【ACM-ICPC 2018 沈阳赛区网络预赛 I】Lattice's basics in digital electronics

【链接】 我是链接,点我呀:)

【题意】

【题解】

每个单词的前缀都不同。
不能更明示了...
裸的字典树。
模拟一下。输出一下就ojbk了。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define lson l,mid,rt<<1
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define rson mid+1,r,rt<<1|1
using namespace std;

const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};

const int NN = 10000;
const int SI = 10;
const int N = 2e5;

int ch[NN+10][2],flag[NN+10];
int tot,m,n;
char si[SI+10];
char s[N+10];
int result[N*4+10],cur;

void ins(int ci){
    int len = strlen(si);
    int now = 1;
    for (int i = 0;i < len;i++){
        if (ch[now][si[i]-'0']==0){
            ch[now][si[i]-'0'] = ++tot;
        }
        now = ch[now][si[i]-'0'];
    }
    flag[now] = ci;
}

void cl(char key){
    cur++;
    int num = 0;
    if (key>='a' && key<='z'){
        key = key-'a'+'A';
    }
    if (key>='A' && key<='Z'){
        num+=key-'A'+10;
    }else num = key-'0';
    for (int i = cur+3;i>=cur;i--){
        result[i] = num&1;
        num/=2;
    }
    cur = cur+3;
}

int main(){
	#ifdef LOCAL_DEFINE
	    freopen("rush_in.txt", "r", stdin);
	#endif
	ios::sync_with_stdio(0),cin.tie(0);
	int T;
    cin >> T;
    while (T--){
        memset(ch,0,sizeof ch);
        memset(flag,255,sizeof flag);
        tot = 1;
        cin >> m >> n;
        for (int i = 1;i <= n;i++){
            int ci;
            cin >> ci >> si;
            ins(ci);
        }
        cin >> s;
        int len = strlen(s);
        cur = 0;
        for (int i = 0;i < len;i++) cl(s[i]);
        int now = 1;
        for (int i = 1;i <= cur;){
            if (i+8>cur) break;
            int cnt = 0;
            for (int j = i;j <= i+7;j++)
                if (result[j]==1) cnt++;
            int odd = result[i+8];
            odd=1-odd;
            if ((cnt&1)==(odd&1)){
                for (int j = 1;j <= 8;j++){
                    result[now+j-1]=result[i+j-1];
                }
                now = now + 8;
            }
            i = i+9;
        }
        int index = 1;
        for (int i = 1;i <= m;i++){
            int now = 1;
            for (int j = index;;j++){
                now = ch[now][result[j]];
                if (flag[now]!=-1) {
                    index = j+1;
                    char key = flag[now];
                    cout<<key;
                    break;
                }
            }
        }
        cout<<endl;
    }
	return 0;
}