Codeforces Round #385 (Div. 2)

A

题意:统计一个字符串的所有循环串的不同个数

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 1e5 + 10;
ll num[qq];

int main(){
	string str;
	map<string , int >  Q;
	cin >> str;
	int c = 1;
	Q[str]++;
	string x;
	for(int i = 1; i < str.size(); ++i){
		for(int j = i; j < str.size(); ++j)
			x += str[j];
		for(int j = 0; j < i; ++j)
			x += str[j];
		if(Q.find(x) == Q.end())	c++;
		Q[x]++;
		x = "";
	}
	cout << c << endl;
	return 0;
}

B

题意:所给图中的所有x都是连在一起的, 问这些x是不是构成了矩形

思路:找出x左上角和右下角的坐标 然后判断是不是矩阵内全是x矩阵外没有x

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 550 + 10;
ll num[qq];
char str[qq][qq];

struct Point{
	int x, y;
}agree1, agree2;

int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < n; ++i)
		cin >> str[i];
	bool f1, f2;
	f1 = f2 = false;
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j)
			if(str[i][j] == 'X'){
				if(!f1)	agree1.x = i, agree1.y = j, f1 = true;
				else{
					if(i < agree1.x){
						agree1.x = i, agree1.y = j;
					}else if(j < agree1.y){
						agree1.x = i, agree1.y = j;
					}
				}
				if(!f2)	agree2.x = i, agree2.y = j, f2 = true;
				else{
					if(i > agree1.x){
						agree2.x = i, agree2.y = j;
					}else if(j > agree1.y){
						agree2.x = i, agree2.y = j;
					}
				}
			}
	bool flag = true;
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j){
			if(i <= agree2.x && i >= agree1.x && j <= agree2.y && j >= agree1.y){
				if(str[i][j] == '.')	flag = false;
			}else{
				if(str[i][j] == 'X')	flag = false;
			}
		}
	if(flag)	cout << "YES" << endl;
	else		cout << "NO" << endl;
	return 0;
}

C

题意: n个城市, 有k个首都, k个首都之间都是互不相连的, 图中有m条边, 问最多还能添加多少条边

思路: 因为首都之间都互不相连, 那么每一个首都都一个联通块, 那么我们取k个首都中最多点的联通块把所有没有连上首都的点都连上去再加上其他首都联通块直接能加的边的综合就是答案

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 1e3 + 10;
int pre[qq];
int vis[qq];
int node[qq];
int map[qq][qq];
ll num[qq];
ll n, m, k;
ll ver, edges;

int find(int x){
	return pre[x] == x?x:pre[x]=find(pre[x]);
}


vector<int> v[qq];

void dfs(int rt){
	int nodes = v[rt].size();
	for(int i = 0; i < nodes; ++i)
		if(!vis[v[rt][i]]){
			ver++;
			vis[v[rt][i]] = 1;
			edges += v[v[rt][i]].size();
			dfs(v[rt][i]);
		}
}

int main(){
	cin >> n >> m >> k;
	memset(vis, 0, sizeof(vis));
	for(int i = 0; i < k; ++i){
		cin >> node[i];
	}
	int a, b;
	for(int i = 0; i < m; ++i){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	if(k == 1){
        cout << n*(n-1)/2-m << endl;
        return 0;
	}
	ll sum = 0;
	ll maxn = 1;
	for(int i = 0; i < k; ++i){
		ver = 1;
        edges = 0;
		edges += v[node[i]].size();
		vis[node[i]] = 1;
		if(v[node[i]].size() == 0)	continue;
		dfs(node[i]);
		edges = edges/2;
		maxn = max(maxn, ver);
		sum += ver*(ver-1)/2 - edges;
	}
	int p = 0;
	for(int i = 1; i <= n; ++i){
		if(vis[i])	continue;
		if(v[i].size() == 0)	continue;
		ver = 1;
		edges = 0;
		edges += v[i].size();
		vis[i] = 1;
		dfs(i);
		edges = edges/2;
		sum += ver*(ver-1)/2 -edges;
		num[p++] = ver;
	}
	for(int i = 0; i < p; ++i){
        sum += maxn*num[i];
        maxn += num[i];
	}
	int emptypoint = 0;
	for(int i = 1; i <= n; ++i){
		if(vis[i])	continue;
		if(v[i].size() == 0)	emptypoint++;
	}
	for(int i = 1; i <= emptypoint; ++i){
        sum += maxn;
        maxn++;
	}
	cout << sum << endl;
	return 0;
}