hihocoder35 模板场

ak完这场之后立刻做去年的北京场，然后就卡铜牌题了

连续训练真的有毒，这场还ak了不少人

A 有歧义的号码

按题意模拟

 #include <bits/stdc++.h>
using namespace std;
map <int, int> mp;

inline void initmp(){
    mp[1] = 1;
    mp[2] = 2;
    mp[5] = 5;
    mp[6] = 9;
    mp[8] = 8;
    mp[9] = 6;
    mp[0] = 0;
}

inline void wtf(int n) {
    int x = 1, cnt = 1, ans = 0;
    for (int i = 1; i <= n; i++) {
        if(i == x * 10) cnt++, x = i;
        int p = i, tmp = 0;
        bool ok = true;
        for(int j = 0; j < cnt; j++, p /= 10) {
            if (mp.find(p%10) == mp.end()) {
                ok = false; break;
            }
            tmp = (tmp*10 + mp[p % 10]);
        }
        if (tmp<=n && ok && tmp!=i && (i%10)>0) {
            printf("%d
", i);
        }
    }
}
int main(){
    initmp();
    for (int n; cin >> n;) {
        wtf(n);
    }
    return 0;
}

B 最短游览路线

bfs，起点先不标记vis，bfs到起点return

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+7;
const int INF = 0x3f3f3f3f;

struct Edge{
    int from, to, nxt;
    Edge(){}
    Edge(int f, int t, int n):from(f), to(t), nxt(n){}
}edges[N];
int n, head[N], E;

inline void init(){
    E = 0;
    memset(head, -1, sizeof head);
}

inline void addEdge(int f, int t){
    edges[E] = Edge(f, t, head[f]);
    head[f] = E++;
}

bool vis[N];
int dist[N];
inline int bfs(int s){
    queue <int> Q;
    memset(vis, false, sizeof vis);
    memset(dist, 0, sizeof dist);
    for (Q.push(s); !Q.empty();){
        int u = Q.front(); Q.pop();
        //printf("u = %d
", u);
        for (int i = head[u]; ~i; i = edges[i].nxt){
            Edge &e = edges[i];
            if (vis[e.to]) continue;
            dist[e.to] = dist[u] + 1;
            //printf("dist[%d] = %d
", e.to, dist[e.to]);
            vis[e.to] = true;
            Q.push(e.to);
            if (e.to == s) return dist[s];
        }
    }
    return dist[s];
}

int main(){
    //freopen("in.txt", "r", stdin);
    for (int m; ~scanf("%d%d", &n, &m);){
        init();
        for (int u, v; m--;){
            scanf("%d%d", &u, &v);
            addEdge(u, v);
        }
        int ans = bfs(1);
        printf("%d
", ans);
    }
    return 0;
}

C 重复字符串匹配

kmp，听说string里的find也能过

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 7;
int Next[N];
char A[N], B[N];
int ans = 0;

void getNext(int m, char b[]){
    int i = 0,j = -1;
    Next[0] = -1;
    while (i < m){
        if (j == -1 || b[i] == b[j]){
            if (b[++i] != b[++j]) Next[i]=j;
            else Next[i] = Next[j];
        }else j = Next[j];
    }
}

//主程序里每组数据需要memset a,b数组!!!
void KMP(int n, char a[], int m, char b[]){
    int i = 0, j = 0;
    getNext(m, b);//这行视情况可以放在main里面
    while (i < n && j < m){
        if (j == -1 || a[i] == b[j]) i++, j++;
        else j = Next[j];
        if (!i && !j)break;
        if (j == m){
            ans = 1;
            j = Next[j];
        }
    }
}

int main(){
    int _;
    scanf("%d", &_);
    for (; _--;) {
        scanf("%s", A);
        scanf("%s", B);
        int lenA = strlen(A);
        int lenB = strlen(B);
        int tmp = (lenB - 1) / lenA;
        for(int i = 0; i < tmp; i++)
            for(int j = 0; j < lenA; j++) A[j + lenA * i] = A[j];
        int LenA = lenA * tmp;
        ans = 0;
        for(int i = 0; i < 3; i++) {
            for(int j = 0; j < lenA; j++) A[j + LenA + lenA * i] = A[j];
            KMP(LenA+lenA*i+lenA, A, lenB, B);
            if(ans == 1) {
                printf("%d
", tmp+i+1);
                break;
            }
        }
        if (!ans) puts("-1");
    }
    return 0;
}

D 缩写命名

二分图，很裸，也没卡匈牙利

#include<bits/stdc++.h>
using namespace std;
const int N = 200 + 7;
const int INF = 0x3f3f3f3f;

struct Hungary{
    vector <int> G[N];
    bool used[N];// main里面记得memset
    int girl[N], n;
    inline void init(int _n){
        n = _n;
        for (int i = 0; i <= n; i++) G[i].clear();
    }
    inline void addEdge(const int &u, const int &v){
        G[u].push_back(v);
        //printf("addEdge %d %d
", u, v);
    }

    bool Find(int x){
        for (int i = 0; i < G[x].size(); i++){    //扫描每个妹子
            int j = G[x][i];
            if (used[j]) continue;
            // 如果有暧昧并且还没有标记过
            // (这里标记的意思是这次查找曾试图改变过该妹子的归属问题，
            // 是没有成功，所以就不用瞎费工夫了）
            used[j] = 1;
            if (girl[j] == -1 || Find(girl[j])) {
                //名花无主或者能腾出个位置来，这里使用递归
                girl[j] = x;
                return true;
            }
        }
        return false;
    }

    inline int hungary(const int &n, const int &m){
        int all = 0;
        memset(girl, -1, sizeof girl);
        for (int i = 0; i < n; i++) {
            memset(used, 0, sizeof(used)); //这个在每一步中清空
            if (Find(i)) all += 1;
        }
        //for (int i = 0; i < m; i++) printf("girl[%d] = %d
", i, girl[i]);
        //printf("all = %d
", all);
        return all;
    }
} hg;

string w[N];

int main(){
    //freopen("in.txt", "r", stdin);
    int _;
    scanf("%d", &_);
    string s;
    for (int n; _--;){
        scanf("%d", &n);
        cin >> s;
        int len = s.length();
        for (int i = 0; i < n; i++) cin >> w[i];
        hg.init(len);
        for (int i = 0; i <= len; i++){
            for (int j = 0; j < n; j++){
                if (w[j].find(s[i]) == string::npos) continue;
                hg.addEdge(i, j);
            }
        }
        int mat = hg.hungary(len, n);
        if (mat >= len) puts("Yes");
        else puts("No");
    }
    return 0;
}