == ch[0]) loc[i] = i;
else loc[i] = loc[i+1];
}
da(r, sa, n + 1, 128LL);
calheight(r, sa, n);
int ans = 0LL;
for(int i = 1; i <= n; i++) {
int tmp = max(loc[sa[i]] - sa[i],  height[i]);
ans += n - sa[i] -  tmp;
}
printf("Case #%I64d: %I64d\n",kas ++,ans);
}
return 0;
}