[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 = 1000;
string s[maxn];
int n;
int ascii[256];

int main() {
    // FRead();
    while(~Rint(n)) {
        Rep(i, n) cin >> s[i];
        sort(s, s+n);
        int ret = 0;
        Rep(k, n) {
            Cls(ascii);
            Rep(j, s[k].length()) ascii[s[k][j]] = 1;
            int cur = 1;
            Rep(i, n) {
                bool flag = 0;
                Rep(j, s[i].length()) {
                    if(ascii[s[i][j]] == 1) {
                        flag = 1;
                        break;
                    }
                }
                if(flag == 0) {
                    Rep(j, s[i].length()) ascii[s[i][j]] = 1;
                    cur++;
                }
            }
            ret = max(ret, cur);
        }
        printf("%d
", ret);
    }
    RT 0;
}

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 = 100010;
typedef struct Node {
    string d;
    int idx;
    Node() { idx = -1; }
    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[100010];

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 = 1;id.cl(); wt.cl(); Cls(vis);
    Rint(n);
    Rep(i, n*10) {
        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]] = 1;
                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 0;
}

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);
            }
        }
    }
}