BZOJ 1562 变换序列(二分图匹配)

时间：2024-10-12 19:34:55

显然每个位置只有两个情况，所以用二分图最大匹配来求解。

如果二分图有完全匹配，则有解。

关键是如何求最小的字典序解。

实际上用匈牙利算法从后面开始找增广路，并优先匹配字典序小的即可。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

//# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Edge{int p, next;}edge[N<<];

int head[N], cnt=, res[N];

int linker[N], uN;

bool used[N];

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}

bool dfs(int u){

    for (int i=head[u]; i; i=edge[i].next) {

        int v=edge[i].p;

        if (!used[v]) {

            used[v]=true;

            if (linker[v]==-||dfs(linker[v])) {linker[v]=u; return true;}

        }

    }

    return false;

}

int hungary(){

    int res=;

    mem(linker,-);

    FO(u,,uN) {

        mem(used,);

        if (dfs(u)) ++res;

    }

    return res;

}

int main ()

{

    int n, x;

    scanf("%d",&n); uN=n;

    FO(i,,n) {

        scanf("%d",&x);

        int u=(i+x)%n, v=(i-x+n)%n;

        if (u<v) swap(u,v);

        if (u!=v) add_edge(i,v);

        add_edge(i,u);

    }

    int ans=hungary();

    if (ans!=n) puts("No Answer");

    else {

        FO(i,,n) res[linker[i]]=i;

        FO(i,,n) printf(i==?"%d":" %d",res[i]);

        putchar('\n');

    }

    return ;

}

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