HDU 4436 str2int

　　把所有串用没出现过的符号相连，建立后缀自动机。记录每个节点所表示的子串对于此问题所求的sum值，再记录这个节点能收到多少个有效子串，于是就能从根结点往后递推sum值了。最后把所有有效节点的sum加起来就是答案。注意：（1）带有前缀零的子串无效；（2）带有分隔符的子串无效。

//2013-02-18 13:07:47Accepted4436250MS12624K1780 BG++Aros#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN = 110000+5;const int MAX_NODE = 220000+5, MAX_CHD = 11;const int MOD = 2012;int nv, chd[MAX_NODE][MAX_CHD], ml[MAX_NODE], fa[MAX_NODE];int N, r[MAX_NODE], sum[MAX_NODE], cnt[MAX_NODE];char s[MAXN];void Add(int u, int _ml, int _fa, int v = -1){ml[u] = _ml; fa[u] = _fa;if (v == -1)memset(chd[u], -1, sizeof(chd[u]));elsememcpy(chd[u], chd[v], sizeof(chd[v]));}void Construct(char *str){nv = 1; Add(0, 0, -1);int cur = 0;for (int i = 0; str[i]; i++){int c = str[i]-'0', p = cur;cur = nv++; Add(cur, i+1, -1);for (; p != -1 && chd[p][c] == -1; p = fa[p])chd[p][c] = cur;if (p == -1)fa[cur] = 0;else{int q = chd[p][c];if (ml[q] == ml[p]+1)fa[cur] = q;else{int r = nv++; Add(r, ml[q], fa[q], q);ml[r] = ml[p]+1; fa[q] = fa[cur] = r;for (; p != -1 && chd[p][c] == q; p = fa[p])chd[p][c] = r;}}}}bool cmp(const int &a, const int &b){return ml[a] < ml[b];}int main(){while (scanf("%d", &N) != EOF){for (int i = 0, n = 0; i < N; i++){if (i)s[n++] = '9'+1;scanf("%s", s+n);n += strlen(s+n);}Construct(s);for (int i = 0; i < nv; i++)r[i] = i;sort(r, r+nv, cmp);int ans = 0;fill(sum, sum+nv, 0);fill(cnt, cnt+nv, 0);cnt[0] = 1;for (int i = 0; i < nv; i++){int u = r[i];if (i && !cnt[u])continue;ans = (ans+sum[u])%MOD;for (int j = 0; j <= 9; j++) if (u+j && chd[u][j] != -1){int v = chd[u][j];sum[v] = (sum[v]+10*sum[u]+cnt[u]*j)%MOD;cnt[v] += cnt[u];}}printf("%d\n", ans);}return 0;}