LightOj 1334 - Genes in DNA

来源：互联网发布：最优化方法好过吗编辑：程序博客网时间：2024/06/08 08:55

很好的KMP题目，正反两次KMP (利用DFA的知识概念，统计每个节点的失败节点到根节点的的距离,即代码中的dis数组) ,接下来的问题就迎刃而解了

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<cstring>#include<vector>#include<bitset>#include<iterator>#include<list>#include<map>#include<cstdlib>#include<ctime>#include<utility>using namespace std;#define L p->ch[0]#define R p->ch[1]#define KeyTree root->ch[1]->ch[0]typedef unsigned long long LL;typedef pair<int, int> PII;typedef vector<int> VI;const int maxn = 1111111;const int INF = 1 << 30;const int mod = (int) (1e8);const LL X = 123;template<typename T>T gcd(T a, T b) {return b ? gcd(b, a % b) : a;}int Next[maxn];int dis[maxn];char s[maxn], t[maxn];int a[maxn], b[maxn];void getnext(char s[], int n) {int i = 0, j;Next[0] = j = -1;dis[0] = 0;while (i < n) {if (j == -1 || s[i] == s[j]) {Next[++i] = ++j;dis[i] = dis[j] + 1;} else {j = Next[j];}}}int KMP(char s[], int n, char t[], int m, int c[]) {int i = 0, j = 0, ans = 0;while (i < n) {if (j == -1 || s[i] == t[j]) {++i, ++j;c[i - 1] = dis[j] - (j == m);} else {j = Next[j];}if (j == m) {++ans;j = Next[j];}}return ans;}int main() {//ios::sync_with_stdio(false);int T, cas;scanf("%d", &T);for (cas = 1; cas <= T; ++cas) {scanf("%s%s", s, t);int n = strlen(s);int m = strlen(t);getnext(t, m);KMP(s, n, t, m, a);reverse(t, t + m);reverse(s, s + n);getnext(t, m);KMP(s, n, t, m, b);LL ans = 0;reverse(b, b + n);ans = 0;a[n] = b[n] = 0;for (int i = 0; i < n; ++i) {ans = ans + 1LL * a[i] * b[i + 1];}printf("Case %d: %lld\n", cas, ans);}return 0;}