BNU 34990 Justice String 2014 ACM-ICPC Beijing Invitational Programming Contest

题目链接:http://acm.bnu.edu.cn/bnuoj/problem_show.php?pid=34990


DEBUG了非常久,还是legal的推断函数写错了...

此题做法。枚举String1的起始位置,对string2的长度进行二分。求出最长公共前缀,然后跳过一个不匹配的地方,然后继续二分匹配,再去掉一个不匹配的地方

//700-800MS   对于hash而言已经算比較快了

以下的是自己又一次写的:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cmath>
#include <map>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdin)

const ull B=31;
const int MAXN = 100000+100;
char a[MAXN],b[MAXN];
ull ah[MAXN],bh[MAXN],base[MAXN];

int n,m;

int Find(int i, int j)
{
    int up=m+1-j,down=0,mid;//二分的是长度////
    ull tmpa,tmpb;
    while(up>down+1)
    {
        mid=(up+down)/2;
        tmpa=(i==0)?

ah[i+mid-1]:ah[i+mid-1]-ah[i-1]*base[mid];/// tmpb=(j==0)?bh[j+mid-1]:bh[j+mid-1]-bh[j-1]*base[mid];/// if(tmpa == tmpb)down=mid; else up=mid; } return down; } int legal(int st) { int prelen=0,j=0,use=0; for(int i=st;;) { prelen=Find(i,j); i+=prelen+1; j+=prelen+1; use++; if(j>=m)return 1; if(use == 2) { if(j>=m)return 1; if(j+Find(i,j)>=m)return 1; return 0; } if(i>=n && j<m)return 0; } } int solve() { ah[0]=a[0],bh[0]=b[0],a[n+1]=0,b[m+1]=0; for(int i=1;i<=m;i++) bh[i]=bh[i-1]*B+b[i]; for(int i=1;i<=n;i++) ah[i]=ah[i-1]*B+a[i]; for(int i=0;i<=n-m;i++) { if(legal(i))return i; } return -1; } int main() { //IN("BNUhash.txt"); int ncase; scanf("%d",&ncase); base[0]=1; rep(i,1,MAXN) base[i]=base[i-1]*B; for(int ic=1;ic<=ncase;ic++) { scanf("%s%s",a,b); n=strlen(a); m=strlen(b); printf("Case #%d: %d\n",ic,solve()); } return 0; }



 以下的legal參考了队友的,。,事实上不该看人家代码太多啊。自己写思路更清晰,

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cmath>
#include <map>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdin)

const ull B=31;
const int MAXN = 100000+100;
char a[MAXN],b[MAXN];
ull ah[MAXN],bh[MAXN],base[MAXN];

int n,m;

int Find(int i, int j)
{
    int up=m+1-j,down=0,mid;//二分的是长度////
    ull tmpa,tmpb;
    while(up>down+1)
    {
        mid=(up+down)/2;
        tmpa=(i==0)?ah[i+mid-1]:ah[i+mid-1]-ah[i-1]*base[mid];///
        tmpb=(j==0)?

bh[j+mid-1]:bh[j+mid-1]-bh[j-1]*base[mid];/// if(tmpa == tmpb)down=mid; else up=mid; } return down; } int legal(int st) { int prelen=0,j=0,use=0; for(int i=st;i<n && use<2 && j<m-1;i++,j++)//i<=n? { prelen=Find(i,j); i+=prelen;// j+=prelen;// use++;//记录二分的次数 if(use>=2 && j<m-1)//又一次写下 { prelen=Find(i+1,j+1); j+=prelen; // if(j>=m-1)return 1; // else return 0; } } return 1;////// } int solve() { ah[0]=a[0],bh[0]=b[0],a[n+1]=0,b[m+1]=0; for(int i=1;i<=m;i++) bh[i]=bh[i-1]*B+b[i]; for(int i=1;i<=n;i++) ah[i]=ah[i-1]*B+a[i]; for(int i=0;i<=n-m;i++) { if(legal(i))return i; } return -1; } int main() { //IN("BNUhash.txt"); int ncase; scanf("%d",&ncase); base[0]=1; rep(i,1,MAXN) base[i]=base[i-1]*B; for(int ic=1;ic<=ncase;ic++) { scanf("%s%s",a,b); n=strlen(a); m=strlen(b); printf("Case #%d: %d\n",ic,solve()); } return 0; }



posted on 2017-05-22 10:53  blfbuaa  阅读(134)  评论(0编辑  收藏  举报