P1948 [USACO08JAN] Telephone Lines S

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典中之典！！

解题思路

可选k条边代价为0
如何决策？

$1.$

将到当前位置选择了几条代价为0的边放入状态，即 $dis[u][cnt],Node(u,cnt,dis)$
若当前状态选的边数小于 $k$ ，则可以进行决策，是否选择当前边，
若选，则 $cnt+1,dis+0$ ，否则 $cnt,Math.max(dis,w_{u->v})$
若当前状态选的边数等于 $k$ ，则只能 $cnt,Math.max(dis,w_{u->v})$

$2.$

考虑分层图，建k层
（实际上就是将上一种方法中多一维记录 $cnt$ 来区别状态），改为用点的标号区分）
对于每条边， $i$ 向 $i+1$ 层连一条相同但代价为0的边
同层连代价为 $w$ 的边
最终输出第k层的 $dis[n]$

for(int i=1;i<=m;++i){int x=read(),y=read(),w=read();for(int j=0;j<=k;++j){Lian(x+j*n,y+j*n,w);Lian(y+j*n,x+j*n,w);if(j==k)continue;Lian(x+j*n,y+(j+1)*n,0);Lian(y+j*n,x+(j+1)*n,0);}}

$3.$

由于代价具有单调性（代价低的，花费更高代价一定也行）
二分答案
对于每条边的边权有小到大排序去重，将边权视为答案进行 $check$
$check$ 时，对大于当前所选答案的边，视作代价为1的边，小于的视作代价为0
判断最终 $dis[n]\leq k$ ，即为可能解，接着往小进行尝试
否则，往大进行尝试

//二分答案
import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;//implements Runnable
public class Main{static long md=(long)20020219;static long Linf=Long.MAX_VALUE/2;static int inf=Integer.MAX_VALUE/2;static int M=10000;static int N=1000;static int n=0;static int m=0;static int k=0;static class Edge{int fr,to,val,nxt;public Edge(int u,int v,int w) {fr=u;to=v;val=w;}}static Edge[] e=new Edge[2*M+1];static int[] head=new int[N+1];static int cnt=0;static void addEdge(int fr,int to,int val) {cnt++;e[cnt]=new Edge(fr, to, val);e[cnt].nxt=head[fr];head[fr]=cnt;}staticclass Node{int x;int y;public Node(int u,int v) {x=u;y=v;}}static boolean Dij(int s,int lim) {int[] dis=new int[N+1];Arrays.fill(dis, inf);boolean[] vis=new boolean[N+1];PriorityQueue<Node> q=new PriorityQueue<Node>((o1,o2)->{return o1.y-o2.y;});q.add(new Node(s, 0));dis[s]=0;while(!q.isEmpty()) {Node now=q.peek();q.poll();int x=now.x;if(vis[x])continue;vis[x]=true;for(int i=head[x];i>0;i=e[i].nxt) {int v=e[i].to;if(vis[v])continue;int w=e[i].val;if(w>lim) {if(dis[x]+1<dis[v]) {dis[v]=dis[x]+1;q.add(new Node(v, dis[v]));}}else {if(dis[x]<dis[v]) {dis[v]=dis[x];q.add(new Node(v, dis[v]));}}}}if(dis[n]<=k)return true;else return false;}public static void main(String[] args) throws Exception{AReader input=new AReader();PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	n=input.nextInt();m=input.nextInt();k=input.nextInt();int[] a=new int[m+1];int aa=0;int[] b=new int[m+1];int bb=0;for(int i=1;i<=m;++i) {int u=input.nextInt();int v=input.nextInt();int w=input.nextInt();addEdge(u, v, w);addEdge(v, u, w);aa++;a[aa]=w;}Arrays.sort(a,1,aa+1);for(int i=1;i<=aa;++i) {if(a[i]!=a[i-1]) {bb++;b[bb]=a[i];}}int l=1,r=bb;int ans=-1;while(l<=r) {int mid=(l+r)>>1;if(Dij(1, b[mid])) {ans=b[mid];r=mid-1;}else l=mid+1;}out.print(ans);out.flush();out.close();}
//	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Main(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}
//	static 
//	class Node{
//		int x;
//		int y;
//		public Node(int u,int v) {
//			x=u;
//			y=v;
//		}
//		@Override
//	    public boolean equals(Object o) {
//	        if (this == o) return true;
//	        if (o == null || getClass() != o.getClass()) return false;
//	        Node may = (Node) o;
//	        return x == may.x && y==may.y;
//	    }
//
//	    @Override
//	    public int hashCode() {
//	        return Objects.hash(x, y);
//	    }
//	}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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