codeforces 512C Fox And Dinner

这里写链接内容
C. Fox And Dinner
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is ai years old.
They will have dinner around some round tables. You want to distribute foxes such that:
1. Each fox is sitting at some table.
2. Each table has at least 3 foxes sitting around it.
3. The sum of ages of any two adjacent foxes around each table should be a prime number.
If k foxes f1, f2, …, fk are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: fi and fi + 1 are adjacent, and f1 and fk are also adjacent.
If it is possible to distribute the foxes in the desired manner, find out a way to do that.
Input
The first line contains single integer n (3 ≤ n ≤ 200): the number of foxes in this party.
The second line contains n integers ai (2 ≤ ai ≤ 104).
Output
If it is impossible to do this, output “Impossible”.
Otherwise, in the first line output an integer m (

): the number of tables.
Then output m lines, each line should start with an integer k -=– the number of foxes around that table, and then k numbers — indices of fox sitting around that table in clockwise order.
If there are several possible arrangements, output any of them.
Examples
input
4
3 4 8 9
output
1
4 1 2 4 3
input
5
2 2 2 2 2
output
Impossible
input
12
2 3 4 5 6 7 8 9 10 11 12 13
output
1
12 1 2 3 6 5 12 9 8 7 10 11 4
input
24
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
output
3
6 1 2 3 6 5 4
10 7 8 9 12 15 14 13 16 11 10
8 17 18 23 22 19 20 21 24
Note
In example 1, they can sit around one table, their ages are: 3-8-9-4, adjacent sums are: 11, 17, 13 and 7, all those integers are primes.
In example 2, it is not possible: the sum of 2+2 = 4 is not a prime number.
首先我们可以通过观察发现只有奇数和偶数互相组合才有可能构成质数 首先%icefox 他发现环一定是偶环所以这应该是一个二分图
所以我把奇数放在左边 偶数放在右边 如果这个奇数加这个偶数可以构成质数 那么我把这个奇数向这个偶数连一条容量为1的边 然后从源点向每个奇数连一条容量为2的边 从每个偶数向汇点连一条容量为2的边表示 然后跑一遍最大流 如果满流 说明说明答案可行 那么我就直接dfs一下把我的环都搜出来 然后输出即可 注意判断素数的时候因为我写的是线性筛所以 至少要筛到两万才可以 注意判断无解情况 比如我奇数和偶数的个数不相同 搜索的话 如果搜索到已经访问过的点 那么那个点就不能搜索下去了

#include<cstdio>#include<vector>#include<cstring>#include<queue>#define N 220#define inf 0x3f3f3f3f#include<algorithm>using namespace std;inline char gc(){    static char now[1<<16],*S,*T;    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}    return *S++;}inline int read(){    int x=0;char ch=gc();    while(ch<'0'||ch>'9') ch=gc();    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}    return x;}struct node{    int y,z,next,x;}data[N*N];int num=1,prime[21000],h[N],T,n,level[N],age[N];bool not_prime[21000],visit[N];vector<int> print[N];inline void insert1(int x,int y,int z){    data[++num].y=y;data[num].z=z;data[num].next=h[x];h[x]=num;data[num].x=x;    data[++num].y=x;data[num].z=0;data[num].next=h[y];h[y]=num;data[num].x=y;}inline bool bfs(){    queue<int>q;memset(level,0,sizeof(level));level[0]=1;q.push(0);    while(!q.empty()){        int x=q.front();q.pop();        for (int i=h[x];i;i=data[i].next){            int y=data[i].y,z=data[i].z;            if(level[y]||!z) continue;level[y]=level[x]+1;q.push(y);if (y==T) return 1;        }    }return  0;}inline int dfs(int x,int s){    if (x==T) return s;int ss=s;    for (int i=h[x];i;i=data[i].next){        int y=data[i].y,z=data[i].z;        if (level[x]+1==level[y]&&z){            int xx=dfs(y,min(s,z));if (!x) level[y]=0;            s-=xx;data[i].z-=xx;data[i^1].z+=xx;if (!s) return ss;        }    }return ss-s;}inline void dfs1(int x,int id,int op){    print[id].push_back(x);visit[x]=1;    for (int i=h[x];i;i=data[i].next){        int y=data[i].y,z=data[i].z;if (visit[y]) continue;        if ((i^op)==0) continue;if (z^op) dfs1(y,id,op^1);    }}int main(){    freopen("cf.in","r",stdin);    n=read();T=n+1;int odd=0,even=0;    for (int i=1;i<=n;++i) {        age[i]=read();if(age[i]%2) insert1(0,i,2),++odd;else insert1(i,T,2),++even;    }int tot=0;if (odd!=even) {printf("Impossible");return 0;}    for (int i=2;i<=20000;++i){        if (!not_prime[i]) prime[++tot]=i;        for (int j=1;prime[j]*i<=2e4;++j){            not_prime[prime[j]*i]=1;            if (i%prime[j]==0) break;        }    }    for (int i=1;i<=n;++i){        if (age[i]%2==0) continue;        for (int j=1;j<=n;++j){            if (i==j) continue;            int sum=age[i]+age[j];            if (!not_prime[sum]){                insert1(i,j,1);            }        }    }    int ans=0;while(bfs()) ans+=dfs(0,inf);    if (ans!=n) {printf("Impossible");return 0;}tot=0;visit[0]=1;visit[T]=1;    //for (int i=2;i<=num;++i) printf("%d %d %d %d\n",i,data[i].x,data[i].y,data[i].z);    for(int i=h[0];i;i=data[i].next){        int y=data[i].y;if (!visit[y]) dfs1(y,++tot,1);    }printf("%d\n",tot);    for (int i=1;i<=tot;++i){        printf("%d ",print[i].size());        for (int j=0;j<print[i].size();++j) printf("%d ",print[i][j]);printf("\n");    }    return 0;}