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%>//Powered by [KawigiEdit] 2.0!

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%>//Powered by [KawigiEdit] 2.0!