E-Escape from the Island (shortest path + dp)

E - Escape from the Island

Big brother problem solving , code wind really loves

State expression: $f(u,j)$ currently at$u$ point,jjhas been drawnThe shortest distance from the end point at step$j$ .
State transition:
take the initiative to draw one step min, then move to the next point$v$
$f(u,j)=f(v,j+1)+1,(u,v)\in E,(u,v)\in E$
drifts max along the water and moves to the next point$v$
$f(u,j)=f(v,0)+1,(u,v)\in E$

Since we know the state of the final point is $f(n,j)=0$ So consider that bfs is updated backwards and
take the initiative to draw a step:$(v,j+1)\to (u,j)$
Drift one step along the water:$(v,0)\to (u,j)$

Very dt is the longest path update, while bfs is the shortest update,
such as from $(u,j)\to (v_1,0),(v_2,0)\dots (v_k,0)$ draws a step. During the bfs process,the point closesttothe end pointwill alwaysbe dequeued first. You might as well set the dequeue order as$(v_1,0),(v_2,0)\dots (v_k,0)$即 默认$f(v_1,0)\le f(v_2,0)\le \cdots \le f(v_k,0)$ , we only need to let the last dequeue point$(v_k,0)$ update$(u,j)$ OK, that is to recorduuThe out degree of$u$ , when the out degree is 0, it is the last point, update it.

For the case of no bounds, we let it update in place

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
const int N=100010,M=200010;
int h[N],e[M],ne[M],idx;
int d[N];
bool nd[N];
void add(int a,int b)
{
    
    
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
int f[N][55];
int n,m,k;
void init()
{
    
    
    memset(h,-1,sizeof(int)*(n+1));idx=0;
    memset(d,0,sizeof(int)*(n+1));
    memset(nd,0,sizeof(bool)*(n+1));
    for(int i=1;i<=n;i++)
        for(int j=0;j<=k;j++)
            f[i][j]=0x3f3f3f3f;
} 
queue<pii> q;
void update(int a,int b,int c,int d)
{
    
    
    if(f[c][d]==0x3f3f3f3f)
        f[c][d]=f[a][b]+1,q.push({
    
    c,d});
}
void bfs()
{
    
    
    for(int i=0;i<=k;i++)
        f[n][i]=0,q.push({
    
    n,i});
    while(q.size())
    {
    
    
        auto [u,t]=q.front();q.pop();
        if(t>0)
            for(int i=h[u];i!=-1;i=ne[i])
                update(u,t,e[i],t-1);
        else
        {
    
       // t==0
            for(int i=h[u];i!=-1;i=ne[i])
            {
    
       //0代表顺着水流 1代表逆这水流
                if(i%2==0) continue;  //倒着更新需要逆着水流 
                --d[e[i]];
                if(!d[e[i]])
                    for(int j=0;j<=k;j++)
                        update(u,t,e[i],j);
            }
            if(nd[u])
                for(int j=0;j<=k;j++)
                    update(u,t,u,j);
        }
    }
}
int main()
{
    
    
    IO;
    int T=1;
    cin>>T;
    for(int ca=1;ca<=T;ca++)
    {
    
    
        cin>>n>>m>>k;
        init();
        for(int i=1;i<=m;i++)
        {
    
    
            int u,v;
            cin>>u>>v;
            add(u,v),add(v,u);
            ++d[u];
        }
        for(int i=1;i<=n;i++)
            if(!d[i]) nd[i]=1;//没有出度
        bfs();
        printf("Case #%d:\n",ca);
        for(int i=1;i<=n;i++)
            printf("%d\n",(f[i][0]==0x3f3f3f3f?-1:f[i][0]));
    }
    return 0;
}

Come on~