ABC319 G - Counting Shortest Paths

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解题思路

按照到 $1$ 的距离远近，进行分层
$1$ 为第一层
分层步骤：用一个集合记录还未定层的点，用 $bfs$ 逐层确定
对于当前点 $x$
与其有连边的（不是删边）且还未确定的点，确定为 $x$ 的下一层，入队列
没连边且还未确定的点，入集合（每次决策建立新集合，用浅拷贝）
直到集合为空
此时，到 $n$ 的最优距离确定
长度为 $dis[n]$ 的方案数即为答案
统计方案数，考虑容斥
逐层统计
对于当前层，不考虑删边，到该层每一个点的最优距离的方案数（dis+1）为上一层的方案数
考虑删边，若 $u\rightarrow v$ 删去，则 $u$ 的方案不会被统计到 $v$ 中（不是最优，+1）， $f[v]=(f[v]-f[x]+md)$
注意无法到达的点距离为 $0$ ，需处理
最终答案为 $f[n]$
对于每层的点不会有重复，使用 $Set<Integer>[] dep=new HashSet[n+1];$
或 $Treeset$ (稍慢，因为有排序)
防止超时（ $Vector$ 太慢）


import java.io.*;
import java.math.BigInteger;
import java.util.*;//implements Runnable
public class Main {static long md=(long)998244353;static long Linf=Long.MAX_VALUE/2;static int inf=Integer.MAX_VALUE/2;static int N=200010;static int n=0;static int m=0;static void solve() throws Exception{AReader input=new AReader();
//        String fileName="C:\\Users\\Lenovo\\Downloads\\055.txt";
//		Scanner input=new Scanner(new FileReader(fileName));//        BufferedReader input = new BufferedReader(new FileReader(fileName));PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));String al="abcdefghijklmnopqrstuvwxyz";char[] ac=al.toCharArray();n=input.nextInt();m=input.nextInt();Set<Integer>[] e=new Set[N+1];for(int i=1;i<=n;++i)e[i]=new HashSet<>();HashSet<Integer> hs=new HashSet<>();for(int i=2;i<=n;++i)hs.add(i);for(int i=1;i<=m;++i){int u=input.nextInt();int v=input.nextInt();e[u].add(v);e[v].add(u);}int[] dis=new int[N+1];Queue<Integer> q=new LinkedList<>();q.add(1);dis[1]=1;while(!q.isEmpty()){HashSet<Integer> now=new HashSet<>();int x=q.poll();for(int v:hs){if(!e[x].contains(v)){dis[v]=dis[x]+1;q.add(v);}else{now.add(v);}}hs=now;}if(dis[n]==0){out.println("-1");}else{Set<Integer>[] dep=new HashSet[n+1];for(int i=0;i<=n;++i)dep[i]=new HashSet<>();for(int i=1;i<=n;++i){dep[dis[i]].add(i);}long[] f=new long[N+1];long sum=1;f[1]=1;for(int i=2;i<=dis[n];++i){for(int x:dep[i]){f[x]=sum;}for(int x:dep[i-1]){for(int v:e[x]){if(dep[i].contains(v)){f[v]=(f[v]+md-f[x])%md;}}}sum=0;for(int x:dep[i]){sum=(sum+f[x])%md;}}out.println(f[n]);}out.flush();out.close();}public static void main(String[] args) throws Exception{solve();}//	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Tx2(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}staticclass AReader{BufferedReader bf;StringTokenizer st;BufferedWriter bw;public AReader(){bf=new BufferedReader(new InputStreamReader(System.in));st=new StringTokenizer("");bw=new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException{return bf.readLine();}public String next() throws IOException{while(!st.hasMoreTokens()){st=new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException{//确定下一个token只有一个字符的时候再用return next().charAt(0);}public int nextInt() throws IOException{return Integer.parseInt(next());}public long nextLong() throws IOException{return Long.parseLong(next());}public double nextDouble() throws IOException{return Double.parseDouble(next());}public float nextFloat() throws IOException{return Float.parseFloat(next());}public byte nextByte() throws IOException{return Byte.parseByte(next());}public short nextShort() throws IOException{return Short.parseShort(next());}public BigInteger nextBigInteger() throws IOException{return new BigInteger(next());}public void println() throws IOException {bw.newLine();}public void println(int[] arr) throws IOException{for (int value : arr) {bw.write(value + " ");}println();}public void println(int l, int r, int[] arr) throws IOException{for (int i = l; i <= r; i ++) {bw.write(arr[i] + " ");}println();}public void println(int a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(int a) throws IOException{bw.write(String.valueOf(a));}public void println(String a) throws IOException{bw.write(a);bw.newLine();}public void print(String a) throws IOException{bw.write(a);}public void println(long a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(long a) throws IOException{bw.write(String.valueOf(a));}public void println(double a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(double a) throws IOException{bw.write(String.valueOf(a));}public void print(char a) throws IOException{bw.write(String.valueOf(a));}public void println(char a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}}
}

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