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

来源：互联网发布：mac电脑中病毒编辑：程序博客网时间：2024/06/13 22:29

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

DEBUG了很久，还是legal的判断函数写错了...

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而言已经算比较快了

下面的是自己重新写的：

#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;}

下面的legal参考了队友的，，，其实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)//重新写下        {            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;}

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