版权声明:本文为博主原创文章,未经博主允许不得转载。 https://cloud.tencent.com/developer/article/1434709
详细代码可以fork下Github上leetcode项目,不定期更新。
练习题如下:
思路:经历了一些波折,首先是考虑DP,DPj,表示在第j个位置所需要的最短的子序列的长度。
更新式还是比较好推的,因为DPj表示到第j个位置所需要的最短子序列长度,那么到下一个位置时,此时的区间长度为si, ti,那么,只要DPj在si和ti之间就可以更新为: DP[ti] =min{DP[ti], DP[j] + 1}, j >= si && j <= ti
物理含义:只要在区间内,1, j能够被排序,那么只要后续区间包含了j,那么自然也就能更新至1,ti.
初始代码如下:
void solve() {
int N = ni();
int M = ni();
Pair[] ps = new Pair[M];
for (int i = 0; i < M; ++i){
ps[i] = new Pair(ni(), ni());
}
int[] dp = new int[N + 16];
Arrays.fill(dp, INF);
dp[1] = 0;
for (int i = 0; i < M; ++i){
int s = ps[i].s;
int e = ps[i].e;
for (int j = 1; j <= N; ++j){
if (j >= s && j <= e){
dp[e] = Math.min(dp[e], dp[j] + 1);
}
}
}
out.println(dp[ps[M - 1].e]);
}
可以观察下for循环中求最小值,此处可以用RMQ替代,无非就是求区间s,e中的最小值,这样时间复杂度降为O(nlogn)O(n\log n)
哈哈,今天更新了最新的算法模版,拿来POJ实测一波,支持codeforce,uva,aoj哟,代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201708/P1769.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
class Pair{
int s;
int e;
public Pair(int s, int e){
this.s = s;
this.e = e;
}
@Override
public String toString() {
return s + " " + e;
}
}
static final int MAX_N = 50000 + 16;
static final int SIZE = (1 << 18) + 1;
static final int INF = 1 << 29;
int[] dat = new int[SIZE];
int[] dp;
void solve() {
int N = ni();
int M = ni();
Pair[] ps = new Pair[M];
for (int i = 0; i < M; ++i){
ps[i] = new Pair(ni(), ni());
}
dp = new int[MAX_N];
Arrays.fill(dp, INF);
init(N);
dp[1] = 0;
update (1, 0);
for (int i = 0; i < M; ++i){
int s = ps[i].s;
int e = ps[i].e;
int min = Math.min(dp[e],query(0, s, e + 1, 0, n_) + 1);
dp[e] = min;
update(e, min);
}
out.println(dp[N]);
}
/*********************以下是RMQ的实现*********************/
/**
* [L, r)
* @param k
* @param l
* @param r
*/
int n_;
public void init(int N){
n_ = 1;
while (n_ < N) n_ *= 2;
for (int i = 0; i < 2 * n_ - 1; ++i) dat[i] = INF;
}
public void update(int k, int val){
k += (n_ - 1);
dat[k] = val;
while (k > 0){
k = (k - 1) / 2;
dat[k] = Math.min(dat[2 * k + 1], dat[k * 2 + 2]);
}
}
public int query(int k, int i, int j, int l, int r){
if (j <= l || i >= r) return INF;
else if (i <= l && j >= r){
return dat[k];
}
else{
int lch = 2 * k + 1;
int rch = 2 * k + 2;
int mid = (l + r) / 2;
int lf = query(lch, i, j, l, mid);
int rt = query(rch, i, j, mid, r);
return Math.min(lf, rt);
}
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = !System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if (!oj) {
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more() {
return in.hasNext();
}
public int ni() {
return in.nextInt();
}
public long nl() {
return in.nextLong();
}
public double nd() {
return in.nextDouble();
}
public String ns() {
return in.nextString();
}
public char nc() {
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext() {
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}
RMQ的实现有一个坑点,其实和《挑战》P188的递归实现还是有一些出入的,原因在与update更新是利用完全二叉树的性质来做的,而如果单纯的递归实现,如果不控制右边界r,那么会导致不匹配,找了很久错误都没找到。
所以init()中,把N转为最近的2次幂,这样一来,在query查询时,可以以完全二叉树的方式来遍历,这样就和update匹配了。
感觉比上一题简单,可以用DIJKSTRA,建立点与边的关系,也可以用DP,同理DPj表示在第j个位置,所需要消耗的最小代价,把上题的dpj + 1,改成dpj + c,其他的都不变。传统DP超时,所以还是用RMQ吧!
代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201708/P3171.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int INF = 1 << 29;
class Cow implements Comparable<Cow>{
int s;
int e;
int c;
public Cow(int s, int e, int c){
this.s = s;
this.e = e;
this.c = c;
}
@Override
public int compareTo(Cow that) {
return this.e == that.e ? this.s - that.s : this.e - that.e;
}
}
void solve() {
int N = ni();
int M = ni();
int E = ni();
Cow[] cows = new Cow[N];
for (int i = 0; i < N; ++i){
cows[i] = new Cow(ni(), ni(), ni());
}
int[] dp = new int[E + 16];
Arrays.fill(dp, INF);
dp[M] = 0;
init(E);
Arrays.sort(cows);
update(M, 0);
for (int i = 0; i < N; ++i){
int s = cows[i].s;
int e = cows[i].e;
s = s == 0 ? 0 : s - 1;
int min = Math.min(dp[e], query(0, s, e, 0, n_) + cows[i].c);
dp[e] = min;
update(e, min);
}
out.println(dp[E] >= INF ? -1 : dp[E]);
}
/*****************RMQ*******************/
static final int MAX_N = (1 << 18) - 1;
int n_;
int[] dat = new int[MAX_N];
public void init(int N){
n_ = 1;
while (n_ < N) n_ *= 2;
for (int i = 0; i < 2 * n_ - 1; ++i) dat[i] = INF;
}
public void update(int k, int val){
k += (n_ - 1);
dat[k] = val;
while (k > 0){
k = (k - 1) / 2;
dat[k] = Math.min(dat[2 * k + 1], dat[2 * k + 2]);
}
}
public int query(int k, int i, int j, int l, int r){
if (j <= l || i >= r) return INF;
else if (i <= l && j >= r) return dat[k];
else{
int lch = 2 * k + 1;
int rch = 2 * k + 2;
int mid = (l + r) / 2;
int lf = query(lch, i, j, l, mid);
int rt = query(rch, i, j, mid, r);
return Math.min(lf, rt);
}
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}