CodeForces-1272E Nearest Opposite

传送门：CodeForces-1272E

思路分析

设两个超级源点A和B
将所有奇数与A连一条有向边，所有偶数与B连一条有向边，$i$和$i+a[i]$,$i-a[i]$连一条有向边
那么题意可以转化为每个奇数点到达B的最短距离以及每个偶数点到达A的最短距离
直接求显然不行的，所以可以建立反向边，跑两次最短路，分别以A,B为起点，就可以得到各个点到A和B的最短距离了

样例输入

10
4 5 7 6 7 5 4 4 6 4

样例输出

1 1 1 2 -1 1 1 3 1 1

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <chrono>
#include <random>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N=2e5+10;

int n;
int dis[N],a[N],ans[N],vis[N];
vector<int>e[N];

void spfa(int s){
    mem(dis,INF);
    mem(vis,0);
    dis[s]=0;
    queue<int>q;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(auto v:e[u]){
            if(dis[v]>dis[u]+1 && !vis[v]){
                q.push(v);
                vis[v]=1;
                dis[v]=dis[u]+1;
            }
        }
    }
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    read(n);
    for(int i=1;i<=n;i++) {
        read(a[i]);
        if(i-a[i]>=1) e[i-a[i]].push_back(i);
        if(i+a[i]<=n) e[i+a[i]].push_back(i);
    }
    for(int i=1;i<=n;i++) {
        if(a[i]&1) e[0].push_back(i);
        else e[n+1].push_back(i);
    }
    spfa(0);
    for(int i=1;i<=n;i++) if(a[i]%2==0) ans[i]=(dis[i]==INF?-1:dis[i]-1);
    spfa(n+1);
    for(int i=1;i<=n;i++) if(a[i]&1) ans[i]=(dis[i]==INF?-1:dis[i]-1);
    for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
    



    return 0;
}