牛客网暑期ACM多校训练营（第一场）菜鸟补题QAQ

　　签到题 J Different Integers（树状数组）

　　题目大意：给一个长为n的数组，每一个询问给两个数字i, j ，询问1~i, j~n这两个区间中有多少不同的数字，真的像是莫队裸题，但是两个区间是分隔的，所以可以考虑将数组延长一倍，即num[i] = num[i+n]，这样就可以变成一段区间中查询了，不过这样会T.....参考了题解的做法还是比较好用的,写法非常巧妙。记录数字i第一次和最后一次出现的位置分别为first[i], last[i]，按每个询问的rhs从小到大排序后，遍历num数组。遍历序号为i，每次遍历处理询问和维护一个树状数组，①只有当某个询问的 rhs == i 时处理该询问②维护树状数组，首先我们可以想到假如某个数x，当last[x] == i 时，下个遍历时下个询问的rhs必然会大于这个last[x]（因为是从1~n遍历），所以当这次遍历过后，剩下的询问rhs全部大于last[x]，所有右边区间rhs~n已经没有x了，这个时候只要看前面lhs是不是小于first[x]就知道这个数在不在询问区间了，这个时候就可以维护一个前缀和，sum[i]表示表示有多少个数的first值是在1~i内的。

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
#include <time.h>
 
#define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#define  lowbit(x) (x&-x)
#pragma warning ( disable : 4996 )
 
using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b)      { return a>b?a:b; }
inline LL LMin(LL a,LL b)      { return a>b?b:a; }
inline int Max(int a,int b)    { return a>b?a:b; }
inline int Min(int a,int b)    { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==0?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod  = 1000000007;
const double eps = 1e-8;
const int inf  = 0x3f3f3f3f;
const int maxk = 1e4+5;
const int maxn = 1e5+10;
 
int n, q, tot;
int first[maxn], last[maxn];
int num[maxn], ans[maxn], _count[maxn];
 
struct node {
    int lhs, rhs, id;
}p[maxn];
 
void add( int x, int d )
{
    while ( x > 0 )
    {
        _count[x] += d;
        x -= lowbit(x);
    }
}
 
int sum(int x)
{
    int s = 0;
    while ( x <= n )
    {
        s += _count[x];
        x += lowbit(x);
    }
    return s;
}
 
bool cmp(const node& a, const node &b)
{ return a.rhs < b.rhs; }
 
void init()
{
    tot = 0;
    memset(first, -1, sizeof(first));
    memset(last, -1, sizeof(last));
    memset(_count, 0, sizeof(_count));
     
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &num[i]);
        if (first[num[i]] == -1)
            { tot++; first[num[i]] = i;}
        last[num[i]] = i;
    }
 
    for (int i = 1; i <= q; i++)
    {
        scanf("%d %d", &p[i].lhs, &p[i].rhs);
        p[i].id = i;
    }
    sort(p+1, p+1+q, cmp);
}
 
 
int main()
{
    while (~scanf("%d %d", &n, &q))
    {
        init();
 
        for (int i = 1, k = 1; i <= n; i++)
        {   //如果有边界到达了某一点i，则可以开始处理左边界
            while (k <= q && p[k].rhs == i)
            {
                ans[p[k].id] = tot - sum(p[k].lhs);
                k++;
            }
 
            if (last[num[i]] == i)
                add(first[num[i]]-1, 1);
        }
 
        for ( int i = 1; i <= q; i++ )
            printf("%d
", ans[i]);
    }
 
    return 0;
}

　　

　　签到题？ D Two Graphs 

　　题目大意：给两个简单图G1(V, E1), G2(V,E2) ,问有多少个G2的子图与G1是同构的，想了好多骚操作，，结果hash去重就可以了

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
#include <set>
#include <string>
#include <algorithm>
#include <time.h>
 
#define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#define  lowbit(x) (x&-x)
#pragma warning ( disable : 4996 )
 
using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b)      { return a>b?a:b; }
inline LL LMin(LL a,LL b)      { return a>b?b:a; }
inline int Max(int a,int b)    { return a>b?a:b; }
inline int Min(int a,int b)    { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==0?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod  = 1000000007;
const double eps = 1e-8;
const int inf  = 0x3f3f3f3f;
const int maxk = 1e4+5;
const int maxn = 1e5+10;
 
struct edge2 {
    int x, y;
}edge[100];
int v, e1, e2;
int g1[10][10], g2[10][10], num[10];
 
void init()
{
    memset(g1, 0, sizeof(g1));
    memset(g2, 0, sizeof(g2));
    for ( int i = 1; i <= v; i++ )
        num[i] = i;
 
    int x, y;
    for ( int i = 1; i <= e1; i++ )
        { scanf("%d %d", &x, &y); g1[x][y] = 1; g1[y][x] = 1; }
    for ( int i = 1; i <= e2; i++ )
    {
        scanf("%d %d", &x, &y);  g2[x][y] = 1; g2[y][x] = 1;
        edge[i].x = x;
        edge[i].y = y;
    }
 
}
 
int main()
{
    while (~scanf("%d %d %d", &v, &e1, &e2))
    {
        init();
 
 
        set<long long> ss;
        do {
            int cnt = 0;
            long long has = 0;
            for (int i = 1; i <= e2; i++)
            {
                if (g1[num[edge[i].x]][num[edge[i].y]])
                {
                    has = has*177 + i;
                    has %= mod;
                    cnt++;
                }
            }
 
            if (cnt == e1)
                ss.insert(has);
 
        }while(next_permutation(num+1, num+1+v));
         
        printf("%zd
", ss.size());
    }
 
    return 0;
}