poj kmp

专题训练ING

今天弄了一天的 KMP ，大且做个总结（poj kmp)。

poj 3167 .3080（另开博文）2752.1961.2406

1961.2406

求一个串能否被其一子串完全不重复覆盖

利用next数组的性质，自我匹配一遍即可，判断i是否整除i-next[i]

 1961

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

char s[1000011];
int next[1000011];
int n,i,t,j,x;
char c;

void Read()
{
    while(c=getchar(),c<'a'||c>'z');
    s[++n]=c;
    while(c=getchar(),c>='a'&&c<='z')s[++n]=c;
}

int main()
{
    while(true){
        scanf("%d",&n);
        t++;
        if(n==0)break;
        n=0;
        Read();
        x=0;
        for(i=2;i<=n;i++){
            while(x!=0&&s[i]!=s[x+1])x=next[x];
            if(s[i]==s[x+1])x++;
            next[i]=x;
        }
        printf("Test case #%d
",t);
        for(i=2;i<=n;i++){
            if(i%(i-next[i])==0&&i/(i-next[i])!=1)printf("%d %d
",i,i/(i-next[i]));
        }
        printf("
");
    }
}

2406

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

char s[1000011];
int next[1000011];
bool p;
int n,i,j,x;
char c;

void Read()
{
    while(c=getchar(),c!='.'&&(c<'a'||c>'z'));
    if(c=='.')p=false;    
    s[++n]=c;
    while(c=getchar(),c>='a'&&c<='z')s[++n]=c;
}

int main()
{
    while(true){
        p=true;
        n=0;
        Read();
        if(p==false)break;
        memset(next,0,sizeof(next));
        x=0;
        for(i=2;i<=n;i++){
            while(x!=0&&s[i]!=s[x+1])x=next[x];
            if(s[i]==s[x+1])x++;
            next[i]=x;
        }
        if(n%(n-next[n])!=0)printf("1
");
        else printf("%d
",n/(n-next[n]));
    }
}

2752

不断next..

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

char s1[400011],s[400011];
int next[400011];
int n,i,j,x,t;
int ans[400011];

int main()
{
    while(scanf("%s",&s)!=EOF){
        n=strlen(s);
        for(i=1;i<=n;i++)s1[i]=s[i-1];
        x=0;
        for(i=2;i<=n;i++){
            while(x!=0&&s1[i]!=s1[x+1])x=next[x];
            if(s1[i]==s1[x+1])x++;
            next[i]=x;
        }
        t=0;
        x=n;
        while(x!=0){
            ans[++t]=x;
            x=next[x];
        }
        for(i=t;i>=2;i--)printf("%d ",ans[i]);
        printf("%d
",ans[1]);
    }
}