135 - ZOJ Monthly, August 2014

A：构造问题，判断序列奇偶性，很容易发现最小值不是1就是0，最大值不是n就是n - 1，注意细节去构造即可

E：dp，dp[i][j]表示长度i，末尾状态为j的最大值，然后每个位置数字取与不取，不断状态转移即可

G：就一个模拟题没什么好说的

H：dfs，每次dfs下去，把子树宽度保存下来，然后找最大值，如果有多个，就是最大值+cnt宽度

I：构造，如果r * 2 > R，肯定无法构造，剩下的就二分底边，按等腰三角形去构造即可

代码：

A:

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int n;

void print(int n) {
    if (n == 3) {
	printf("3 1 2");
	return;
    }
    if (n % 2) {
	int len = (n - 3) / 2;
	printf("%d %d", n, n - len);
	for (int i = n - 1; i > n - len; i--)
	    printf(" %d %d", i, i - len);
	printf(" 3 1 2");
    }
    else {
	int len = n / 2;
	printf("%d %d", n, n - len);
	for (int i = n - 1; i > n - len; i--)
	    printf(" %d %d", i, i - len);
    }
}

void print2(int n) {
    print(n - 2);
    printf(" %d %d", n - 1, n);
}

void solve() {
    if (n == 1) {
	printf("1 1\n1\n1\n");
	return;
    }
    if (n == 2) {
	printf("1 1\n1 2\n2 1\n");
	return;
    }
    if (n == 3) {
	printf("0 2\n3 1 2\n1 2 3\n");
	return;
    }
    if (n % 2 == 0) {
	if (n / 2 % 2) {
	    printf("1 %d\n", n - 1);
	    print2(n); printf("\n");
	    print2(n - 1);
	    printf(" %d\n", n);
	}
	else {
	    printf("0 %d\n", n);
	    print(n); printf("\n");
	    print(n - 1); printf(" %d\n", n);
	}
    }
    else {
	if ((n + 1) / 2 % 2) {
	    printf("1 %d\n", n);
	    print(n - 2); printf(" %d %d\n", n - 1, n);
	    print(n - 1); printf(" %d\n", n);
	}
	else {
	    printf("0 %d\n", n - 1);
	    print(n); printf("\n");
	    print2(n - 1); printf(" %d\n", n);
	}
    }
}

int main() {
    while (~scanf("%d", &n)) {
	solve();
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;

const int INF = 0x3f3f3f3f;
int t, n;

map<int, int> dp[2];
map<int, int>::iterator it;

int lowbit(int x) {
    return (x&(-x));
}

int solve() {
    dp[0].clear();
    int pre = 1, now = 0;
    int num;
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
	scanf("%d", &num);
	num /= 2;
	swap(pre, now);
	dp[now].clear();
	for (it = dp[pre].begin(); it != dp[pre].end(); it++) {
	    int s = it->first;
	    if (dp[now].count(s) == 0) dp[now][s] = dp[pre][s];
	    else dp[now][s] = max(dp[now][s], dp[pre][s]);
	    int next;
	    if (s % num) {
		next = num;
		if (dp[now].count(next) == 0) dp[now][next] = dp[pre][s] + num * 2;
		else dp[now][next] = max(dp[now][next], dp[pre][s] + num * 2);
	    }
	    else {
		next = s + num;
		int add = (s % lowbit(next) * 2 + num) * 2;
		if (dp[now].count(next) == 0) dp[now][next] = dp[pre][s] + add;
		else dp[now][next] = max(dp[now][next], dp[pre][s] + add);
	    }
	}
    }
    int ans = 0;
    for (it = dp[now].begin(); it != dp[now].end(); it++)
	ans = max(ans, it->second);
    return ans;
}

int main() {
    scanf("%d", &t);
    while (t--) {
	scanf("%d", &n);
	printf("%d\n", solve());
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 55;
const int d[8][2] = {{1, 0}, {1, 1}, {1, -1}, {0, 1}, {0, -1}, {-1, 0}, {-1, 1}, {-1, -1}};

typedef pair<int, int> pii;

int t;
int n, m, f, k;
int g[N][N];
int gg[N][N];
char str[55];
vector<pii> go[1005];

void solve() {
    for (int ti = 1; ti <= f; ti++) {
	memset(gg, 0, sizeof(gg));
	for (int i = 1; i <= n; i++) {
	    for (int j = 1; j <= m; j++) {
		if (g[i][j] == 1) {
		    for (int k = 0; k < 8; k++) {
			int xx = i + d[k][0];
			int yy = j + d[k][1];
			if (xx <= 0 || xx > n || yy <= 0 || yy > m) continue;
			gg[xx][yy]++;
		    }
		}
	    }
	}

	for (int i = 1; i <= n; i++)
	    for (int j = 1; j <= m; j++) {
		if (g[i][j] == 2) continue;
		else if (g[i][j] == 0) {
		    if (gg[i][j] == 3) g[i][j] = 1;
		}
		else {
		    if (gg[i][j] < 2 || gg[i][j] > 3) g[i][j] = 0;
		}
	    }
	for (int i = 0; i < go[ti].size(); i++) {
	    g[go[ti][i].first][go[ti][i].second] = 2;
	}
    }
    for (int i = 1; i <= n; i++) {
	for (int j = 1; j <= m; j++) {
	    if (g[i][j] == 2) printf("X");
	    else printf("%d", g[i][j]);
	}
	printf("\n");
    }
}

int main() {
    scanf("%d", &t);
    while (t--) {
	scanf("%d%d%d%d", &n, &m, &f, &k);
	for (int i = 1; i <= f; i++)
	    go[i].clear();
	for (int i = 1; i <= n; i++) {
	    scanf("%s", str + 1);
	    for (int j = 1; j <= m; j++) {
		g[i][j] = str[j] - '0';
	    }
	}
	int ti, x, y;
	while (k--) {
	    scanf("%d%d%d", &ti, &x, &y);
	    go[ti].push_back(make_pair(x, y));
	}
	solve();
    }
    return 0;
}

H:

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 10005;

int n;
vector<int> g[N];

int dfs(int u) {
    int sz = g[u].size();
    vector<int> save;
    for (int i = 0; i < sz; i++)
	save.push_back(dfs(g[u][i]));
    sort(save.begin(), save.end());
    sz = save.size();
    int cnt = 0;
    int ans = 1;
    for (int i = sz - 1; i >= 0; i--) {
	if (i != sz - 1 && save[i] != save[i + 1]) break;
	ans = save[i] + cnt;
	cnt++;
    }
    return ans;
}

int main() {
    while (~scanf("%d", &n)) {
	for (int i = 1; i <= n; i++)
	    g[i].clear();
	int v;
	for (int i = 2; i <= n; i++) {
	    scanf("%d", &v);
	    g[v].push_back(i);
	}
	printf("%d\n", dfs(1));
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>

double r, R;

double h, x;

double cal(double a) {
    double d = a / 2;
    h = sqrt(R * R - d * d) + R;
    x = sqrt(h * h + d * d);
    return a * x * x / (2 * R * (a + x + x));
}

void solve() {
    double lx = 0, rx = sqrt(3.0) * R;
    double mid;
    for (int i = 0; i < 1000; i++) {
	mid = (lx + rx) / 2;
	double tmp = cal(mid);
	if (tmp > r) rx = mid;
	else lx = mid;
    }
    cal((lx + rx) / 2);
    printf("%.10lf %.10lf %.10lf\n", mid, x, x);
}

int main() {
    while (~scanf("%lf%lf", &r, &R)) {
	if (r * 2 > R) printf("NO Solution!\n");
	else solve();
    }
    return 0;
}

135 - ZOJ Monthly, August 2014