codeforces-510E Fox And Dinner(带限制的二分图多重匹配+奇偶建图+打印路径)

题意：

有n个fox，分别有一个年龄为a[i]。 现在要将n个fox分配入座，但要求两两相邻的fox的年龄和为质数。一桌(环形桌)至少要有三个fox。求要分几桌才能满足上述要求，并打印每桌有几个fox以及每个fox的具体id。(n <= 200, 2 <= a[i] <= 1e4)

思路：

由于年龄最小是2，所以两两相加为质数只可能是一奇一偶相加所得，所以安排座位就是一个奇数两旁是偶数，一个偶数两旁是奇数，一个桌子上的fox个数是偶数个。所以我们可以把奇数和偶数分到二分图的两部，同样得到特判：当且仅当两部的元素个数相等有解。我们将源点S向奇数部建有向边容量为2，将偶数部向汇点T建有向边容量为2，然后就是奇偶建边，如果两个数之和是质数就建一条从奇数到偶数的有向边容量为1。然后如果求得最大流等于n则有解，否则没有。为什么能用最大流做呢，最大流==n就代表A部的每个点恰好匹配B部的两个点，B部的每个点也恰好匹配A部的两个点。然后根据网络流流量的性质就可以得出所有的路径，打印出来即可。

代码：

#include <algorithm>    #include <iostream>    #include <string.h>  #include <vector>  #include <cstdio>    #include <queue>    using namespace std;    const int inf = 0x3f3f3f3f;    const int maxn = 205;    const int maxm = maxn*maxn;  struct node{int w; int v, next;} edge[maxm];    int pre[maxn], rec[maxn], head[maxn], gap[maxn], now[maxn];    int dis[maxn];  int n, no, up;    int S, T;   queue<int> q;  inline void add(int u, int v, int w)    {        edge[no].v = v; edge[no].w = w;        edge[no].next = head[u]; head[u] = no++;        edge[no].v = u; edge[no].w = 0;        edge[no].next = head[v]; head[v] = no++;    }    inline void pre_init()    {        no = 0;      memset(head, -1, sizeof head);    }    void init(int S, int T)    {        memset(gap, 0, sizeof gap);        memset(dis, 0x3f, sizeof dis);        for(int i = 0; i <= up; ++i)         now[i] = head[i];      while(!q.empty()) q.pop();        dis[T] = 0; q.push(T);        while(!q.empty())       {            int tp = q.front(); q.pop();            ++gap[dis[tp]];            int k = head[tp];            while(k != -1)            {                if(dis[edge[k].v] == inf && edge[k^1].w)                {                    dis[edge[k].v] = dis[tp]+1;                    q.push(edge[k].v);                }                k = edge[k].next;            }        }    }    int SAP(int S, int T)    {      int ans = 0, flow = inf;      int top = S;        pre[S] = S; init(S, T);        while(dis[S] < up)      {          if(top == T)            {                ans += flow;                while(top != S)              {                    edge[rec[top]].w -= flow;                    edge[rec[top]^1].w += flow;                    top = pre[top];                }                flow = inf;            }            int k = now[top];            while(k != -1)            {                int v = edge[k].v;                if(edge[k].w && dis[top] == dis[v]+1)                {                    flow = min(flow, edge[k].w);                    pre[v] = top; rec[v] = k;                    now[top] = k; top = v;                    break;                }                k = edge[k].next;            }            if(k == -1)            {                int mind = up;                if(--gap[dis[top]] == 0) break;              int k = now[top] = head[top];              while(k != -1)                {                    if(edge[k].w && mind>dis[edge[k].v]) mind = dis[edge[k].v];                    k = edge[k].next;                }                ++gap[dis[top] = mind+1];                top = pre[top];            }        }        return ans;    }  int a[maxn];bool isprime[20005], vis[maxn];  vector<int> g[maxn], vt[maxn];void prime_init(){memset(isprime, 0, sizeof isprime);isprime[1] = true;for(int i = 2; i*i <= 20000; ++i)if(!isprime[i])for(int j = i*i; j <= 20000; j+=i)isprime[j] = true;}bool mapping()  { int ttx = 0;    for(int i = 1; i <= n; ++i)        {            scanf("%d", &a[i]);        if(a[i]&1) add(S, i, 2), ++ttx;        else add(i, T, 2);        for(int j = 1; j < i; ++j)        if(!isprime[a[i]+a[j]])        {        if(a[i]&1) add(i, j, 1);        else add(j, i, 1);}    }      return ttx*2 == n;}  void dfs(int id, int u){vis[u] = true;vt[id].push_back(u);for(int i = 0; i < g[u].size(); ++i){int v = g[u][i];if(vis[v]) continue;dfs(id, v);}}void work(){memset(vis, 0, sizeof vis);for(int i = 1; i <= n; ++i) g[i].clear();for(int i = 1; i <= n; ++i){if(!(a[i]&1)) continue;for(int k = head[i]; k+1; k = edge[k].next){int v = edge[k].v;if(v == S) continue;if(!edge[k].w && !(k&1)){g[i].push_back(v);g[v].push_back(i);}}}int cnt = 0;for(int i = 1; i <= n; ++i){if(vis[i]) continue;++cnt; vt[cnt].clear();dfs(cnt, i);}printf("%d\n", cnt);for(int i = 1; i <= cnt; ++i){printf("%d", vt[i].size());for(int j = 0; j < vt[i].size(); ++j)printf(" %d", vt[i][j]);printf("\n");}}int main()    {      while(~scanf("%d", &n))        {            up = n+2, S = 0, T = n+1;          pre_init(); prime_init();           if(!mapping()) puts("Impossible");        else if(SAP(S, T) != n) puts("Impossible");        else work();     }    return 0;    }

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