题目
输入格式
输出格式
输入样例
5
1 1 2 2 1
输出样例
1 2 4 0 3
提示
30%的数据中N≤50；
60%的数据中N≤500；
100%的数据中N≤10000。
题解
每个位置可以和两种数匹配，显然是一个二分图匹配问题
但要求字典序最小，我们就按字典序存边
由于在KM算法中，后来着具有更高的优先级，我们倒序匹配即可
#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<\' \'; puts("");
using namespace std;
const int maxn = 10005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == \'-\') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int lk[maxn],vis[maxn],ans[maxn];
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
}
bool find(int u){
Redge(u) if (!vis[to = ed[k].to]){
vis[to] = true;
if (lk[to] == -1 || find(lk[to])){
lk[to] = u;
return true;
}
}
return false;
}
int n,a[maxn],b[maxn];
int main(){
memset(lk,-1,sizeof(lk));
n = read(); int tmp,a,b;
for (int i = 0; i < n; i++){
tmp = read();
a = (i + tmp) % n;
b = (i - tmp + n) % n;
if (a < b) swap(a,b);
build(i,a); build(i,b);
}
for (int i = n - 1; i >= 0; i--){
memset(vis,0,sizeof(vis));
if (!find(i)){
puts("No Answer");
return 0;
}
}
for (int i = 0; i < n; i++) ans[lk[i]] = i;
for (int i = 0; i < n; i++){
printf("%d",ans[i]);
if (i < n - 1) printf(" ");
}
return 0;
}