好久没AC题目了..

普通的题做不出来,只好回去做点模版题什么的

模版题还能WA一次自己还是太弱了..

求后缀数组代码是O(n log^2n)的..

void dfs(int t,int n)
{
    if (t>n) return;
    int i;
    for (i=0;i<n;i++)
    {
        if ((i==0)||(b[i-1]<b[i]))
        {
            c[b[i].id]=i;
        }
        else
        {
            c[b[i].id]=c[b[i-1].id];
        }
    }
    for (i=0;i<n;i++)
    {
        b[i].x=c[b[i].id];
    }
    for (i=0;i<n;i++)
    {
        if (b[i].id+t<n)
        {
            b[i].y=c[b[i].id+t];
        }
        else
        {
            b[i].y=-1;
        }
    }
    sort(b,b+n);
    dfs(t*2,n);
}

代码:

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<time.h>
#include<math.h>
#include<memory>
#include<vector>
#include<bitset>
#include<fstream>
#include<stdio.h>
#include<utility>
#include<sstream>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
    int x;
    int y;
    int id;
    friend bool operator < (const node &a,const node &b)
    {
        if (a.x!=b.x) return a.x<b.x;
        return a.y<b.y;
    }
};
node b[200005];
char a[200005];
int c[200005];
void dfs(int t,int n)
{
    if (t>n) return;
    int i;
    for (i=0;i<n;i++)
    {
        if ((i==0)||(b[i-1]<b[i]))
        {
            c[b[i].id]=i;
        }
        else
        {
            c[b[i].id]=c[b[i-1].id];
        }
    }
    for (i=0;i<n;i++)
    {
        b[i].x=c[b[i].id];
    }
    for (i=0;i<n;i++)
    {
        if (b[i].id+t<n)
        {
            b[i].y=c[b[i].id+t];
        }
        else
        {
            b[i].y=-1;
        }
    }
    sort(b,b+n);
    dfs(t*2,n);
}
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    scanf("%s",a);
    int n=strlen(a);
    int i;
    for (i=0;i<n;i++)
    {
        a[i+n]=a[i];
    }
    for (i=0;i<2*n;i++)
    {
        b[i].x=a[i];
        b[i].y=-1;
        b[i].id=i;
    }
    sort(b,b+2*n);
    dfs(1,2*n);
    for (i=0;i<2*n;i++)
    {
        if (b[i].id<n)
        {
            printf("%c",a[b[i].id+n-1]);
        }
    }
    printf("\n");
    return 0;
}

卡常版:

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<time.h>
#include<math.h>
#include<memory>
#include<vector>
#include<bitset>
#include<fstream>
#include<stdio.h>
#include<utility>
#include<sstream>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
    int x;
    int y;
    int id;
    friend bool operator < (const node &a,const node &b)
    {
        //if (a.x!=b.x) return a.x<b.x;
        return a.y<b.y;
    }
};
bool spj(const node &a,const node &b)
{
    return a.x<b.x;
}
node b[200005];
char a[200005];
int c[200005];
void dfs(int t,int n)
{
    if (t>n) return;
    int i;
    for (i=0;i<n;i++)
    {
        if ((i==0)||(b[i-1].x<b[i].x)||((b[i-1].x==b[i-1].x)&&(b[i-1].y<b[i].y)))
        {
            c[b[i].id]=i;
        }
        else
        {
            c[b[i].id]=c[b[i-1].id];
        }
    }
    for (i=0;i<n;i++)
    {
        b[i].x=c[b[i].id];
    }
    for (i=0;i<n;i++)
    {
        if (b[i].id+t<n)
        {
            b[i].y=c[b[i].id+t];
        }
        else
        {
            b[i].y=-1;    
        }
    }
    int last=0;
    for (i=1;i<=n;i++)
    {
        if ((i==n)||(b[i-1].x<b[i].x))
        {
            sort(b+last,b+i);
            last=i;
        }
    }
    dfs(t*2,n);
}
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    scanf("%s",a);
    int n=strlen(a);
    int i;
    for (i=0;i<n;i++)
    {
        a[i+n]=a[i];
    }
    for (i=0;i<2*n;i++)
    {
        b[i].x=a[i];
        b[i].y=-1;
        b[i].id=i;
    }
    sort(b,b+2*n,spj);
    dfs(1,2*n);
    for (i=0;i<2*n;i++)
    {
        if (b[i].id<n)
        {
            printf("%c",a[b[i].id+n-1]);
        }
    }
    printf("\n");
    return 0;
}