題目連結:hdu 4436 str2int
解題思路
将所有字元串拼接在一起,連接配接處用不會出現的字元10代替。然後根據拓撲序,維護每個節點的sum 取模後的和,cnt 以該節點為終止位置的字元串個數。
代碼
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = ;
const int SIGMA_SIZE = ;
const int mod = ;
struct SAM {
int sz, last;
int g[maxn<<][SIGMA_SIZE], pre[maxn<<], step[maxn<<];
void newNode (int s) {
step[++sz] = s;
pre[sz] = ;
memset(g[sz], , sizeof(g[sz]));
}
void init () {
sz = , last = ;
newNode();
}
void insert(int v) {
newNode(step[last] + );
int p = last, np = sz;
while (p && !g[p][v]) {
g[p][v] = np;
p = pre[p];
}
if (p) {
int q = g[p][v];
if (step[q] == step[p] + )
pre[np] = q;
else {
newNode(step[p] + );
int nq = sz;
for (int j = ; j < SIGMA_SIZE; j++) g[nq][j] = g[q][j];
pre[nq] = pre[q];
pre[np] = pre[q] = nq;
while (p && g[p][v] == q) {
g[p][v] = nq;
p = pre[p];
}
}
} else
pre[np] = ;
last = np;
}
int cnt[maxn<<], sum[maxn<<], du[maxn<<];
int solve () {
int ret = ;
for (int i = ; i <= sz; i++)
cnt[i] = sum[i] = du[i] = ;
for (int i = ; i <= sz; i++) {
for (int j = ; j < SIGMA_SIZE; j++)
du[g[i][j]]++;
}
queue<int> que;
que.push();
cnt[] = ;
while (!que.empty()) {
int u = que.front();
que.pop();
ret = (ret + sum[u]) % mod;
for (int i = ; i < SIGMA_SIZE; i++) {
int v = g[u][i];
du[v]--;
if (du[v] == && v) que.push(v);
if (v == || (i == && u == )) continue;
if (i == ) continue;
cnt[v] = (cnt[u] + cnt[v]) % mod;
sum[v] = (sum[v] + sum[u] * % mod + cnt[u] * i % mod) % mod;
}
}
return ret;
}
}SA;
int N;
char str[maxn];
int main () {
while (scanf("%d", &N) == ) {
SA.init();
while (N--) {
scanf("%s", str);
int n = strlen(str);
for (int i = ; i < n; i++) SA.insert(str[i] - '0');
SA.insert();
}
printf("%d\n", SA.solve());
}
return ;
}