问题一:
题目描述
对于一个字符串,将其后缀子串进行排序,例如grain 其子串有: grain rain ain in n 然后对各子串按字典顺序排序,即: ain,grain,in,n,rain
输入描述:
每个案例为一行字符串。
输出描述:
将子串排序输出
示例1
输入
grain
输出
ain
grain
in
n
rain
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args){
try {
BufferedReader br = new BufferedReader(new BufferedReader(new InputStreamReader(System.in)));
String str = br.readLine();
int len = str.length();
String[] subs = new String[len];
for(int i = 0; i < len; i++) {
subs[i] = str.substring(i);
}
Arrays.parallelSort(subs);
for(int i = 0; i < len; i++) {
System.out.println(subs[i]);
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
问题二:
题目描述
N个城市,标号从0到N-1,M条道路,第K条道路(K从0开始)的长度为2^K,求编号为0的城市到其他城市的最短距离
输入描述:
第一行两个正整数N(2<=N<=100)M(M<=500),表示有N个城市,M条道路
接下来M行两个整数,表示相连的两个城市的编号
输出描述:
N-1行,表示0号城市到其他城市的最短路,如果无法到达,输出-1,数值太大的以MOD 100000 的结果输出。
示例1
输入
4 4
1 2
2 3
1 3
0 1
输出
8
9
11
Prim算法
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import java.util.*;
import java.io.*;
import java.text.* ;
public class Main
{
public static void main(String[] args) throws ParseException{
try {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
String str;
while((str = br.readLine()) != null) {
String[] parts = str.split(" ");
int n = Integer.parseInt(parts[0]);
int len = Integer.parseInt(parts[1]);
int[][] dis = new int[n][n];
int[] tree = new int[n];
int lens = 1;
int i;
for(i = 0; i < n; i++) {
tree[i] = -1;
for(int j = 0; j < n; j++) dis[i][j] = -1;
dis[i][i] = 0;
}
int x, y;
for(i = 0; i < len; i++) {
parts = br.readLine().split(" ");
int a = Integer.parseInt(parts[0]);
int b = Integer.parseInt(parts[1]);
x = findRoot(a, tree);
y = findRoot(b, tree);
if(i>0){
lens = (lens*2)%100000;
}
if(x != y) {
for(int j = 0; j < n; j++) {
if(x == findRoot(j, tree)) {
for(int k = 0; k < n; k++) {
if(y == findRoot(k, tree)){
dis[j][k]=(dis[j][a]+dis[b][k]+lens)%100000;
dis[k][j]=dis[j][k];
}
}
}
}
tree[y] = x;
}
}
for(i = 1; i < n; i++)
System.out.println(dis[0][i]);
}
} catch (IOException e) {
e.printStackTrace();
}
}
public static int findRoot(int x, int[] tree) {
if(tree[x] == -1) return x;
else {
tree[x] = findRoot(tree[x], tree);
return tree[x];
}
}
}
问题三:
题目描述
对于一个不存在括号的表达式进行计算
输入描述:
存在多种数据,每组数据一行,表达式不存在空格
输出描述:
输出结果
示例1
输入
6/2+3+3*4
输出
18
计算中缀表达式
import java.util.*;
import java.io.*;
import java.text.* ;
public class Main
{
public static int[] pos = {0};
public static char[] ch;
public static int len;
public static int i;
public static void main(String[] args) throws ParseException{
try {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
ch = br.readLine().toCharArray();
System.out.println((int)compute());
} catch (IOException e) {
e.printStackTrace();
}
}
//转换小数
public static double translation() {
double integer = 0.0;
double remainder = 0.0;
while(i < len && ch[i] >= '0' && ch[i] <= '9') {
integer *= 10;
integer += (ch[i] - '0');
i++;
}
if(i < len && ch[i] == '.') {
i++;
double c = 0.1;
while(ch[i] >= '0' && ch[i] <= '9') {
double t = ch[i] -'0';
t *= c;
c *= 0.1;
remainder += t;
i++;
}
}
return integer + remainder;
}
//计算优先级
public static int getLevel(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '(':
return 0;
};
return -1;
}
public static double operate(double a1, char op, double a2)
{
switch (op)
{
case '+':
return a1 + a2;
case '-':
return a1 - a2;
case '*':
return a1 * a2;
case '/':
return a1 / a2;
};
return 0;
}
public static double compute() {
Stack<Character> optr = new Stack<>();
Stack<Double> opnd = new Stack<>();
len = ch.length;
boolean isMinus = true; // 判断'-'是减号还是负号, true表示负号
for(i = 0; i < len;) {
if(ch[i] == '-' && isMinus) {
opnd.push(0.0);
optr.push('-');
i++;
}
else if(ch[i] == ')') {
isMinus = false;
i++;
while(optr.peek() != '(') {
double a2 = opnd.pop();
double a1 = opnd.pop();
char op = optr.pop();
double result = operate(a1, op, a2);
opnd.push(result);
}
optr.pop();
}
else if(ch[i] >= '0' && ch[i] <= '9') {
isMinus = false;
opnd.push(translation());
}
else if(ch[i] == '(') {
isMinus = true;
optr.push(ch[i]);
i++;
}
else {
while(!optr.empty() && getLevel(ch[i]) <= getLevel(optr.peek())) {
double a2 = opnd.pop();
double a1 = opnd.pop();
char op = optr.pop();
double result = operate(a1, op, a2);
opnd.push(result);
}
optr.push(ch[i]);
i++;
}
}
while(!optr.empty()) {
double a2 = opnd.pop();
double a1 = opnd.pop();
char op = optr.pop();
double result = operate(a1, op, a2);
opnd.push(result);
}
return opnd.pop();
}
}
后缀表达式计算
import java.util.*;
import java.io.*;
import java.text.* ;
public class Main
{
public static char[] ch;
public static int len;
public static int i;
public static void main(String[] args) throws ParseException{
try {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
ch = br.readLine().toCharArray();
System.out.println((int)compute());
} catch (IOException e) {
e.printStackTrace();
}
}
public static double operate(double a1, char op, double a2) {
switch(op) {
case '+': return a1+a2;
case '-': return a1-a2;
case '*': return a1*a2;
case '/': return a1/a2;
};
return 0;
}
public static double compute() {
Stack<Double> opnd = new Stack<>();
len = ch.length;
for(i = 0; i < len;) {
if(ch[i] >= '0' && ch[i] <= '9') {
opnd.push((double)(ch[i]-'0'));
i++;
}
else {
double a2 = opnd.pop();
double a1 = opnd.pop();
char op = ch[i];
double result = operate(a1, op, a2);
opnd.push(result);
i++;
}
}
return opnd.pop();
}
}
前缀表达式转中缀表达式
import java.util.*;
import java.io.*;
import java.text.* ;
public class Main
{
public static char[] ch;
public static int len;
public static int i;
public static void main(String[] args) throws ParseException{
try {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
ch = br.readLine().toCharArray();
System.out.println(compute());
} catch (IOException e) {
e.printStackTrace();
}
}
//计算优先级
public static int getLevel(char ch) {
switch(ch) {
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '(':
return 0;
};
return -1;
}
public static StringBuilder compute() {
StringBuilder res = new StringBuilder();
Stack<Character> optr = new Stack<>();
Stack<Character> opnd = new Stack<>();
len = ch.length;
boolean isMinus = true;
for(i = 0; i < len;) {
System.out.println(res);
if(ch[i] == '-' && isMinus) {
opnd.push('0');
optr.push('-');
i++;
}
else if(ch[i] == ')') {
isMinus = false;
i++;
while(optr.peek() != '(') {
char a2 = opnd.pop();
char a1 = opnd.pop();
char op = optr.pop();
if(a1 != ' ') res.append(a1);
if(a2 != ' ') res.append(a2);
res.append(op);
opnd.push(' ');
}
optr.pop();
}
else if(ch[i] >= '0' && ch[i] <= '9') {
isMinus = false;
opnd.push(ch[i]);
i++;
}
else if(ch[i] == '(') {
isMinus = true;
optr.push(ch[i]);
i++;
}
else {
while(!optr.empty() && getLevel(ch[i]) <= getLevel(optr.peek())) {
char a2 = opnd.pop();
char a1 = opnd.pop();
char op = optr.pop();
if(a1 != ' ') res.append(a1);
if(a2 != ' ') res.append(a2);
res.append(op);
opnd.push(' ');
}
optr.push(ch[i]);
i++;
}
}
while(!optr.empty()) {
char a2 = opnd.pop();
char a1 = opnd.pop();
char op= optr.pop();
if(a1 != ' ') res.append(a1);
if(a2 != ' ') res.append(a2);
res.append(op);
opnd.push(' ');
}
return res;
}
}
题目描述
一个N*M的矩阵,找出这个矩阵中所有元素的和不小于K的面积最小的子矩阵(矩阵中元素个数为矩阵面积)
输入描述:
每个案例第一行三个正整数N,M<=100,表示矩阵大小,和一个整数K
接下来N行,每行M个数,表示矩阵每个元素的值
输出描述:
输出最小面积的值。如果出现任意矩阵的和都小于K,直接输出-1。
示例1
输入
4 4 10
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
输出
1
暴力解法:
import java.util.*;
import java.io.*;
import java.text.* ;
public class Main
{
static int n;
static int m;
static int[][] matrix;
static int res = Integer.MAX_VALUE;
static int k;
public static void main(String[] args){
try {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
String[] parts = br.readLine().split(" ");
n = Integer.parseInt(parts[0]);
m = Integer.parseInt(parts[1]);
k = Integer.parseInt(parts[2]);
matrix = new int[n][m];
for(int i = 0; i < n; i++) {
String[] lines = br.readLine().split(" ");
for(int j = 0; j < m; j++) {
matrix[i][j] = Integer.parseInt(lines[j]);
}
}
System.out.println(matrixSum());
} catch (IOException e) {
e.printStackTrace();
}
}
public static int matrixSum() {
int ans = m*n+1;
for(int i = 0; i < n; i++) {
int[] tmp = new int[m];
for(int j = i; j < n; j++) {
for(int l = 0; l < m; l++)
tmp[l] += matrix[j][l];
int len = getLen(tmp);
ans = Math.min(ans, len*(j-i+1));
}
}
if(ans == m*n+1) return -1;
return ans;
}
private static int getLen(int[] a) {
int low = 0, high = 0;
int len = m*n+1;
int s = 0;
while ( high<m ) {
while ( s<k && high<m ) {
s += a[high];
high ++;
}
while ( s>=k && low<high) {
len = Math.min(len, high-low);
s -= a[low];
low ++;
}
//先往后扩,再往后缩
}
return len;
}
}