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news 2025/7/24 5:26:26

P5357 【模板】AC 自动机 - 洛谷

主要是构建fail树

我们可以知道的是，当访问一个点x时，接下来需要跳转其fail[x]，以此类推，如果在某个fail[x]上出现了一个字符串，那么相应的统计次数应该加1，然后当访问到fail[x]时，又要继续访问fail[fail[x]]造成了一个点多次访问的情况

那么我们可以看出如果我们访问到了一个点x,那么接下来他的fail点都会继续访问到，那么我们其实可以构造一个fail树，当匹配主串的时候仅让siz[u]++,而不跳转他的fail[u],然后再遍历完主串的时候我们得到了每个点访问的次数，构建fail树，根据树的特性遍历求出哪些点访问过了多少次，这样就能把时间控制在O(S + T + n)内，S是模式串的总长度，T是主串长度，n是模式串个数

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 1e6 + 10, INF = 0x3f3f3f3f;
int n;
int cnt[N], ne[N], ch[N][30], idx;
string s[N], t;
int match[N]; // 统计第几个字符串出现在哪个idx上
int siz[N];  // 每个点访问过的次数
vector<int> vec[N];  // 构建fail树void insert(string x, int id)
{int p = 0;rep(0, x.size() - 1){int j = x[i] - 'a';if(!ch[p][j]) ch[p][j] = ++idx;p = ch[p][j];}cnt[p]++;match[id] = p;
}void built()
{queue<int> q;rep(0, 25)if(ch[0][i])q.push(ch[0][i]);while(!q.empty()){auto u = q.front();q.pop();for(int i = 0; i <= 25; i++){int v = ch[u][i];if(v) ne[v] = ch[ne[u]][i], q.push(v);elsech[u][i] = ch[ne[u]][i];}}
}void dfs(int u)
{for(auto it : vec[u]){dfs(it);siz[u] += siz[it];}
}void query(string x)
{for (int k = 0, i = 0; k < x.size(); k++){i = ch[i][x[k] - 'a'];++siz[i];}rep(1, idx)vec[ne[i]].pb(i);dfs(0);rep(1, n)cout << siz[match[i]] << '\n';
}void solve()
{cin >> n;rep(1, n){cin >> s[i];insert(s[i], i);}built();cin >> t;query(t);
}signed main()
{IOS;int T = 1;// cin >> T;while(T --)solve();return 0;
}

https://codeforces.com/problemset/problem/2109/B

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 2e5 + 10, INF = 0x3f3f3f3f;
int n, m, a, b;void solve()
{cin >> n >> m >> a >> b;if(n == 2 && m == 2){cout << 2 << '\n';return;}int t1 = ceil(log2(n)), t2 = ceil(log2(m));int c1 = ceil(log2(min(a, n - a + 1))), c2 = ceil(log2(min(b, m - b + 1)));cout << min(t1 + c2, c1 + t2) + 1 << '\n';
}signed main()
{IOS;int T = 1;cin >> T;while(T --)solve();return 0;
}

https://codeforces.com/problemset/problem/2108/B

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 2e5 + 10, INF = 0x3f3f3f3f;
int n, x;void solve()
{cin >> n >> x;int cnt = 0;int tx = x;if(n == 1 && x == 0){cout << -1 << '\n';return;}while(tx){if(tx & 1) cnt++;tx /= 2;}if(n <= cnt){cout << x << '\n';return;}else{if((n - cnt) % 2 == 0)cout << x + (n - cnt) << '\n';else{if(x > 1) cout << x + (n - cnt) + 1 << '\n';elsecout << n + 3 << '\n';}}
}signed main()
{IOS;int T = 1;cin >> T;while(T --)solve();return 0;
}

Contest Login

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 2e5 + 10, INF = 0x3f3f3f3f;
int n, x, y;int count(int a)
{int ans = 0;while(a){if(a & 1) ans ++;a /= 2;}return ans;
}void solve()
{cin >> n >> x >> y;if(x == y){cout <<  0 << '\n';return;}if(count(x) == count(y) || lowbit(x) == lowbit(y)){cout << 1 << "\n";return;}cout << 2 << '\n';
}signed main()
{IOS;int T = 1;cin >> T;while(T --)solve();return 0;
}

Contest Login

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 2e5 + 10, INF = 0x3f3f3f3f;void solve()
{int n;scanf("%d", &n);printf("%.4f\n", 1.0 * 3 * n / 2);
}signed main()
{// IOS;int T = 1;cin >> T;while(T --)solve();return 0;
}

二维偏序问题

Contest Login

/*
题目的意思就是对于一个数在a[]数组中的位置为i,在b[]数组中的位置为j，那么我们的结果就是(i - 1) + (j - 1) + 前面重复的部分
主要就是重复的部分怎么求呢？如果有重复的部分，那么是不是一个数x在a[]和b[]中的下标分别小于i, j。这就是一个典型的二维偏序问题。经典的做法就是树状数组
我们对于a[]原封不动，对于b[]数组，记录下b[]元素所在b[]数组中的位置。我们开始遍历a[]数组，得到a[]数组的(i - 1) + (j - 1)这个值。根据遍历a[]数组的顺序就可以知道，a[]数组后面的元素所要加的值肯定有a[]数组前面的元素，那么我们在把a[i]在b[]数组中出现的位置用BIT统计上，这样如果我们访问a[]的下一个元素的时候，如果这个元素在b[]中的位置在j之后，那么因为我们刚才更新了BIT就能直接通过BIT减去重复的，如果在j之前的话，那么就没有影响。。。以此类推
*/


#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int> TUP;const int N = 1e6 + 10, INF = 0x3f3f3f3f;
int n;
int a[N], b[N], p[N], ans[N];struct BIT
{int s[N];BIT(){init();}void init(){memset(s, 0, sizeof s);}void update(int poi, int val){for (int i = poi; i <= n; i += lowbit(i))s[i] += val;}int query(int poi){int ans = 0;for (int i = poi; i > 0; i -= lowbit(i))ans += s[i];return ans;}
};BIT bit;void solve()
{cin >> n;bit.init();rep(1, n) cin >> a[i];rep(1, n){cin >> b[i];p[b[i]] = i;}rep(1, n){int res = (i - 1) + (p[a[i]] - 1);res -= bit.query(p[a[i]]);// cout << ans << " \n"[i == n];ans[a[i]] = res;bit.update(p[a[i]], 1);}rep(1, n) cout << ans[i] << " \n"[i == n];
}signed main()
{IOS;int T = 1;cin >> T;while(T --)solve();return 0;
}

P2163 [SHOI2007] 园丁的烦恼 - 洛谷

核心思想就是因为我们统计的是每一个矩形框内的点的个数，那么难免会用到二维前缀和，但是有时候我们的二维前缀和的数组需要开太大题目不支持。我们在计算二维前缀和的时候会用到一个公式，那么我们就就可以根据扫描线的原理，先将所有的点按照x从大到小的顺序排序，然后将满足x条件的点全加进BIT中，但是加进来的是这个点的纵坐标，然后因为我们将访问一个矩阵变成了4个矩阵的而访问和，那么我们在对每一个访问矩阵的操作的时候就可以直接计算满足其y的点的个数，本质还是通过BIT的离线处理

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int, int> TUP;const int N = 1e7 + 5, INF = 0x3f3f3f3f;
int n, m;struct BIT
{int s[N];BIT(){init();}void init(){memset(s, 0, sizeof s);}void update(int poi, int val){for (int i = poi; i < N; i += lowbit(i))s[i] += val;}int query(int poi){int ans = 0;for (int i = poi; i > 0; i -= lowbit(i))ans += s[i];return ans;}
} bit;vector<PII> vec;
vector<TUP> qu;
int ans[(int)5e5 + 10];void solve()
{cin >> n >> m;rep(1, n){int a, b;cin >> a >> b;++a, ++b;vec.pb({a, b});}rep(1, m){int x1, x2, y1, y2;cin >> x1 >> y1 >> x2 >> y2;++x1, ++y1, ++x2, ++y2;qu.pb({x2, y2, 1, i});qu.pb({x1 - 1, y2, -1, i});qu.pb({x2, y1 - 1, -1, i});qu.pb({x1 - 1, y1 - 1, 1, i});}sort(vec.begin(), vec.end());sort(qu.begin(), qu.end());int cur = 0;for(auto [x, y, c, id] : qu){while(cur < n && vec[cur].first <= x)bit.update(vec[cur].second, 1), cur++;ans[id] += bit.query(y) * c;}rep(1, m) cout << ans[i] << '\n';
}signed main()
{IOS;int T = 1;// cin >> T;while(T --)solve();return 0;
}

P3755 [CQOI2017] 老C的任务 - 洛谷

加了离散化的，就是关键是一定要注意树状数组的大小问题。

#include<bits/stdc++.h>
using namespace std;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lowbit(x) (x & (-x))
#define int long long
#define rep(l, r) for(int i = l; i <= r; i++)
#define pb push_backtypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, double> PID;
typedef tuple<int, int, int, int> TUP;const int N = 1e5 + 10, INF = 0x3f3f3f3f;
int n, m;
int maxlen;
vector<array<int, 3>> vec;
vector<TUP> que;vector<int> vy;
map<int, int> mp;
int ans[N];struct BIT
{int s[N * 10];BIT(){init();}void init(){memset(s, 0, sizeof s);}void update(int poi, int val){for (int i = poi; i <= maxlen; i += lowbit(i))s[i] += val;}int query(int poi){int ans = 0;for (int i = poi; i > 0; i -= lowbit(i))ans += s[i];return ans;}
} bit;void solve()
{cin >> n >> m;rep(1, n){int a, b, c;cin >> a >> b >> c;vec.pb({a, b, c});vy.pb(b);}rep(1, m){int x1, x2, y1, y2;cin >> x1 >> y1 >> x2 >> y2;que.pb({x1 - 1, y1 - 1, 1, i});que.pb({x2, y2, 1, i});que.pb({x1 - 1, y2, -1, i});que.pb({x2, y1 - 1, -1, i});vy.pb(y2), vy.pb(y1 - 1);}sort(vec.begin(), vec.end());sort(vy.begin(), vy.end());sort(que.begin(), que.end());vy.erase(unique(vy.begin(), vy.end()), vy.end());maxlen = vy.size();rep(0, vy.size() - 1)mp[vy[i]] = i + 1;int cur = 0;for(auto [x, y, c, id] : que){while(cur < n && vec[cur][0] <= x){bit.update(mp[vec[cur][1]], vec[cur][2]);cur++;}ans[id] += c * bit.query(mp[y]);}rep(1, m) cout << ans[i] << '\n';
}signed main()
{IOS;int T = 1;// cin >> T;while(T --)solve();return 0;
}

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