AtCoder Grand Contest 007 题解

A - Shik and Stone

直接模拟判断就好了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n, m;
 
char str[20][20];
 
int main()
{
	gii(n, m);
	for (int i = 1; i <= n; ++i) scanf("%s", str[i] + 1);
	int x = 1, y = 1;
	if (str[x][y] == '.')
	{
		puts("Impossible");
		return 0;
	}
	int cnt = 0;
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) cnt += (str[i][j] == '#');
	for (int i = 1; i < cnt; ++i)
	{
		if (str[x + 1][y] == '#') ++x;
		else if (str[x][y + 1] == '#') ++y;
		else 
		{
			puts("Impossible");
			return 0;
		}
	}
	puts("Possible");
	return 0;
}

　　

B - Construct Sequences

我们可以让a[p[i]]=20000*p[i]+i，b[i]=20000*(n-p[i]+1)+i，这样子就好了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 20010;
 
int n, p[N], a[N], b[N];
 
int main()
{
	gi(n);
	for (int i = 1; i <= n; ++i) gi(p[i]);
	for (int i = 1; i <= n; ++i) a[i] = a[i - 1] + 20000;
	for (int i = n; i; --i) b[i] = b[i + 1] + 20000;
	for (int i = 1; i <= n; ++i) a[p[i]] += i, b[p[i]] += i;
	for (int i = 1; i <= n; ++i) printf("%d ", a[i]); puts("");
	for (int i = 1; i <= n; ++i) printf("%d ", b[i]); puts("");
	return 0;
}

　　

C - Pushing Balls

我们用整体法，发现每做一次操作，递归下去依然是一个等差序列。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n;
 
double a, x, ans;
 
int main() 
{
	giii(n, a, x);
	for(int i = n; i; --i) 
	{
		double m = 2 * i;
		double v = (a + (a + (m - 1) *x)) / m * i;
		ans += v;
		a = ((m + 2) * a + 5 * x) / m;
		x = (m + 4) / m * x;
	}
	printf("%.15lf
", ans);
	return 0;
}

　　

D - Shik and Game

列出dp方程，分类讨论max，用线段树优化一下复杂度即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int MaxN = 1e5 + 10;
 
int N, E, T, x[MaxN];
 
long long f[MaxN];
 
int find(int i)
{
	int j = i;
	for (int k = 18; ~k; --k)
		if ((j - (1 << k)) >= 0 && 2 * (x[i] - x[j - (1 << k) + 1]) <= T)
			j -= 1 << k;
	return j;
}
 
struct zkw
{
	long long s[MaxN << 2];
 
	int cnt;
 
	void build()
	{
		cnt = 1;
		for (; cnt < N + 3; cnt <<= 1); --cnt;
		for (int i = 1; i <= N + 1; ++i) s[cnt + i] = 1e18;
		for (int i = cnt; i; --i) s[i] = 1e18;
	}
 
	void modify(int pos, long long x)
	{
		++pos;
		pos += cnt;
		s[pos] = min(s[pos], x);
		for (pos >>= 1; pos; pos >>= 1) s[pos] = min(s[pos << 1], s[pos << 1 | 1]);
	}
 
	long long query(int l, int r)
	{
		long long ans = 1e18;
		++l, ++r;
		for (l += cnt - 1, r += cnt + 1; l ^ r ^ 1; l >>= 1, r >>= 1)
		{
			if (~l & 1) ans = min(ans, s[l ^ 1]);
			if (r & 1) ans = min(ans, s[r ^ 1]);
		}
		return ans;
	}
} s1, s2;
 
 
 
int main()
{
	giii(N, E, T);
	for (int i = 1; i <= N; ++i) gi(x[i]), f[i] = 1e18;
	s1.build(), s2.build();
	s1.modify(0, -2 * x[1]);
	s2.modify(0, 0);
	for (int i = 1; i <= N; ++i)
	{
		int k = find(i);
		f[i] = min(f[i], s1.query(0, k - 1) + 3LL * x[i]);
		/*for (int j = 0; j < k; ++j)
			f[i] = min(f[i], 3LL * x[i] + f[j] - x[j] - 2 * x[j + 1]));*/
		f[i] = min(f[i], s2.query(k, i - 1) + x[i] + T);
		s1.modify(i, f[i] - x[i] - 2 * x[i + 1]);
		s2.modify(i, f[i] - x[i]);
		/*for (int j = k; j < i; ++j)
			f[i] = min(f[i], f[j] + x[i] - x[j] + T);*/
	}
	//cerr << f[2] << endl;
	f[N] += E - x[N];
	printf("%lld
", f[N]);
	return 0;
}

　　

E - Shik and Travel

二分答案，每个点存很多pair(x,y)表示第一次走x，最后一次走y其它都小于等于二分答案的值能不能构造出来，转移直接合并即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
typedef pair<int64, int64> PII;
typedef vector<int> VI;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 1 << 20;
 
int a[N], v[N], n;
 
vector<PII> p[N];
 
int son[N][2];
 
bool level[N];
 
vector<PII> A, B;
 
bool work(int x, int64 lim)
{
	p[x].clear();
	if (level[x])
	{
		p[x].pb(mp(0LL, 0LL));
		return 1;
	}
	int ls = son[x][0], rs = son[x][1];
	if (!SZ(p[ls]) || !SZ(p[rs])) return 0;
	if (p[ls].size() > p[rs].size()) swap(ls, rs);
	A.clear(); B.clear();
	int64 L = lim - v[ls] - v[rs];
	int64 mi = 1LL << 35;
	for (int i = 0, j = 0; i < SZ(p[ls]); ++i)
	{
		while (p[ls][i].se > L - mi && j < SZ(p[rs])) mi = min(mi, p[rs][j].fi), ++j;
		if (p[ls][i].se > L - mi) break;
		if (j) A.pb(mp(p[ls][i].fi + v[ls], p[rs][j - 1].se + v[rs]));
	}
	mi = 1LL << 35;
	for (int j = 0, i = 0; j < SZ(p[rs]); ++j)
	{
		while (p[rs][j].se > L - mi && i < SZ(p[ls])) mi = min(mi, p[ls][i].fi), ++i;
		if (p[rs][j].se > L - mi) break;
		if (i) 
		{
			if (SZ(B))
			{
				if (B[SZ(B) - 1].se == p[ls][i - 1].se + v[ls])
					B[SZ(B) - 1].fi = min(B[SZ(B) - 1].fi, p[rs][j].fi + v[rs]);
				else B.pb(mp(p[rs][j].fi + v[rs], p[ls][i - 1].se + v[ls]));
			}
			else B.pb(mp(p[rs][j].fi + v[rs], p[ls][i - 1].se + v[ls])); 
		}
	}
	int i = 0, j = 0;
	while (i < SZ(A) || j < SZ(B))
	{
		if (j >= SZ(B)) p[x].pb(A[i++]);
		else if (i >= SZ(A) || A[i].se > B[j].se) p[x].pb(B[j++]);
		else p[x].pb(A[i++]);
	}
	if (!SZ(p[x])) return 0;
	return 1;
}
 
bool check(int64 lim)
{
	for (int i = n; i; --i) if (!work(i, lim)) return 0;
	return 1;
}
 
main()
{
	gi(n);
	for (int i = 1; i <= n; ++i) level[i] = 1;
	for (int i = 2; i <= n; ++i) 
	{
		gii(a[i], v[i]);
		level[a[i]] = 0;
		if (son[a[i]][0]) son[a[i]][1] = i;
		else son[a[i]][0] = i; 
	}
	int64 x = -1;
	for (int i = 35; ~i; --i)
		if (!check(x + (1LL << i)))
			x += (1LL << i);
	printf("%lld
", x + 1);
	return 0;
} 

　　

F - Shik and Copying String

贪心取，用个队列维护互相影响的点的最大值，就是答案了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 1e6 + 10;
 
int n;
 
char s[N], t[N];
 
int q[N];
 
int main()
{
	int l = 1, r = 0;
	gi(n);
	scanf("%s", s + 1);
	scanf("%s", t + 1);
	if (!strcmp(s + 1, t + 1))
	{
		puts("0");
		return 0;
	}
	int ans = 0, cnt = 0;
	for (int i = n, j = n; i; --i)
	{
		if (t[i] == t[i - 1]);
		else
		{
			j = min(j, i);
			while (j && t[i] != s[j]) --j;
			if (!j) return puts("-1"), 0;
			while (l <= r)
			{
				if (q[l] - (r - l) > i) ++l;
				else break;
			}
			q[++r] = j;
			if (i != j) ans = max(ans, r - l + 1);
		}
	}
	printf("%d
", ++ans);
	return 0;
}