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CF Rd478 Div2 A Aramic script

题意：给定几个字符串，去重后，求种类

思路：直接map乱搞

#include<bits/stdc++.h>

using namespace std;

string b;
bool vis[100005];
map<string, int> M;

int main() {
	int n; scanf("%d", &n);
	int ans = 0;
	while(n--) {
		memset(vis, 0, sizeof vis);
		string a;
		cin >> a;
		b.clear();
		int len = a.size();
		 
		sort(a.begin(), a.end());
		for (int i = 0; i < len; ++i) {
			if(a[i] == a[i + 1]) vis[i] = true;
		}
		for (int i = 0; i < len; ++i) {
			if(!vis[i]) b += a[i];
		}
		if(!M[b]) {
			ans++;
			M[b] = true;
		}
	} printf("%d\n", ans);
}

CF Rd482 Div2 A Pizza, Pizza, Pizza!!!

题意：一个披萨，要切成n个同样的块，求最小刀数

思路：偶数直接除以2，奇数直接输出

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

ll N;

int main() {
	scanf("%lld", &N);
	if(N == 0) printf("0\n");
	else if((N + 1) & 1) printf("%lld\n", N + 1);
	else printf("%lld\n", (N + 1) / 2);
	return 0;
}

CF Rd482 Div2 B Treasure Hunt

题意：三个字符串，给定n轮,每轮必须改变一个字符，问最后重复的单个字符最多的字符串是哪个

思路：每次改变出现次数最多的就好

#include<bits/stdc++.h>

using namespace std;
const int maxn = 3005;

int n;
string a, b, c;
int num[maxn];

int calc(string a) {
	memset(num, 0, sizeof num);
	int now = 0;
	int len = a.size();
	for (int i = 0; i < len; ++i) {
		num[a[i]]++;
	}
	for (int i = 0; i <= 300; ++i) {
		if(num[i] > now) now = num[i];
	}
	if(now == a.size()) return now - (n == 1);
	else return min(now + n, len);
} 

int main() {
	cin >> n;
	cin >> a; cin >> b; cin >> c;
	int ans1 = calc(a);
	int ans2 = calc(b);
	int ans3 = calc(c);
	if(ans1 > ans2 && ans1 > ans3) printf("Kuro\n");
	else if(ans2 > ans1 && ans2 > ans3) printf("Shiro\n");
	else if(ans3 > ans1 && ans3 > ans2) printf("Katie\n");
	else printf("Draw");
	return 0;
}

CF Rd482 Div2 C Kuro and Walking Route

题意：给定一颗树，给定x,y求不依次经过x -- y的路径对数为多少

思路：首先，全部点对数为n * (n - 1),dfs求出每个点的子节点个数，则y的子节点与x上面的节点组成的路径对数需要减去

即为n * (n - 1) - son[y] * (n - son[x])

#include<bits/stdc++.h>

using namespace std;
const int maxn = 3 * 100005;

int head[maxn << 1], cnt = 1;
int n, x, y; 
struct Node{
	int v, nxt;
} G[maxn << 1];
bool vis[maxn << 1];
int f[maxn << 1], siz[maxn << 1];
typedef long long ll;

void insert(int u, int v) {
	G[cnt] = (Node) {v, head[u]}; head[u] = cnt++;
}
void dfs(int x, int fa) {
	f[x] = fa; siz[x] = 1;
	for (int i = head[x]; i; i = G[i].nxt) {
		int v = G[i].v;
		if(v == fa) continue;
		dfs(v, x);
		siz[x] += siz[v]; 
	}
}

int main() {
	scanf("%d%d%d", &n, &x, &y);
	for (int i = 1; i <= n - 1; ++i) {
		int a, b; scanf("%d%d", &a, &b);
		insert(a, b); insert(b, a);
	} dfs(x, 0);
	int z = y;
	while(f[z] != x) z = f[z];
	ll ans = (ll)n * (ll)(n - 1) - (ll)(n - siz[z]) * (ll)siz[y];
	printf("%lld\n", ans);
	return 0;
}

CF Rd480 Div2 E The Number Games

题意：给定一棵树，节点i有权值2^i，求删除k个点后，剩下的点需要联通，并且还要使剩下的点的权值和最大

思路：首先有个贪心思想，2^i比sum(2^1 + 2^2 + ......2^(i - 1))还要大，因此想到要保留大的，首先最大的点肯定要保留，

然后以这个点为根，将点号从大到小枚举，依次检查每个点是否满足题意(加入的点不超过n - k)