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hdu 4436 str2int(后缀自动机)

2015-11-06 21:13 429 查看
题目链接:hdu 4436 str2int

解题思路

将所有字符串拼接在一起,连接处用不会出现的字符10代替。然后根据拓扑序,维护每个节点的sum 取模后的和,cnt 以该节点为终止位置的字符串个数。

代码

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 100005;
const int SIGMA_SIZE = 11;
const int mod = 2012;

struct SAM {
int sz, last;
int g[maxn<<1][SIGMA_SIZE], pre[maxn<<1], step[maxn<<1];

void newNode (int s) {
step[++sz] = s;
pre[sz] = 0;
memset(g[sz], 0, sizeof(g[sz]));
}

void init () {
sz = 0, last = 1;
newNode(0);
}

void insert(int v) {
newNode(step[last] + 1);
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] + 1)
pre[np] = q;
else {
newNode(step[p] + 1);
int nq = sz;
for (int j = 0; 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] = 1;
last = np;
}

int cnt[maxn<<1], sum[maxn<<1], du[maxn<<1];

int solve () {
int ret = 0;
for (int i = 1; i <= sz; i++)
cnt[i] = sum[i] = du[i] = 0;

for (int i = 1; i <= sz; i++) {
for (int j = 0; j < SIGMA_SIZE; j++)
du[g[i][j]]++;
}

queue<int> que;
que.push(1);
cnt[1] = 1;

while (!que.empty()) {
int u = que.front();
que.pop();

ret = (ret + sum[u]) % mod;

for (int i = 0; i < SIGMA_SIZE; i++) {
int v = g[u][i];

du[v]--;
if (du[v] == 0 && v) que.push(v);

if (v == 0 || (i == 0 && u == 1)) continue;
if (i == 10) continue;
cnt[v] = (cnt[u] + cnt[v]) % mod;
sum[v] = (sum[v] + sum[u] * 10 % mod + cnt[u] * i % mod) % mod;
}
}
return ret;
}
}SA;

int N;
char str[maxn];

int main () {
while (scanf("%d", &N) == 1) {
SA.init();
while (N--) {
scanf("%s", str);
int n = strlen(str);
for (int i = 0; i < n; i++) SA.insert(str[i] - '0');
SA.insert(10);
}
printf("%d\n", SA.solve());
}
return 0;
}
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