Nearest Opposite Parity最短路径+超级源点

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are given an array a consisting of n integers. In one move, you can jump from the position i to the position i−ai (if 1≤i−ai) or to the position i+ai (if i+ai≤n).

For each position i from 1 to n you want to know the minimum the number of moves required to reach any position j such that aj has the opposite parity from ai (i.e. if ai is odd then aj has to be even and vice versa).

input

10
4 5 7 6 7 5 4 4 6 4

output

1 1 1 2 -1 1 1 3 1 1 

/// spfa最最短路，建立两个超级源点，一个连接奇数点一个链接偶数点，在建边的时候，边权设置为 1

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e6 + 7;
typedef int itn;
int n;
int a[maxn];
struct Node{
	int u;
	int v;
	int next;
	int w;
}edge[maxn]; 
int cnt;
int head[maxn];
int dis[maxn];
itn ans[maxn];
itn vis[maxn];
void _Init(){
	cnt = 0;
	for(int i=0;i<maxn;i++) head[i] = -1;
}
void _Add(int u,int v){
	edge[cnt].u = u;
	edge[cnt].v = v;
	edge[cnt].w = 1;
	edge[cnt].next = head[u];
	head[u] = cnt ++;
}
void _SPFA(int u){
	for(int i=1;i<maxn;i++){
		dis[i] = 0x3f3f3f3f;
		vis[i] = 0;
	}
	dis[u] = 0;
	vis[u] = 1;
	queue<int> que;
	///cout << que.size() <<endl;
	que.push(u);
	while(que.size()){
		int t = que.front();
		que.pop();
		vis[t] = 0;
		for(int i = head[t]; ~i;i = edge[i].next){
			int to = edge[i].v;
			if(dis[to] > dis[t] + edge[i].w){
				dis[to] = dis[t] + edge[i].w;
				if(vis[to] == 0){
					que.push(to);
					vis[to] = 1;
				} 
			}
			///puts("ok");
		}
	} 
}
int main(){
	cin >> n;
	_Init();
	for(int i=1;i<=n;i++) cin >> a[i];
	for(int i=1;i<=n;i++){
		if(i - a[i] >= 1) _Add(i - a[i],i);
		if(i + a[i] <= n) _Add(i + a[i],i);
		if(a[i] % 2 == 0) _Add(n+1,i);
		else _Add(n+2,i);
	}
	
	_SPFA(n + 1);
	///puts("okay");
	for(int i=1;i<=n;i++){
		if(a[i] % 2) ans[i] = dis[i];
	}
	_SPFA(n + 2);
	for(int i=1;i<=n;i++){
		if(a[i] % 2 == 0) ans[i] = dis[i];
	}
	for(int i=1;i<=n;i++){
		if(ans[i] == 0x3f3f3f3f) printf("-1 ");
		else printf("%d ",ans[i] - 1);
	}
	///cout << 0x3f3f3f3f <<endl;
	return 0;
}