[HIHO]hihoCoder太阁最新面经算法竞赛7

题目链接：http://hihocoder.com/contest/hihointerview12

期末完事了，终于有时间成套刷题了。这套题比较简单，难度上感觉和上一套差不多。除了最后一个题是看了讨论版说数据水才敢写的。

A Word Construction

解法：正解应该是dfs+剪枝，我的思路是贪心，竟然过了。先把字符串按字典序排序，然后枚举每一个串作为起始，记下串内有什么字符，再从头到尾枚举所有字符串，看看每一个是否符合条件。就那么100个字符串，数据弱得很。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <cstdlib>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 string s[maxn];
 int n;
 int ascii[];

 int main() {
     // FRead();
     while(~Rint(n)) {
         Rep(i, n) cin >> s[i];
         sort(s, s+n);
         int ret = ;
         Rep(k, n) {
             Cls(ascii);
             Rep(j, s[k].length()) ascii[s[k][j]] = ;
             int cur = ;
             Rep(i, n) {
                 bool flag = ;
                 Rep(j, s[i].length()) {
                     if(ascii[s[i][j]] == ) {
                         flag = ;
                         break;
                     }
                 }
                 if(flag == ) {
                     Rep(j, s[i].length()) ascii[s[i][j]] = ;
                     cur++;
                 }
             }
             ret = max(ret, cur);
         }
         printf("%d\n", ret);
     }
     RT ;
 }

A

B Email Merge

解法：STL+并查集，这题很好想，但是写起来比较麻烦。对于字符串数组，想要做并查集操作的话，需要map<string, int>这样从字符串到整型的映射支持。读入数据按照邮箱为根节点，儿子是用户名，构建一个森林。这个时候同时更新并查集，也就是合并同一个树根下的所有用户名字符串。接着遍历所有用户名，扔到对应的belong中。我开的belong是堆，因为输出要求按照出现的顺序，所以堆序就是输入的顺序，也就是之前赋值的id号。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <cstdlib>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 typedef struct Node {
     string d;
     int idx;
     Node() { idx = -; }
     Node(string dd, int ii) : d(dd), idx(ii) {}
     friend bool operator<(Node a, Node b) { return a.idx > b.idx; }
 }Node;

 string mail, user;
 bool vis[maxn];
 map<string, int> id;
 map<string, set<string> > wt;
 priority_queue<Node> belong[];

 int n, m, cnt;
 int pre[maxn];

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

 void unite(int x, int y) {
     x = find(x); y = find(y);
     if(x < y) pre[y] = x;
     else pre[x] = y;
 }

 int main() {
     // FRead();
     cnt = ;id.cl(); wt.cl(); Cls(vis);
     Rint(n);
     Rep(i, n*) {
         pre[i] = i;
         while(!belong[i].empty()) belong[i].pop();
     }
     Rep(i, n) {
         cin >> user;
         cin >> m;
         id[user] = cnt++;
         Rep(j, m) {
             cin >> mail;
             if(id.find(mail) == id.end()) id[mail] = cnt++;
             wt[mail].insert(user);
             unite(id[user], id[mail]);
         }
     }
     map<string, set<string> >::iterator it;
     set<string>::iterator each;
     for(it = wt.begin(); it != wt.end(); it++) {
         for(each = it->second.begin(); each != it->second.end(); each++) {
             if(!vis[id[*each]]) {
                 vis[id[*each]] = ;
                 belong[find(id[*each])].push(Node(*each, id[*each]));
             }
         }
     }
     Rep(i, cnt) {
         if(!belong[i].empty()) {
             while(!belong[i].empty()) {
                 cout << belong[i].top().d << " ";
                 belong[i].pop();
             }
             cout << endl;
         }
     }
     RT ;
 }

B

C Matrix Sum

解法：这题毒瘤题…数据里有负数，不同语言对负数取模运算的结果不一样，C++和Java是一个负数对一个正数取模是一个负数，而Python是正数。这题的正解应当是二维线段树or树状数组（树套树），分别维护自己列的值。而我直接维护前缀和了。写了各种语言的解法，改来改去最后改了java。

 import java.lang.reflect.Array;
 import java.util.Arrays;
 import java.util.Comparator;
 import java.util.Scanner;

 import javax.swing.text.GapContent;

 public class Main {
     public static void main(String[] args) {
         Scanner in = new Scanner(System.in);
         int n, m;
         int[][] dp = new int[1010][1010];
         String cmd = new String();
         for(int i = 0; i < 1010; i++) {
             for(int j = 0; j < 1010; j++) {
                 dp[i][j] = 0;
             }
         }
         n = in.nextInt(); m = in.nextInt();
         int x1, x2, y1, y2, num;
         while(m-- > 0) {
             cmd = in.next();
             if(cmd.charAt(0) == 'A') {
                 x1 = in.nextInt();
                 y1 = in.nextInt();
                 num = in.nextInt();
                 for(int i = y1; i < n; i++) {
                     dp[x1][i] = (dp[x1][i] + num) % 1000000007;
                 }
             }
             else {
                 x1 = in.nextInt();
                 x2 = in.nextInt();
                 y1 = in.nextInt();
                 y2 = in.nextInt();
                 int ret = 0;
                 for(int i = x1; i <= y1; i++) {
                     if(x2 == 0) ret = (ret + dp[i][y2]) % 1000000007;
                     else ret = (ret + dp[i][y2] - dp[i][x2-1]) % 1000000007;
                 }
                 System.out.println((ret + 1000000007) % 1000000007);
             }
         }
     }
 }

C