hihocoder35 模板场

来源：互联网发布：淘宝店如何实名认证编辑：程序博客网时间：2024/06/01 22:17

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\n", 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\n", 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\n", 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\n", 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\n", 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\n", 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\n", i, girl[i]);        //printf("all = %d\n", 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;}

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