题意：
思路：
因为最终状态是不确定的，考虑反向建图，这样如果可以到达一个偶数点，那么求得的距离肯定是所有能够到达他的奇数点的距离最小的。
代码：
///#pragma GCC optimize(3) ///#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline") ///#pragma GCC optimize(2) #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<ll,ll>PLL; typedef pair<int,int>PII; typedef pair<double,double>PDD; #define I_int ll inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } char F[200]; inline void out(I_int x) { if (x == 0) return (void) (putchar('0')); I_int tmp = x > 0 ? x : -x; if (x < 0) putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0) putchar(F[--cnt]); //cout<<" "; } ll ksm(ll a,ll b,ll p){ll res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;} const ll inf = 0x3f3f3f3f3f3f3f3f; const int maxn=1e6+100,N=1e6+100; const double PI = atan(1.0)*4; const double eps=1e-6; int h[maxn],idx; struct node{ int e,ne; }edge[maxn*2]; void add(int u,int v){ edge[idx].e=v,edge[idx].ne=h[u],h[u]=idx++; } int a[maxn]; int n; int res[maxn],dp[maxn]; int vis[maxn]; void bfs1(){///奇数变为偶数 queue<int>q; for(int i=1;i<=n;i++){ if(a[i]%2) q.push(i),dp[i]=0,vis[i]=0; else vis[i]=-1,dp[i]=-1; } while(!q.empty()){ int t=q.front();q.pop(); for(int i=h[t];~i;i=edge[i].ne){ int j=edge[i].e; if(vis[j]==-1){ q.push(j); vis[j]=1; dp[j]=dp[t]+1; } } } } int dp1[maxn]; int vis1[maxn]; void bfs2(){///偶数能否bfs到奇数 queue<int>q; for(int i=1;i<=n;i++){ if(a[i]%2==0) q.push(i),dp1[i]=0,vis1[i]=0; else vis1[i]=-1,dp1[i]=-1; } while(!q.empty()){ int t=q.front();q.pop(); for(int i=h[t];~i;i=edge[i].ne){ int j=edge[i].e; if(vis1[j]==-1){ dp1[j]=dp1[t]+1; vis1[j]=1; q.push(j); } } } } int main(){ memset(res,0x3f,sizeof res); memset(h,-1,sizeof h); n=read(); for(int i=1;i<=n;i++){ a[i]=read(); int tmp=i+a[i]; if(tmp<=n&&tmp>=1) add(tmp,i); tmp=i-a[i]; if(tmp<=n&&tmp>=1) add(tmp,i); } bfs1();bfs2(); for(int i=1;i<=n;i++) if(a[i]%2) cout<<dp1[i]<<" "; else cout<<dp[i]<<" "; return 0; }
