SRM 584div2

昨天做TC手一抖浪费了涨rating的大好机会， 本来三题都会的500pt写挂了， 有个小bug， 1000pt本来就打算试试的最后没调试完， 其实再有一分钟就能debug完的， 最后只出了一题， 悲剧啊。

250pt: 水， 不解释

500pt:第一次在SRM上遇上图论， 最后没A真可惜， 这题我的做法就是先判断所有人形成的联通分量个数如果大于1就输出-1（两个没关系的人的差值可以为INF），否则考虑任意两个人的最大差距， 不难发现和两人间的最短路有关， 用Floyd求个最短路就解了。

#include <vector>
#include <cstring>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class Egalitarianism {
public:
	int maxDifference(vector <string>, int);
};

const int N = 55;
const int INF = 1 << 29;
int dis[N][N];
bool vis[N];

void dfs(int u, int n) {
	vis[u] = 1;
	for (int i = 0; i < n; i++) {
		if (!vis[i] && dis[u][i] == 1)
			dfs(i, n);
	}
}

int Egalitarianism::maxDifference(vector<string> vec, int d) {
	int n = vec.size();

	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			dis[i][j] = INF;

	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			if (vec[i][j] == 'Y') {
				dis[i][j] = 1;
				dis[j][i] = 1;
			}
	memset(vis, 0, sizeof(vis));

	int cnt = 0;

	for (int i = 0; i < n; i++)
		if (!vis[i]) {
			dfs(i, n);
			cnt++;
		}

	if (cnt > 1) return -1;

	for (int k = 0; k < n; k++)
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);

	int ans = 0;

	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			if (dis[i][j] != INF && i != j)
				ans = max(ans, dis[i][j]);

	return ans * d;
}

<%:testing-code%>
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1000pts:是个组合计数题目题意很难懂， 最后只想到了个类似暴力的dfs解法估计会T的没想到过了看了下时间最慢的一组跑了1.4秒, 应该不是正解。。。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class Excavations2 {
public:
	long long count(vector <int>, vector <int>, int);
};

typedef long long LL;
const int N = 55;

LL C[N][N];
int n;
int sum[N];
LL res;

void init() {
	for (int i = 0; i < N; i++) {
		C[i][0] = C[i][i] = 1;
		for (int j = 1; j < i; j++) 
			C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
	}
	memset(sum, 0, sizeof(sum));
}


void dfs(int d, int k, LL tmp, int cnt) {
	if (d == n - 1) {
		if (cnt < k) {
			res += tmp * C[sum[d]][k - cnt];
		}
		return;
	}
	for (int i = 1; i <= sum[d]; i++) {
		if (i + cnt >= k) break;
		dfs(d + 1, k, tmp * C[sum[d]][i], cnt + i);
	}
}

long long Excavations2::count(vector<int> kind, vector<int> found, int K) {
	init();
	n = found.size();
	for (int i = 0; i < kind.size(); i++) {
		int typ = kind[i];
		for (int j = 0; j < n; j++)
			if (found[j] == typ) {
				sum[j]++;
				break;
			}	
	}
	res = 0LL;
	dfs(0, K, 1, 0);
	return res;
}

<%:testing-code%>
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