图论练习3
内容：过程中视条件改变边权，利用树状数组区间加处理
卯酉东海道
题目链接
题目大意
个点，
条有向边，每条边有颜色
和费用
- 总共有
种颜色 - 若当前颜色与要走的边颜色相同，则花费为
- 若当前颜色与要走的边颜色不同，则花费为
，且颜色变为边的颜色 - 出发时可以自定义颜色
- 问
的最小花费
解题思路
- 选边
时，进行判断
- 对于初始自定义颜色，且
，则跑趟最短路
import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main{
static
class Node{
int id;
int col;
long dis;
public Node(int I,int C,long D) {
id=I;
col=C;
dis=D;
}
}
static int cnt=0;
static int[] head;
static Edge[] e;
static
class Edge{
int fr;
int to;
int nxt;
long val;
int col;
}
static void addEdge(int fr,int to,long val,int col) {
cnt++;
e[cnt]=new Edge();
e[cnt].fr=fr;
e[cnt].to=to;
e[cnt].val=val;
e[cnt].nxt=head[fr];
e[cnt].col=col;
head[fr]=cnt;
}
public static void main(String[] args) throws IOException{
AReader input=new AReader();
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int n=input.nextInt();
int m=input.nextInt();
int l=input.nextInt();
int base=input.nextInt();
e=new Edge[m<<1|1];
head=new int[n+1];
for(int i=1;i<=m;++i) {
int u=input.nextInt();
int v=input.nextInt();
int col=input.nextInt();
long w=input.nextLong();
addEdge(u, v, w,col);
}
PriorityQueue que=new PriorityQueue((o1,o2)->{
if(o1.dis-o2.dis>0)return 1;
else if(o1.dis-o2.dis<0)return -1;
else return 0;
});
boolean[] vis=new boolean[n+1];
long[] dis=new long[n+1];
long ans=Long.MAX_VALUE;
for(int o=1;o<=l;++o) {
Node s=new Node(1,o,0);
Arrays.fill(vis, false);
Arrays.fill(dis, Long.MAX_VALUE);
dis[1]=0;
que.add(s);
while(!que.isEmpty()) {
Node now=que.peek();
que.poll();
int u=now.id;
if(vis[u])continue;
long disu=now.dis;
int colu=now.col;
vis[u]=true;
for(int i=head[u];i>0;i=e[i].nxt) {
int v=e[i].to;
int colv=e[i].col;
long w=e[i].val;
if(colv!=colu) {
w=base*w;
}
if(dis[v]>disu+w) {
dis[v]=disu+w;
que.add(new Node(v, colv, dis[v]));
}
}
}
ans=Math.min(ans, dis[n]);
}
if(ans==Long.MAX_VALUE) {
out.println(-1);
}else {
out.println(ans);
}
out.flush();
out.close();
}
static
class 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();
}
}
}
AtCoder abc338 D - Island Tour
题目链接
题目大意
- 个点，条无向边依次连接，第条连接
- 给定序列
，从
依次访问 
的路径长度，定义为经过的边的个数- 问删除一条边后，完成访问的最短路径长度
解题思路
- 先不管删边
- 有两种方式，一种绕一圈，一种直接顺着走
- 两种方式种的较小值加入答案
- 但是有删边，所以删去边后可能导致取到较小值的路径无法通过，需要撤回，改为另一种
- 若删除这条边，则通过它的所有较小值均要改变，即加上差值
- 由于边标号连续，考虑树状数组区间加
- 对于当前较小值经过的所有边，加上差值，作为撤回代价
- 在不考虑删边的情况下统计完答案后，单点查询每条边，取撤回代价最小的边为删边
import java.io.*;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class Main{
static int cnt=0;
static int[] head;
static
class Edge{
int fr;
int to;
int nxt;
long val;
}
static Edge[] e;
static void addEdge(int fr,int to,long val) {
cnt++;
e[cnt]=new Edge();
e[cnt].fr=fr;
e[cnt].to=to;
e[cnt].nxt=head[fr];
head[fr]=cnt;
}
static
class Node{
int x;
long dis;
long h;
public Node(int X,long D,long H) {
x=X;
dis=D;
h=H;
}
}
static
class BIT{
int size;
long[] tr;
public BIT(int n) {
size=n;
tr=new long[n+2];
}
public int lowbit(int x) {
return x&(-x);
}
public void update(int x,long y) {
for(int i=x;i<=size;i+=lowbit(i)) {
tr[i]+=y;
}
}
public long query(int x) {
long res=0;
for(int i=x;i>0;i-=lowbit(i)) {
res+=tr[i];
}
return res;
}
}
public static void main(String[] args) throws IOException{
AReader input=new AReader();
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int n=input.nextInt();
int m=input.nextInt();
BIT Tree=new BIT(n);
int[] X=new int[m+1];
for(int i=1;i<=m;++i) {
X[i]=input.nextInt();
}
long ans=0;
for(int i=1;itor) {
ans+=tor;
cha=tol-tor;
Tree.update(x, cha);
Tree.update(y, -cha);
}else {
ans+=tol;
cha=tor-tol;
Tree.update(1, cha);
Tree.update(x,-cha);
Tree.update(y, cha);
Tree.update(n+1, -cha);
}
}else {
tol=x-y;
tor=n-x+y;
if(tol>tor) {
ans+=tor;
cha=tol-tor;
Tree.update(x, cha);
Tree.update(n+1,-cha);
Tree.update(1, cha);
Tree.update(y, -cha);
}else {
ans+=tol;
cha=tor-tol;
Tree.update(y, cha);
Tree.update(x, -cha);
}
}
}
long shan=Long.MAX_VALUE/2;
for(int i=1;i<=n;++i) {
shan=Math.min(shan, Tree.query(i));
}
// out.println(ans);
out.println(ans+shan);
out.flush();
out.close();
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}
AtCoder abc338 E - Chords
题目链接
题目大意
- 在一个圆上等距离放置了
个点，从某个点开始顺时针编号为
到 - 同时在圆上有条弦，第i条弦连接点
和
。保证所有的
不同 - 判断这些弦是否有交
解题思路
- 集合划分
- 若两个不同集合之间有弦，则有交
- 由于点标号连续，对于每个弦，考虑树状数组区间加
-
之间的点，区间加，表示一个新集合 - 端点无所谓，反正保证所有的不同
- 若
，则
不在一个集合且有弦，满足有交
import java.io.*;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class Main{
static int cnt=0;
static int[] head;
static
class Edge{
int fr;
int to;
int nxt;
long val;
}
static Edge[] e;
static void addEdge(int fr,int to,long val) {
cnt++;
e[cnt]=new Edge();
e[cnt].fr=fr;
e[cnt].to=to;
e[cnt].nxt=head[fr];
head[fr]=cnt;
}
static
class Node{
int x;
long dis;
long h;
public Node(int X,long D,long H) {
x=X;
dis=D;
h=H;
}
}
static
class BIT{
int size;
long[] tr;
public BIT(int n) {
size=n;
tr=new long[n+2];
}
public int lowbit(int x) {
return x&(-x);
}
public void update(int x,long y) {
for(int i=x;i<=size;i+=lowbit(i)) {
tr[i]+=y;
}
}
public long query(int x) {
long res=0;
for(int i=x;i>0;i-=lowbit(i)) {
res+=tr[i];
}
return res;
}
}
public static void main(String[] args) throws IOException{
AReader input=new AReader();
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int n=input.nextInt();
BIT Tree=new BIT(2*n);
boolean ok=true;
for(int i=1;i<=n;++i) {
int x=input.nextInt();
int y=input.nextInt();
if(x>y) {
int t=x;
x=y;
y=t;
}
if(!ok)continue;
long fx=Tree.query(x);
long fy=Tree.query(y);
if(fx!=fy) {
ok=false;
}else {
Tree.update(x, 1);
Tree.update(y, -1);
}
}
if(ok) {
out.println("No");
}else {
out.println("Yes");
}
out.flush();
out.close();
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}