The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

题目链接

第一场网络赛打得还不错。。搞了5题
A，B，C，H，J

A：就一个签到题，找一遍即可

B：构造问题，我的构造方法是这样的，先铺满第一排，然后从第二排开始到一半，就每次拐1长度去铺，这样保证都不出重复，然后一半之后用拐2长度的的去铺，然后最后会剩下一个空位，正好能放下一个，这个拐的长度为(n + 1) / 2，一开始还wa了不知道为什么，原来是n = 6的时候，最后一块和拐2长度的一块会重复，于是手动构造个6个就过了

C：dfs，首先l < k就是错误，不连通就是错误，从每个侦探器进去dfs，每次搜到其他侦探器就停下，并把侦探器标记清空，走过的点就标记掉无须走第二次，然后如果有一个侦探器（第一个除外）进入的时候没被清空，就是错误

H：贪心，dfs，用类似数位dp的方法去dfs贪心构造串，每次前面放一个数字，后面也跟着放一些数字，要加个剪枝，就是如果当前放的前几位比已经找过的答案要小，那就没必要搜下去了

J：这题也算水题，由于字符串长度不大，暴力乱搞即可

代码：

A：

#include <cstdio>
#include <cstring>

const int N = 55;

int t, n, H[N], ans;

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			scanf("%d", &H[i]);
		ans = 0;
		for (int i = 2; i <= n - 1; i++) {
			if (H[i] > H[i - 1] && H[i] > H[i + 1])
				ans++;
		}
		printf("%d\n", ans);
	}
	return 0;
}

B：

#include <cstdio>
#include <cstring>

const int N = 105;

int t, n;
char g[N][N];
char col[2];

void solve() {
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			g[i][j] = 'B';
		}
	for (int i = 0; i < n; i++)
		g[0][i] = 'Y';
	for (int i = 0; i < (n - 1) / 2; i++) {
		char c = col[i % 2];
		for (int j = i + 1; j < n; j++)
			g[j][i] = c;
		for (int j = 1; j <= i + 1; j++)
			g[j][i + 1] = c;
	} 
	for (int i = (n - 1) / 2; i < n - 1; i++) {
		char c = col[i % 2];
		for (int j = i + 2; j < n; j++)
			g[j][i] = c;
		g[i + 2][i + 1] = c;
		for (int j = 2; j <= i + 2; j++)
			g[j][i + 2] = c;
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++)
			printf("%c", g[i][j]);
		printf("\n");
	}
}

int main() {
	col[0] = 'G';
	col[1] = 'R';
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		if (n == 1) printf("Y\n");
		else if (n <= 4) printf("No solution!\n");
		else if (n == 6) {
			printf("YYYYYY\n");
			printf("GGRGRR\n");
			printf("GRRGRB\n");
			printf("GRGGRB\n");
			printf("GRGRRB\n");
			printf("GRGBBB\n");
		}
		else {
			solve();
		}
	}
	return 0;
}

C：

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

const int N = 100005;

int t, n, m, l, k, s[N];
bool vis[N], isch[N];
vector<int> g[N];

int dfs2(int u) {
	vis[u] = true;
	int ans = 1;
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (vis[v]) continue;
		ans += dfs2(v);
	}
	return ans;
}

void dfs(int u) {
	vis[u] = true;
	if (isch[u]) {
		isch[u] = false;
		return;
	}
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (vis[v]) continue;
		dfs(v);
	}
}

bool solve() {
	if (l < k) return false;
	if (dfs2(1) != n) return false;
	memset(vis, false, sizeof(vis));
	isch[s[0]] = false;
	for (int i = 0; i < l; i++) {
		if (isch[s[i]]) return false;
		dfs(s[i]);
	}
	return true;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d", &n, &m, &k);
		for (int i = 1; i <= n; i++)
			g[i].clear();
		int tmp;
		memset(vis, false, sizeof(vis));
		memset(isch, false, sizeof(isch));
		for (int i = 0; i < k; i++) {
			scanf("%d", &tmp);
			isch[tmp] = true;
		}
		int u, v;
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
			g[v].push_back(u);
		}
		scanf("%d", &l);
		for (int i = 0; i < l; i++)
			scanf("%d", &s[i]);
		printf("%s\n", solve() ? "Yes" : "No");
	}
	return 0;
}

H：

#include <cstdio>
#include <cstring>

char A[35];
int t, n, num[35], ans[35];

bool judge(int s) {
	for (int i = 0; i <= s; i++) {
		if (ans[i] > num[i]) return false;
		else if (ans[i] < num[i]) return true;
	}
	return true;
}

bool check() {
	for (int i = 0; i < n; i++) {
		if (num[i] > ans[i])
			return true;
		else if (num[i] < ans[i]) return false;
	}
	return false;
}

void dfs(int s, int e, int flag1, int flag2, int flag3) {
	if (s > e) {
		if (flag1 == 0 && flag2) return;
		if (check()) {
			for (int i = 0; i < n; i++)
				ans[i] = num[i];
		}
		return;
	}
	int end = A[s] - '0';
	if (flag1) end = 9;
	for (int i = end; i >= 0; i--) {
		if (flag1 == 0 && i < end) flag1 = 1;
		if (!flag3 && i == 0) continue;
		num[s] = i;
		if (!judge(s)) break;
		if (e != n - 1 && num[e + 1] == i)
			dfs(s + 1, e, flag1, flag2, 1);
		for (int j = e; j >= s; j--) {
			num[j] = i;
			if (num[j] > A[j] - '0') flag2 = 1;
			if (num[j] < A[j] - '0') flag2 = 0;
			dfs(s + 1, j - 1, flag1, flag2, 1);
		}
	}
	if (!judge(s)) return;
	if (!flag3) {
		num[s] = 0;
		dfs(s + 1, e, flag1, flag2, flag3);
	}
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%s", A);
		n = strlen(A);
		for (int i = 0; i < n; i++)
			ans[i] = 0;
		dfs(0, n - 1, 0, 1, 0);
		int s = 0;
		while (s < n - 1 && ans[s] == 0) {s++;}
		for (int i = s; i < n; i++)
			printf("%d", ans[i]);
		printf("\n");
	}
	return 0;
}

J:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;

int t;
string str;

void handle() {
	string tmp = "";
	for (int i = 0; i < str.length(); i++)
		if ((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z')) {
			tmp += str[i];
		}
	str = tmp;
}

int n;
bool solve1() {
	for (int i = 0; i < n; i++) {
		string A = "";
		for (int j = 0; j <= i; j++)
			A += str[j];
		int bl = n - (i + 1) * 3;
		if (bl <= 0 || bl % 2) continue;
		string B = "";
		for (int j = i + 1; j < i + 1 + bl / 2; j++)
			B += str[j];
		if (A != B && A + B + A + B + A == str) return true;
	}
	return false;
}

bool judge(string AB, string C) {
	for (int i = 0; i < AB.length(); i++) {
		string A = "", B = "";
		for (int j = 0; j <= i; j++)
			A += AB[j];
		for (int j = i + 1; j < AB.length(); j++)
			B += AB[j];
		if (A == "" || B == "" || C == "") continue;
		if (A == B || A == C || C == B) continue;
		return true;
	}
	return false;
}

bool solve2() {
	for (int i = 0; i < n; i++) {
		string AB = "";
		for (int j = 0; j <= i; j++)
			AB += str[j];
		int cl = n - (i + 1) * 3;
		if (cl <= 0) continue;
		string C = "";
		for (int j = 2 * i + 2; j < 2 * i + 2 + cl; j++)
			C += str[j];
		if (judge(AB, C) && AB + AB + C + AB == str) return true;
	}
	return false;
}

int main() {
	scanf("%d%*c", &t);
	while (t--) {
		getline(cin, str);
		handle();
		n = str.length();
		if (solve1() || solve2()) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}