题意
构造一个字典序最小的序列T，使得 Dis（i， Ti） = di，其中i是从0开始的，Dis(x,y)=min{∣x−y∣,N−∣x−y∣} ，di由题目给定。
思路
二分图匹配，把左边的看成i，右边看成Ti，对于固定的i和d，Ti是由两种可能的，连上有向边即可。
至于字典序要最小，怎么做呢，我们可以反着跑匈牙利算法，就是从n-1跑到0，这样小一点的i，可以直接拿走大一点的i刚匹配的较小的值。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 来了老弟 ＼('ㅅ') > ⌒ヽ / へ＼ / / ＼＼ ﾚ ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / 'ノ ) Lﾉ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 1e4+9; struct E{ int v,nxt; }edge[4*maxn]; int head[maxn],gtot; void addedge(int u,int v){ edge[gtot].v = v; edge[gtot].nxt = head[u]; head[u] = gtot++; } int vis[maxn],pt[maxn],py[maxn]; bool dfs(int u){ for(int i=head[u]; ~i; i = edge[i].nxt){ int v = edge[i].v; if(vis[v] == 0){ vis[v] = 1; if(pt[v] == 0 || dfs(pt[v])){ pt[v] = u; py[u] = v; return true; } } } return false; } int main(){ int n; scanf("%d", &n); memset(head, -1, sizeof(head)); priority_queue<int>que; while(!que.empty()) que.pop(); for(int i=0; i<n; i++){ int d; scanf("%d", &d); int a = i-d; if(a>=0&&a<n) que.push(a); a = i+d; if(a>=0&&a<n) que.push(a); int b = i - (n-d); if(b>=0&&b<n) que.push(b); b = i + (n-d); if(b>=0&&b<n) que.push(b); while(!que.empty()){ int t = que.top(); que.pop(); addedge(i, t); } } int res = n; for(int i=n-1; i>=0; i--){ memset(vis, 0, sizeof(vis)); if(dfs(i)) res = i; else break; } if(res == 0) { for(int i=0; i<n; i++) printf("%d ", py[i]); puts(""); } else puts("No Answer"); return 0; }