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<bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int MAXN = 2e5+10;
vector<int> G[MAXN];
int a[MAXN], ans[MAXN], dis[MAXN], vis[MAXN];
int n;
void SPFA(int s)
{
queue<int> 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;
}
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