Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals)

A

这次的A比之前几次都简单、 因为不用想 hhhh

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

using namespace std;
#define LL long long
#define pill pair<int, int>
#define pb push_back
#define mk make_pair
#define REP(i, x, n)	for(int i = x; i < n; ++i)
const int qq = 1e5 + 10;
int num[qq];
int n;
int main(){
	scanf("%d", &n);
	int maxn = 0;
	int x;
	int cnt = 0;
	for(int i = 0; i < 2 * n; ++i){
		scanf("%d", &x);
		num[x]++;
		if(num[x] == 1)	cnt++;
		else if(num[x] == 2)	cnt--;
		maxn = max(maxn, cnt);
	}
	printf("%d\n", maxn);
	return 0;
}

B

这题是直接二分了、

题目中有句英文说明了， 不一定在整数点相遇， 很显然我们可以算在一定时间内每个人能达到的区间， n个人的区间有交集则true

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define REP(i, x, n)	for(int i = x; i < n; ++i)
const int qq = 60000 + 10;
const double eps = 1e-7;
double x[qq], v[qq];
int n;
bool check(double k){
	double l = x[0] - k * v[0];
	double r = x[0] + k * v[0];
	for(int i = 1; i < n; ++i){
		double a = x[i] - k * v[i];
		double b = x[i] + k * v[i];
		if(b < l)	return false;
		if(a > r)	return false;
		l = max(l, a);
		r = min(r, b);
	}
	return true;
}
int main(){
	scanf("%d", &n);
	for(int i = 0; i < n; ++i){
		scanf("%lf", x + i);
	}
	for(int i = 0; i < n; ++i){
		scanf("%lf", v + i);
	}
	double l = 0, r = 1000000000.;
	while(l + eps < r){
		double mid = (l + r) / 2.;
		if(check(mid))	r = mid - eps;
		else	l = mid + eps;
	}
	printf("%.8lf\n", l);
	return 0;
}

C

题意：如果a、b、c之间a能到b 且 b能到c， 则a b c三个顶点不同色， 求每个点的颜色编号以及需要最小的颜色数

我们可以发现需要最少的颜色数是等于图中顶点的度数最大的那个去加上1， 就是k值

然后处理每一个点的颜色， 以1为根节点， 对于每一个有父节节点的顶点，我们可以直接对它的孩子节点着色， 着色方式是用map统计当前没用过的颜色

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

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define REP(i, x, n)	for(int i = x; i < n; ++i)
const int qq = 2e5 + 10;
int deg[qq];
vector<int> vt[qq];
int color[qq];
int vis[qq];
int n;
map<int, bool> mp;
void dfs(int rt, int a, int b){
	mp.clear();
	int sz = (int)vt[rt].size();
	mp[a] = true, mp[b] = true;
	int cnt = 1;
	while(mp[cnt] == true)	cnt++;
	for(int j = 0; j < sz; ++j)if(!vis[vt[rt][j]]){
		int v = vt[rt][j];
		if(!color[v]){
			color[v] = cnt;
			mp[cnt] = true;
			while(mp[cnt] == true)	cnt++;
		}
	}
	for(int j = 0; j < sz; ++j)if(!vis[vt[rt][j]]){
		int v = vt[rt][j];
		vis[v] = 1;
		dfs(v, b, color[v]);
	}
}
int main(){
	scanf("%d", &n);
	int a, b;
	for(int i = 1; i < n; ++i){
		scanf("%d%d", &a, &b);
		deg[a]++, deg[b]++;
		vt[a].pb(b);
		vt[b].pb(a);
	}
	int k = 0;
	for(int i = 1; i <= n; ++i)
		k = max(k, deg[i] + 1);
	int cnt = 2;
	vis[1] = 1;
	color[1] = 1;
	for(int i = 0; i < (int)vt[1].size(); ++i){
		color[vt[1][i]] = cnt++;
		vis[vt[1][i]] = 1;
	}
	for(int i = 0; i < (int)vt[1].size(); ++i){
		dfs(vt[1][i], color[1], color[vt[1][i]]);
	}
	printf("%d\n", k);
	for(int i = 1; i <= n; ++i)
		printf("%d ", color[i]);
	puts("");
	return 0;
}