Codeforces Round #290 (Div. 1) A, B, C

A：对于每个名字，和上一个人比较一下，可以知道哪些字母应该在哪些字母前面，然后拓扑排序判一下是否有环，要注意判断是否存在字符串等于前一个字符串的前缀，有环就是矛盾，没环就输出拓扑序即可

B：其实只要选一些数字gcd能满足1，就是可以构造无限多的数字（这个跟辗转相除法有关系），然后题目就转换成，选一些数字使得gcd为1的最小代价，那么进行背包即可，dp[i][j]表示选到i个数字，gcd为j的最小代价，由于数字挺大的，但是实际要存的数字不多，用map存即可

C：神奇的网络流，首先题目不会加出2这个素数，所以加出来的素数都是奇数，那必然是一个奇数一个偶数相邻，这样的话，可以构造一个二分图，左边奇数，右边偶数，由于一个奇数要匹配两个偶数，一个偶数匹配两个奇数，所以只要源点连向奇数容量2，偶数连向汇点容量2，然后满足的奇数偶数连一个边容量1，跑一下最大流，判断流量是否为n即可，然后这题还要输出路径，那么就把匹配边找出来，就可以知道奇数偶数分别要匹配哪两个，然后去搜一遍即可

代码：

A：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
using namespace std;

int n, flag = 0, du[30];
char str[105][105];
vector<int> g[30];
int out[30], on = 0;

void gao() {
    queue<int> Q;
    for (int i = 0; i < 26; i++)
        if (!du[i]) Q.push(i);
    while (!Q.empty()){
        int u = Q.front();
        out[on++] = u;
        Q.pop();
        for (int j = 0; j < g[u].size(); j++) {
            int v = g[u][j];
            du[v]--;
            if (du[v] == 0) Q.push(v);
        }
    }
    for (int i = 0; i < 26; i++)
        if (du[i]) flag = 1;
    if (flag) printf("Impossible\n");
    else {
        for (int i = 0; i < on; i++)
            printf("%c", out[i] + 'a');
        printf("\n");
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%s", str[i]);
        if (i == 0) continue;
        int len1 = strlen(str[i - 1]);
        int len2 = strlen(str[i]);
        int u1 = 0, u2 = 0;
        while (u1 != len1 && u2 != len2) {
            if (str[i - 1][u1] == str[i][u2]) {
                u1++; u2++;
            } else {
                g[str[i - 1][u1] - 'a'].push_back(str[i][u2] - 'a');
                du[str[i][u2] - 'a']++;
                break;
            }
        }
        if (u2 == len2 && len1 > len2) flag = 1;
    }
    gao();
    return 0;
}
           

B：

#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

const int N = 305;

map<int, int> dp[2];
map<int, int>::iterator it;

int n, l[N], c[N];

int gcd(int a, int b) {
    if (!b) return a;
    return gcd(b, a % b);
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &l[i]);
    for (int i = 0; i < n; i++) scanf("%d", &c[i]);
    int now = 0, pre = 1;
    dp[now].clear();
    dp[now][0] = 0;
    for (int i = 0; i < n; i++) {
        swap(now, pre);
        dp[now].clear();
        for (it = dp[pre].begin(); it != dp[pre].end(); it++) {
            int u = it->first, w = it->second;
            int v = gcd(u, l[i]), vw = w + c[i];
            if (!dp[now].count(u)) dp[now][u] = w;
            else dp[now][u] = min(dp[now][u], w);
            if (!dp[now].count(v)) dp[now][v] = vw;
            else dp[now][v] = min(dp[now][v], vw);
        }
    }
    if (dp[now].count(1)) printf("%d\n", dp[now][1]);
    else printf("-1\n");
    return 0;
}
           

C：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 205 * 2;
const int MAXEDGE = 100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

const int N = 205;

int n, a, odd[N], even[N], oddid[N], evenid[N];
int on, en;
int vis[20005], prime[20005], pn = 0;

struct Edge {
    int u, v;
    Type cap, flow;
    Edge() {}
    Edge(int u, int v, Type cap, Type flow) {
        this->u = u;
        this->v = v;
        this->cap = cap;
        this->flow = flow;
    }
};

struct Dinic {
    int n, m, s, t;
    Edge edges[MAXEDGE];
    int first[MAXNODE];
    int next[MAXEDGE];
    bool vis[MAXNODE];
    Type d[MAXNODE];
    int cur[MAXNODE];
    int match[MAXNODE][2], mn[MAXNODE];
    int vv[MAXNODE];
    vector<int> cut;

    void init(int n) {
        this->n = n;
        memset(first, -1, sizeof(first));
        memset(mn, 0, sizeof(mn));
        m = 0;
    }
    void add_Edge(int u, int v, Type cap) {
        edges[m] = Edge(u, v, cap, 0);
        next[m] = first[u];
        first[u] = m++;
        edges[m] = Edge(v, u, 0, 0);
        next[m] = first[v];
        first[v] = m++;
    }

    bool bfs() {
        memset(vis, false, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = true;
        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            for (int i = first[u]; i != -1; i = next[i]) {
                Edge& e = edges[i];
                if (!vis[e.v] && e.cap > e.flow) {
                    vis[e.v] = true;
                    d[e.v] = d[u] + 1;
                    Q.push(e.v);
                }
            }
        }
        return vis[t];
    }

    Type dfs(int u, Type a) {
        if (u == t || a == 0) return a;
        Type flow = 0, f;
        for (int &i = cur[u]; i != -1; i = next[i]) {
            Edge& e = edges[i];
            if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[i^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    Type Maxflow(int s, int t) {
        this->s = s; this->t = t;
        Type flow = 0;
        while (bfs()) {
            for (int i = 0; i < n; i++)
                cur[i] = first[i];
            flow += dfs(s, INF);
        }
        return flow;
    }

    void solve() {
        for (int i = 0; i < m; i += 2) {
            if (edges[i].u == n - 2 || edges[i].v == n - 1) continue;
            if (edges[i].flow == 0) continue;
            match[oddid[edges[i].u]][mn[oddid[edges[i].u]]++] = evenid[edges[i].v - on];
            match[evenid[edges[i].v - on]][mn[evenid[edges[i].v - on]]++] = oddid[edges[i].u];
        }
        memset(vv, 0, sizeof(vv));
        int out[205][205], hn = 0, on[205];
        memset(on, 0, sizeof(on));
        for (int i = 1; i <= n - 2; i++) {
            if (vv[i]) continue;
            on[hn] = 0;
            int s = i;
            while (1) {
                vv[s] = 1;
                out[hn][on[hn]++] = s;
                if (!vv[match[s][0]]) {
                    s = match[s][0];
                    continue;
                }
                if (!vv[match[s][1]]) {
                    s = match[s][1];
                    continue;
                }
                break;
            }
            hn++;
        }
        printf("%d\n", hn);
        for (int i = 0; i < hn; i++) {
            printf("%d", on[i]);
            for (int j = 0; j < on[i]; j++)
                printf(" %d", out[i][j]);
            printf("\n");
        }
    }
} gao;

int main() {
    for (int i = 2; i < 20005; i++) {
        if (vis[i]) continue;
        prime[pn++] = i;
        for (int j = i * i; j < 20005; j += i) {
            vis[j] = 1;
        }
    }
    scanf("%d", &n);
    gao.init(n + 2);
    on = en = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a);
        if (a % 2) {
            oddid[on] = i;
            odd[on++] = a;
        }
        else {
            evenid[en] = i;
            even[en++] = a;
        }
    }
    for (int i = 0; i < on; i++) gao.add_Edge(n, i, 2);
    for (int i = 0; i < en; i++) gao.add_Edge(i + on, n + 1, 2);
    for (int i = 0; i < on; i++)
        for (int j = 0; j < en; j++) {
            if (!vis[odd[i] + even[j]])
                gao.add_Edge(i, on + j, 1);
    }
    if (gao.Maxflow(n, n + 1) != n) printf("Impossible\n");
    else {
        gao.solve();
    }
    return 0;
}