【AtCoder】AISing Programming Contest 2019

本来以为是1199rated的。。仔细一看发现是1999，所以就做了一下
这场涨分很轻松啊。。。为啥又没打
等pkuwc考完我一定打一场atcoder（咕咕咕，咕咕咕，咕咕咕咕咕咕咕~）
但是其实我思维速度上真的有点不行。。。

A - Bulletin Board

输出((N - W + 1)(N - H + 1))

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N,H,W;
void Solve() {
	read(N);read(H);read(W);
	out(1LL * (N - H + 1) * (N - W + 1));enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

B - Contests

统计这三个区间的个数，输出最小值

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;
int P[105],A,B;
int cnt[3];
void Solve() {
	read(N);read(A);read(B);
	for(int i = 1 ; i <= N ; ++i) {
		read(P[i]);
		if(P[i] <= A) ++cnt[0];
		else if(P[i] <= B) ++cnt[1];
		else cnt[2]++;
	}
	out(min(min(cnt[0],cnt[1]),cnt[2]));enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

C - Alternating Path

相邻的黑白格子之间有边，答案是每个联通块里的黑白点个数乘积之和

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int H,W,a[405][405],cnt[2];
char s[405][405];
bool vis[405][405];
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
bool in_range(int x,int y) {
	return x >= 1 && x <= H && y >= 1 && y <= W;
}
void dfs(int x,int y) {
	cnt[a[x][y]]++;vis[x][y] = 1;
	for(int k = 0 ; k < 4 ; ++k) {
		int tx = x + dx[k],ty = y + dy[k];
		if(in_range(tx,ty)) {
			if(!vis[tx][ty] && a[tx][ty] != a[x][y]) {
				dfs(tx,ty);
			}
		}
	}
}
void Solve() {
	read(H);read(W);
	for(int i = 1 ; i <= H ; i++) {
		scanf("%s",s[i] + 1);
		for(int j = 1 ; j <= W ; ++j) {
			if(s[i][j] == '#') a[i][j] = 1;
			else a[i][j] = 0;
		}
	}
	int64 ans = 0;
	for(int i = 1 ; i <= H ; ++i) {
		for(int j = 1 ; j <= W ; ++j) {
			if(!vis[i][j]) {
				cnt[1] = 0;cnt[0] = 0;
				dfs(i,j);
				ans += 1LL * cnt[1] * cnt[0];
			}
		}
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

D - Nearest Card Game

先二分一个k，使得([x - k,x + k])区间里的数的个数小于等于((x + k,infty))里的数，且(k)最大，先把([x - k,x + k])这些区间里这么多数两个人分别取走
然后从这个状态暴力模拟（大于x的数不会超过两个了），使得所有的数都小于x，剩下的就是交替一个一个取了，可以用一个前缀和预处理出来

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N,Q;
int64 A[MAXN],sum[MAXN],b[MAXN];
pii Calc(int64 l,int64 r) {
	int a = lower_bound(A + 1,A + N + 1,l) - A;
	int b = upper_bound(A + 1,A + N + 1,r) - A - 1;
	return mp(a,b);
}
int64 Process(int64 x) {
	if(x >= A[N]) return sum[N];
	int64 L = 0,R = A[N];
	int p = lower_bound(A + 1,A + N + 1,x) - A;
	while(L < R) {
		int64 mid = (L + R + 1) >> 1;
		pii t = Calc(x - mid,x + mid);
		int rem = N - p + 1 - (t.se - p + 1);
		if(rem >= t.se - t.fi + 1) L = mid;
		else R = mid - 1;
	}
	pii t = Calc(x - L,x + L);
	int len = t.se - t.fi + 1;
	int64 res = b[N] - b[N - len];
	int q = N - len,h0 = t.fi - 1,h1 = t.se + 1;
	while(q >= h1) {
		res += A[q];--q;
		if(h1 > q && h0 < 1) break;
		int now;
		if(h1 > q) now = h0;
		else if(h0 < 1) now = h1;
		else {
			if(x - A[h0] <= A[h1] - x) now = h0;
			else now = h1;
		}
		if(now == h1) ++h1;
		if(now == h0) --h0;
	}
	res += sum[h0];
	return res;
}
void Solve() {
	read(N);read(Q);
	for(int i = 1 ; i <= N ; ++i) read(A[i]);
	for(int i = 1 ; i <= N ; ++i) {
		sum[i] = A[i];
		if(i >= 2) sum[i] += sum[i - 2];
		b[i] = A[i] + b[i - 1];
	}
	int64 x;
	for(int i = 1 ; i <= Q ; ++i) {
		read(x);
		out(Process(x));enter;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

E - Attack to a Tree

这个直接dp就好了，(dp[u][j][0/1])表示以(u)为根，砍断了(j)条边，0表示没有电脑，1表示有电脑，dp里的值就是和u联通的联通块值最小值是多少
用一个bool数组辅助记录一下这个dp状态可不可以被达到
转移的话就是树背包，(dp[u][j][a] + dp[v][h][b] ightarrow dp[u][j + h][a | b])
然后如果有父亲的话用(dp[u][j][0])和(dp[u][j][1])更新(dp[u][j + 1][0])
没有的话就直接记录答案就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 5005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
struct node {
	int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
bool vis[MAXN][MAXN][2],used[MAXN][2];
int64 dp[MAXN][MAXN][2],A[MAXN],g[MAXN][2];
int N,siz[MAXN],ans;
void add(int u,int v) {
	E[++sumE].to = v;
	E[sumE].next = head[u];
	head[u] = sumE;
}
void dfs(int u,int fa) {
	if(A[u] < 0) {
		vis[u][0][1] = 1;dp[u][0][1] = A[u];
	}
	if(A[u] > 0) {
		vis[u][0][0] = 1;dp[u][0][0] = A[u];
	}
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != fa) {
			dfs(v,u);
		}
	}
	memset(used,0,sizeof(used));
	siz[u] = 1;
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != fa) {
			for(int j = 0 ; j <= siz[u] + siz[v] ; ++j) used[j][0] = used[j][1] = 0;
			for(int j = 0 ; j <= siz[u] ; ++j) {
				for(int a = 0 ; a <= 1 ; ++a) {
					if(!vis[u][j][a]) continue;
					for(int h = 0 ; h <= siz[v] ; ++h) {
						for(int b = 0 ; b <= 1 ; ++b) {
							if(!vis[v][h][b]) continue;
							if(!used[j + h][a | b]) {
								g[j + h][a | b] = dp[u][j][a] + dp[v][h][b];
								used[j + h][a | b] = 1;
							}
							else g[j + h][a | b] = min(g[j + h][a | b],dp[u][j][a] + dp[v][h][b]);
						}
					}
				}
			}
			for(int j = 0 ; j <= siz[u] + siz[v] ; ++j) {
				for(int a = 0 ; a <= 1 ; ++a) {
					vis[u][j][a] = used[j][a];
					dp[u][j][a] = g[j][a];
				}
			}
			siz[u] += siz[v];
		}
	}
	if(fa) {
		for(int j = siz[u] ; j >= 0 ; --j) {
			if(vis[u][j][1] && dp[u][j][1] < 0) {dp[u][j + 1][0] = 0;vis[u][j + 1][0] = 1;}
			if(vis[u][j][0]) {dp[u][j + 1][0] = 0;vis[u][j + 1][0] = 1;}
		}
	}
	else {
		for(int j = 0 ; j <= siz[u] ; ++j) {
			if(vis[u][j][1] && dp[u][j][1] < 0) {ans = j;break;}
			if(vis[u][j][0]) {ans = j;break;}
		}
	}
}
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) read(A[i]);
	int u,v;
	for(int i = 1 ; i < N ; ++i) {
		read(u);read(v);
		add(u,v);add(v,u);
	}
	dfs(1,0);
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}