• [JSOI2012]玄武密码


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    这道题就是一个SAM的水题(所以这也是一篇很水的博客)。
    我们建完SAM后,把每一个串放上去跑就行了。因为题目要求前缀匹配,所以一旦失配,就直接返回了。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<vector>
    #include<queue>
    #include<assert.h>
    #include<ctime>
    using namespace std;
    #define enter puts("") 
    #define space putchar(' ')
    #define Mem(a, x) memset(a, x, sizeof(a))
    #define In inline
    #define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
    typedef long long ll;
    typedef double db;
    const int INF = 0x3f3f3f3f;
    const db eps = 1e-8;
    const int maxn = 1e7 + 5;
    const int maxs = 4;
    In ll read()
    {
    	ll ans = 0;
    	char ch = getchar(), las = ' ';
    	while(!isdigit(ch)) las = ch, ch = getchar();
    	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
    	if(las == '-') ans = -ans;
    	return ans;
    }
    In void write(ll x)
    {
    	if(x < 0) x = -x, putchar('-');
    	if(x >= 10) write(x / 10);
    	putchar(x % 10 + '0');
    }
    In void MYFILE()
    {
    #ifndef mrclr
    	freopen(".in", "r", stdin);
    	freopen(".out", "w", stdout);
    #endif
    }
    
    int n, m, H[105];
    char s[maxn];
    struct Sam
    {
    	int tra[maxn << 1][maxs], len[maxn << 1], link[maxn << 1], cnt, las;
    	In void init() {link[cnt = las = 0] = -1;}
    	In void insert(int c)
    	{
    		int now = ++cnt, p = las; Mem(tra[now], 0);
    		len[now] = len[las] + 1;
    		while(~p && !tra[p][c]) tra[p][c] = now, p = link[p];
    		if(p == -1) link[now] = 0;
    		else
    		{
    			int q = tra[p][c];
    			if(len[q] == len[p] + 1) link[now] = q;
    			else
    			{
    				int clo = ++cnt; 
    				len[clo] = len[p] + 1;
    				memcpy(tra[clo], tra[q], sizeof(tra[q]));
    				link[clo] = link[q], link[q] = link[now] = clo;
    				while(~p && tra[p][c] == q) tra[p][c] = clo, p = link[p];
    			}
    		}
    		las = now;
    	}
    	In int solve(char* s)
    	{
    		int ret = 0, n = strlen(s);
    		for(int i = 0, p = 0, l = 0; i < n && tra[p][H[s[i]]]; ++i) ++ret, p = tra[p][H[s[i]]];
    		return ret;
    	}
    }S;
    
    int main()
    {
    //	MYFILE();
    	n = read(), m = read();
    	scanf("%s", s);
    	H['E'] = 0, H['W'] = 1, H['S'] = 2, H['N'] = 3;
    	S.init();
    	for(int i = 0; i < n; ++i) S.insert(H[s[i]]); 
    	for(int i = 1; i <= m; ++i)
    	{
    		scanf("%s", s);
    		write(S.solve(s)), enter;
    	}
    	return 0;	
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/13781419.html
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