Number Sequence HDU - 1711（KMP匹配）
【摘要】
题目：
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a...
题目：
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.
2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1
6 -1
代码：
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#include<iostream>
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#include<string.h>
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using namespace std;
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int listn[1000001]; //长串
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int listm[10001]; //短串
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int next_[10001];
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void getnext(int *t,int m, int *next) //m是短串t的长度
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{
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int i = 1, j = 0;
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next[1] = 0;
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while (i < m)
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{
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if (j == 0 || t[i] == t[j])
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{
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i++;
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j++;
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if (t[i] != t[j])next[i] = j;
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else next[i] = next[j];
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}
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else j = next[j];
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}
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}
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int main()
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{
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ios_base::sync_with_stdio(false); //减小输入输出的时间
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int t, n, m;
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cin >> t;
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while (t--)
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{
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cin >> n >> m;
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for (int i = 1; i <= n; i++)cin >> listn[i];
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for (int i = 1; i <= m; i++)cin >> listm[i];
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listm[0] = 2000000;
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next_[0] = 0;
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next_[1] = 0;
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int i,j;
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getnext(listm, m, next_);
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i = 1;
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j = 1;
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while (i <= n)
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{
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if (listn[i] == listm[j])
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{
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i++;
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j++;
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if (j > m)break;
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}
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else
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{
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if (next_[j])j = next_[j];
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else
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{
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i++;
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j = 1;
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}
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}
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}
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if (j > m)cout << i - m;
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else cout << -1;
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cout << endl;
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}
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return 0;
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}
文章来源: blog.csdn.net，作者：csuzhucong，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/nameofcsdn/article/details/79691556
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