标签：
两个题目意思差不多，都是让求最长公共子串，只不过poj那个让输出长度，而URAL那个让输出一个任意的最长的子串。
解体思路：
| Time Limit: 4000MS | Memory Limit: 131072K | |
| Total Submissions: 22313 | Accepted: 9145 | |
| Case Time Limit: 1000MS | ||
Description
Input
Output
Sample Input
yeshowmuchiloveyoumydearmotherreallyicannotbelieveit yeaphowmuchiloveyoumydearmother
Sample Output
27
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 200050;
int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];
int cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int *r, int *sa, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;
for(j = 1, p = 1; p < n; j <<= 1, m = p)
{
for(p = 0, i = n-j; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++)
if(sa[i] >= j) y[p++] = sa[i]-j;
for(i = 0; i < n; i++) wv[i] = x[y[i]];
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[wv[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
}
}
int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(int i = 0; i < n; height[rank[i++]] = k)
for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
return ;
}
int dp[maxn][30];
void RMQ(int len)
{
for(int i = 1; i <= len; i++)
dp[i][0] = height[i];
for(int j = 1; 1<<j <= maxn; j++)
{
for(int i = 1; i+(1<<j)-1 <= len; i++)
dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
}
}
int lg[maxn];
int querry(int l, int r)
{
int k = lg[r-l+1];
return min(dp[l][k], dp[r-(1<<k)+1][k]);
}
void init()
{
lg[0] = -1;
for (int i = 1; i < maxn; ++i)
lg[i] = lg[i>>1] + 1;
}
int seq[2*maxn];
char str1[maxn], str2[maxn];
void Del(int n, int len1, int len2)
{
int xp = 0;
for(int i = 2; i <= n; i++)
{
if(xp < height[i])
{
int fx = sa[i-1];
int fy = sa[i];
int xx = max(fx, fy);
int yy = min(fx, fy);
if(xx > len1 && yy < len1) xp = height[i];
}
}
printf("%d\n", xp);
}
int main()
{
///init();
while(~scanf("%s %s",str1, str2))
{
int len1 = strlen(str1);
int len2 = strlen(str2);
int n = 0;
for(int i = 0; i < len1; i++) seq[n++] = str1[i];
seq[n++] = 1;
for(int i = 0; i < len2; i++) seq[n++] = str2[i];
seq[n] = 0;
da(seq, sa, n+1, 130);
calheight(seq, sa, n);
Del(n, len1, len2);
}
return 0;
}
POJ 2774 Long Long Message && URAL 1517. Freedom of Choice(求最长重复子序列)
标签：
原文地址：http://blog.csdn.net/xu12110501127/article/details/43161985
