标签:字符数组 char message 题意 cpp turn long iostream ini
题意:给定两个字符串A 和B。求最长公共子串。
思路:将两个字符串连接起来中间用一个没出现过的符号切割。
所以答案为满足后缀在不同的串中且height值最大的height值
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#define eps 1e-6
#define LL long long
using namespace std;
const int maxn = 200000 + 200;
//const int INF = 0x3f3f3f3f;
char str1[100010], str2[100010];
struct SuffixArray {
int s[maxn]; /// 原始字符数组(最后一个字符应必须是0,而前面的字符必须非0)
int sa[maxn]; // 后缀数组,sa[0]一定是n-1,即最后一个字符
int rank[maxn]; // 名次数组
int height[maxn]; // height数组
int t[maxn], t2[maxn], c[maxn]; // 辅助数组
int n; // 字符个数
void clear() { n = 0; memset(sa, 0, sizeof(sa)); }
/// m为最大字符值加1。!!! 调用之前需设置好s和n
void build_sa(int m) {
int i, *x = t, *y = t2;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[i] = s[i]]++;
for(i = 1; i < m; i++) c[i] += c[i-1];
for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
for(int k = 1; k <= n; k <<= 1) {
int p = 0;
for(i = n-k; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[y[i]]]++;
for(i = 0; i < m; i++) c[i] += c[i-1];
for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1; x[sa[0]] = 0;
for(i = 1; i < n; i++)
x[sa[i]] = y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k] ? p-1 : p++;
if(p >= n) break;
m = p;
}
}
void build_height() {
int i, j, k = 0;
for(i = 0; i < n; i++) rank[sa[i]] = i;
for(i = 0; i < n; i++) {
if(k) k--;
j = sa[rank[i]-1];
while(s[i+k] == s[j+k]) k++;
height[rank[i]] = k;
}
}
} sa;
void init() {
sa.clear();
int len1 = strlen(str1), len2 = strlen(str2);
for(int i = 0; i < len1; i++) sa.s[i] = str1[i] - ‘a‘ + 1;
for(int i = 0; i < len2; i++) sa.s[len1+i+1] = str2[i] - ‘a‘ + 1;
sa.s[len1] = 27; sa.s[len1+len2+1] = 0;
sa.n = len1 + len2 + 2;
sa.build_sa(30);
sa.build_height();
}
void solve() {
int ans = 0;
int len1 = strlen(str1);
for(int i = 1; i < sa.n; i++) {
if(sa.height[i] > ans)
if(sa.sa[i-1]<len1&&sa.sa[i]>len1 || sa.sa[i]<len1&&sa.sa[i-1]>len1)
ans = sa.height[i];
}
cout << ans << endl;
}
int main() {
//freopen("input.txt", "r", stdin);
while(scanf("%s%s", str1, str2) == 2) {
init();
solve();
}
return 0;
}
poj 2774 Long Long Message(后缀数组)
标签:字符数组 char message 题意 cpp turn long iostream ini
原文地址:http://www.cnblogs.com/jzdwajue/p/6943949.html