Codeforces Round #605 (Div. 3) E. Nearest Opposite Parity（最短路）

链接：

https://codeforces.com/contest/1272/problem/E

题意：

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).

思路：

写了好久DFS发现有环不能处理。。看了大佬博客才懂。
考虑从a开始的最短路，计算的是a到x的最短路。
反向建图，跑出来的最短路，就是他能到达的点往自己的最短路。
建立两个新点，分别连奇数点和偶数点，跑最短路。

代码：

#include
using namespace std;
const int INF = 1e9;
const int MAXN = 2e5+10;

vector G[MAXN];
int a[MAXN], ans[MAXN], dis[MAXN], vis[MAXN];
int n;

void SPFA(int s)
{
    queue que;
    for (int i = 1;i <= n+2;i++)
        dis[i] = INF, vis[i] = 0;
    que.push(s);
    dis[s] = 0;
    vis[s] = 1;
    while(!que.empty())
    {
        int u = que.front();
        que.pop();
        vis[u] = 0;
        for (int i = 0;i < (int)G[u].size();++i)
        {
            int node = G[u][i];
            if (dis[node] > dis[u]+1)
            {
                dis[node] = dis[u]+1;
                if (vis[node] == 0)
                {
                    que.push(node);
                    vis[node] = 1;
                }
            }
        }
    }
}

int main()
{
    cin >> n;
    for (int i = 1;i <= n;++i)
        cin >> a[i];
    for (int i = 1;i <= n;++i)
    {
        if (i-a[i] >= 1)
            G[i-a[i]].push_back(i);
        if (i+a[i] <= n)
            G[i+a[i]].push_back(i);
    }
    for (int i = 1;i <= n;++i)
    {
        if (a[i]%2 == 1)
            G[n+1].push_back(i);
        else
            G[n+2].push_back(i);
    }
    SPFA(n+1);
    for (int i = 1;i <= n;++i)
        if (a[i]%2 == 0) ans[i] = dis[i];
    SPFA(n+2);
    for (int i = 1;i <= n;++i)
        if (a[i]%2 == 1) ans[i] = dis[i];
    for (int i = 1;i <= n;++i)
        cout << ((ans[i] == INF) ? -1 : ans[i]-1) << ' ' ;
    cout << endl;

    return 0;
}