2017 UESTC Training for Search Algorithm & String

2017 UESTC Training for Search Algorithm & String

A   next[]数组应用

题意：求一个字符串所有前缀出现的次数和。

tags：  dp[i-1] = dp[next[i]] + 1。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 1000005, mod = 10007;

int nex[N], n;
void getnext(char *B, int lenb)
{
    mes(nex, 0);    nex[0]=-1;
    for(int i=0,k=-1; i<lenb; )
    {
        if(k==-1 || B[i]==B[k]) nex[++i]=++k;
        else k=nex[k];
    }
}
char A[N];
ll  dp[N], ans;
int main()
{
    scanf("%d", &n);
    scanf("%s", A+1);
    getnext(A+1, n);
    rep(i,0,n) {
        if(nex[i]-1>=0) dp[i-1] += dp[nex[i]-1];
        ++dp[i-1];
        dp[i-1] %= mod;
        ans += dp[i-1],  ans %= mod;
    }
    --ans;
    printf("%lld
", (ans+mod)%mod);

    return 0;
}

D   字符串next[]数组

题意：求给出字符串的最短循环节。

tags：看了题解，原来next[]数组可以这样搞。。

思考KMP的next数组的定义：next[j] = i 代表 s[1...(i-1)] = s[j - i, j-1]。 并且i~j之间已经不可能有以j结尾的子串是前缀了，不然next[j]就不是 i 了。那么很显然，字符串的最短循环节长度为 len - next[len]。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<bitset>
#include<vector>
#include<set>
#include<list>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
typedef long long ll;
const int N = 1000005;

int nex[N];
void getnext(char *B, int lenb)
{
    mes(nex, 0);   nex[0]=-1;
    for(int i=0,k=-1; i<lenb; )
    {
        if(k==-1 || B[i]==B[k]) nex[++i]=++k;
        else k=nex[k];
    }
}
char A[N];
int main()
{
    scanf("%s", A);
    int len=strlen(A);
    getnext(A, len);
    printf("%d
", len-nex[len]);

    return 0;
}

I   dfs，九皇后，水题，但调了好久，还是不熟练啊

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

int n, sta[20], top;
bool vis1[20], vis2[20], vis3[20], vis4[20];
void dfs(int x, int y)
{
    sta[++top]=y;
    if(top==9)
    {
        rep(i,1,top) printf("%d ", sta[i]);
        puts("");
        --top;
        return ;
    }
    vis1[x]=vis2[y]=vis3[x+y]=vis4[y+n-x]=1;
    rep(i,x+1,n) rep(j,1,n)
        if(!vis1[i] && !vis2[j] && !vis3[i+j] && !vis4[j+n-i])
        {
            dfs(i, j);
        }
    --top;
    vis1[x]=vis2[y]=vis3[x+y]=vis4[y+n-x]=0;
}
int main()
{
    scanf("%d", &n);
    printf("352
");
    rep(j,1,n) dfs(1, j);

    return 0;
}