2020牛客国庆集训派对day1

题目来源： ICPC Central Europe Regional Contest 2019

链接

A. AAB

找最长后缀回文串

思路：

KMP，正串和逆串匹配

#include<bits/stdc++.h>
using namespace std;

string s,p;
int n;
const int maxn = 4e5+10;
int nxt[maxn];
int main(){
    ios::sync_with_stdio(0);
    cin >> n;
    cin >> s;
    p = s;
    reverse(p.begin(),p.end());
    nxt[0] = -1;
	int j = 0, k = -1;
    int m = n;
	while (j < m) {
		if (k == -1 or s[j] == s[k]) nxt[++j] = ++k;
		else k = nxt[k];
	}
    int i = 0;
	j = 0;
    int ans = 1;
	while (i < n) {
		if (j == -1 or s[i] == p[j]) {
			i++; j++;
			if(i == n){
                ans = max(ans,j);
            }
		}
		else {
			j = nxt[j];
		}
	}
	cout << n - ans << endl;
    
}

也可以用马拉车，注意要开两倍空间

#include<bits/stdc++.h>
using namespace std;

string s,t;
int n;
const int maxn = 1e6+10;
int p[maxn];
int main(){
    ios::sync_with_stdio(0);
    cin >> n;
    cin >> s;
    t = "$";
    for(int i = 0;i < n;i++){
        t += "#" + s.substr(i,1);
    }
    t += "#@";
    n = t.length();
    int ans = 1;
    int c = 0,r = 0;
    for(int i = 1;i < n - 1;i++){
        p[i] = i < r ? min(p[2*c-i],r-i) : 1;
        while(t[i-p[i]] == t[i+p[i]])p[i]++;
        if(i+p[i] > r){
            r = p[i] + i;
            c = i;
        }
        if( (i - p[i] + 1) / 2 + p[i] - 1 == s.size()){
            ans = max(ans,p[i]-1);
        }
    }
    cout << s.size() - ans << endl;
}

C. Bob in Wonderland

拆链，思维题

队友的代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//typedef __int128 lll;
#define print(i) cout << "debug: " << i << endl
#define close() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mem(a, b) memset(a, b, sizeof(a))
const ll mod = 1e9 + 7;
const int maxn = 2e6;
const int inf = 0x3f3f3f3f;
int d[maxn];

int main()
{
    int n; cin >> n;
	for(int i = 1; i < n; i++)
	{
		int a, b; scanf("%d%d", &a, &b);
		d[a]++, d[b]++;
	}
	int res = 0;
	for(int i = 1; i <= n; i++)
		if(d[i] >= 3)
			res += (d[i] - 2);
	cout << res << endl;
}

D. Deep800080

计算几何

转换成线段的交，注意排序规则

不知道为什么用 %d 输入再强制类型转换过不去？？一直卡 98.94%

#include<bits/stdc++.h>
using namespace std;

const int N = 3e5 + 10;
int n;
double r, a, b;
const double pi = acos(-1.0);
const double eps = 1e-8;

int sgn(double x) {
	if (fabs(x) < eps)return 0;
	if (x > 0)return 1;
	else return -1;
}
struct Point {
	double x, y;
	int val;
	Point() {}
	Point(double _x, double _y) {
		x = _x;
		y = _y;
		val = 0;
	}
	double distance(Point p) {
		return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
	}
	bool operator <(const Point& b)const {
		//左下小
		if (sgn(x - b.x) == 0) {
			if (sgn(y - b.y) == 0) {
				return val > b.val;
			}
			return y < b.y;
		}
		return x < b.x;
	}
	Point operator +(const Point& b)const {
		return Point{ x + b.x, y + b.y };
	}
	Point operator *(const double& k)const {
		return Point{ x * k, y * k };

	}
	Point operator /(const double& k)const {
		return Point{ x / k, y / k };

	}
	Point operator -(const Point& b)const {
		return Point{ x - b.x,y - b.y };
	}
	double operator *(const Point& b)const {
		return x * b.x + y * b.y;
	}
	double len() {
		return hypot(x, y);
	}
	double len2() {
		return x * x + y * y;
	}
};

struct Line {
	Point s, e;
	Line() {}
	Line(Point _s, Point _e) {
		s = _s;
		e = _e;
	}
	Point lineProg(Point p) {
		return s + (((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));
	}
	double angle() {
		double k = atan2(e.y - s.y, e.x - s.x);
		if (sgn(k) < 0) k += pi;
		if (sgn(k - pi) == 0) k -= pi;
		return k;
	}
};
vector<Point>A;
int main() {
    //scanf("%d%d%d%d") ???? %d 输入再强转就wa????
	scanf("%d%lf%lf%lf", &n, &r, &a, &b);
	Line l = Line(Point(0, 0), Point(a, b));
	double angle = l.angle();
	for (int i = 0; i < n; i++) {
		double x, y;
		scanf("%lf%lf", &x, &y);
		Point u = l.lineProg(Point{ x,y });
		double d = u.distance(Point{ x,y });
		if (sgn(d - r) > 0)continue;
		double t = sqrt(r * r - d * d);
		Point a = Point{ u.x + cos(angle) * t,u.y + sin(angle) * t };
		Point b = Point{ u.x - cos(angle) * t,u.y - sin(angle) * t };
		if (a < b) { a.val = 1; b.val = -1; }

        else { a.val = -1; b.val = 1; }
		A.push_back(a);
		A.push_back(b);
	}
	sort(A.begin(), A.end());
	int ans = 0, sum = 0;
	for (Point x : A) {
		sum += x.val;
		ans = max(ans, sum);
	}
	printf("%d
", ans);
}

E. Zeldain Garden

(d(n)) 前缀和，不用筛，简单分块

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
ll solve(ll n){
    ll ans = 0;
    for(ll l = 1,r;l <= n;l = r + 1){
        r = n / (n / l);
        ans += (r-l+1)*(n/l);
    }
    return ans;
}
int main(){
    ll x,y;
    cin >> x >> y;
    cout << solve(y) - solve(x-1) << endl;
}

F. Light Emitting Hindenburg

二进制，简单模拟

题目的锅，是 (2^{30-L}) ，其实就是二进制表示

其实就是 (n) 个数里面挑 (m) 个数，使得他们相与（&）最大

从高位往低位挑就可以

#include<bits/stdc++.h>
using namespace std;


int n,k,x;
const int N = 2e5+10;
int bt[40][N];
int vis[40][N];

int main(){
    scanf("%d%d",&n,&k);
    for(int i = 1;i <= n;i++){
        scanf("%d",&x);
        for(int j = 0;j < 31;j++){
            if(x & (1 << j)){
                bt[j][i] = 1;
            }
        }
    }
    int ans = 0;
    int t = 1;
    for(int j = 30;j >= 0;j--){
        int cnt = 0;
        for(int i = n;i >= 1;i--){
            if(vis[j+1][i] == t - 1 and bt[j][i] == 1){
                cnt++;
            }
            vis[j][i] = vis[j+1][i];
        }
        if(cnt >= k){
            ans += (1<<j);
         
            for(int i = n;i >= 1;i--){
                if(vis[j+1][i] == t - 1 and bt[j][i] == 1){
                    vis[j][i] = t;
                }
            }
            t++;
        }
    }
    printf("%d
",ans);
}

G. K == S

AC自动机 ，矩阵快速幂

#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N = 7e5 + 10;
queue<int>q;
const int mod = 1e9 + 7;

struct Mat {
	int m[500][500], n;
	Mat(int _n,int v) {
		n = _n;
		memset(m, 0, sizeof m);
		for (int i = 0; i < n; i++)m[i][i] = v;
	}
	Mat operator *(const Mat& b)const {
		Mat res = Mat(b.n,0);
		int n = b.n;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				for (int k = 0; k < n; k++) {
					res.m[i][j] = (res.m[i][j] + m[i][k] * b.m[k][j]) % mod;
				}
			}
		}
		return res;
	}
};
struct {
	int c[N][26], fail[N], val[N], cnt;
	void insert(char* s) {
		int len = strlen(s); int now = 0;
		for (int i = 0; i < len; i++) {
			int v = s[i] - 'a';
			if (!c[now][v])c[now][v] = ++cnt;
			now = c[now][v];
		}
		//val[now]++;
        val[now] = 1;
	}
	void getFail() {
		for (int i = 0; i < 26; i++) {
			if (c[0][i])fail[c[0][i]] = 0, q.push(c[0][i]);
			//***
			else c[0][i] = 0;
		}
		
		
		while (!q.empty()) {
			int u = q.front(); q.pop();

			//***
			if (val[fail[u]] == 1) {
				val[u] = 1;
			}

			for (int i = 0; i < 26; i++) {
				if (c[u][i]) {
					fail[c[u][i]] = c[fail[u]][i];
					q.push(c[u][i]);
				}
				else c[u][i] = c[fail[u]][i];
			}
		}
	}
	int query(char* s) {
		int len = strlen(s); int now = 0, ans = 0;
		for (int i = 0; i < len; i++) {
			now = c[now][s[i] - 'a'];
			for (int t = now; t && val[t] != -1; t = fail[t]) {
				ans += val[t];
				val[t] = -1;
			}
		}
		return ans;
	}
	Mat getMat() {
        //这里 cnt 也能过，但是上面的POJ会wa，这里数据的问题，应该是cnt+1
		Mat res = Mat(cnt + 1, 0);
		for (int i = 0; i <= cnt; i++) {
			for (int j = 0; j < 26; j++) {
				if (!val[c[i][j]]) {
					res.m[i][c[i][j]]++;
				}
			}
		}
		return res;
	}

}Ac;

Mat qpow(Mat a, int p) {
	Mat res = Mat(a.n, 1);
	while (p) {
		if (p & 1) res = a * res;
		a = a * a;
		p >>= 1;
	}
	return res;
}

int n;
char p[N];

signed main() {
	int n, m, x;
	scanf("%lld%lld", &n, &m);
	for (int i = 0; i < m; i++) {
		scanf("%lld%s", &x, p);
		Ac.insert(p);
	}
	Ac.getFail();
	Mat mat = Ac.getMat();
	mat = qpow(mat, n);

	int ans = 0;
	for (int i = 0; i < mat.n; i++) {
		ans = (ans + mat.m[0][i]) % mod;
	}
	printf("%lld
", ans);
}

H. Ponk Warshall

拆环

这份dfs的代码MLE 了，就改成了暴力拆

/*
* Status： MLE
*/
#include<bits/stdc++.h>
using namespace std;

map<char, int>mp;
int G[5][5];
int out[5];
const int N = 1e6+10;
char s[N],t[N];
int ans;

void dfs(int u, int len, int target) {
	if (u == target and len != -1) {
		ans += len;
		return;
	}
	for (int i = 1; i <= 4; i++) {
		if (G[u][i] > 0) {
			dfs(i, len + 1, target);
			G[u][i]--;
			out[u]--;
		}
	}
}
int main() {
	mp['A'] = 1; mp['C'] = 2;
	mp['G'] = 3; mp['T'] = 4;

	scanf("%s",s);
    scanf("%s",t);
	int n = strlen(s);
	for (int i = 0; i < n; i++) {
		G[mp[s[i]]][mp[t[i]]]++;
		out[mp[s[i]]]++;
	}

	for (int i = 1; i <= 4; i++) {
		while (out[i] != 0) {
			dfs(i, -1, i);
		}
	}
	printf("%d
",ans);
}

#include<bits/stdc++.h>
using namespace std;

map<char, int>mp;
int G[5][5];
int out[5];
const int N = 1e6 + 10;
char s[N], t[N];
int ans;

int main() {
	mp['A'] = 1; mp['C'] = 2;
	mp['G'] = 3; mp['T'] = 4;

	scanf("%s", s);
	scanf("%s", t);
	int n = strlen(s);
	for (int i = 0; i < n; i++) {
		G[mp[s[i]]][mp[t[i]]]++;
	}
	for (int i = 1; i <= 4; i++) {
		while (G[i][i])G[i][i]--;
	}

	for (int i = 1; i <= 4; i++) {
		for (int j = i + 1; j <= 4; j++) {
			while (G[i][j] and G[j][i]) {
				G[i][j]--; G[j][i]--;
				ans++;
			}
		}
	}

	for (int i = 1; i <= 4; i++) {
		for (int j = 1; j <= 4; j++) {
			for (int k = 1; k <= 4; k++) {
				if (i == j or j == k or k == i)continue;
				while (G[i][j] and G[j][k] and G[k][i]) {
					G[i][j]--; G[j][k]--; G[k][i]--;
					ans += 2;
				}
			}
		}
	}
	for (int i = 1; i <= 4; i++) {
		for (int j = 1; j <= 4; j++) {
			if (i == j)continue;
			for (int k = 1; k <= 4; k++) {
				if (k == i or k == j)continue;
				for (int u = 1; u <= 4; u++) {
					if (u == i or u == j or u == k)continue;
					while (G[i][j] and G[j][k] and G[k][u] and G[u][i]) {
						G[i][j]--; G[j][k]--; G[k][u]--; G[u][i]--;
						ans += 3;
					}
				}
			}
		}
	}
	printf("%d
", ans);
}

J. The Bugs

稍微复杂一点的模拟+RMQ