痛苦的数据结构OJ续
数据结构的OJ题目为什么更新这么频繁。。。。
下面的代码有我写的，有搜索之后修改的，。。。但是这么难的题，大部分都还是修改的(那些代码超级长的题)。
Let the Balloon Rise
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
struct node {
char x;
int i;
};
bool operator<(node a, node b) {
return a.x > b.x;
}
class Set{
private:
char s[1050][30];
char n;
int num[1050];
public:
Set() {
for (int i = 0; i < 1050; i++) {
strcpy(s[i], "");
num[i] = 0;
}
n = 0;
}
int find(char *name) {
for (int i = 0; i < n; i++)
if (! strcmp(s[i], name))
return i;
return -1;
}
void insert(char *name) {
int i = find(name);
if (i >= 0)
num[i]++;
else
strcpy(s[n++], name);
}
void show() {
int max = 0;
priority_queue<node> q;
node x;
for (int i = 0; i < n; i++) {
if (max < num[i]) {
max = num[i];
while (! q.empty()) {
q.pop();
}
x.x = s[i][0];
x.i = i;
q.push(x);
} else if (max == num[i]){
x.x = s[i][0];
x.i = i;
q.push(x);
}
}
while (! q.empty()) {
cout << s[q.top().i] << endl;
q.pop();
}
}
};
int main() {
int n;
char name[30];
while ((cin >> n) && n) {
Set s;
while (n--) {
cin >> name;
s.insert(name);
}
s.show();
}
return 0;
}
Elevator
#include <iostream>
using namespace std;
int main()
{
int x[10000]={0},t,m;
while(cin>>m)
{
if(m!=0)
{
t=0;
for(int i=1;i<=m;i++)
{
cin>>x[i];
if(x[i-1]<x[i])
{
t+=(x[i]-x[i-1])*6;
}
if(x[i-1]>x[i])
{
t+=(x[i-1]-x[i])*4;
}
t+=5;
}
cout<<t<<endl;
}
else
break;
}
return 0;
}
彩色字符串
#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;
int fun(char x) {
if (x == '(')
return 1;
if (x == '<')
return 2;
if (x == '[')
return 3;
if (x == ']' || x == ')' || x == '>')
return -1;
return 0;
}
int ans[5];
int main() {
int n,idx;
char x;
cin >> n;
getchar();
while (n--) {
stack<char> s;
ans[1] = ans[2] = ans[3] = 0;
while (1) {
x = getchar();
if (x == '\n') {
cout << ans[1] << " " << ans[2] << " " << ans[3] << endl;
break;
}
if (! fun(x))
ans[idx]++;
else if (fun(x) > 0) {
s.push(x);
idx = fun(x);
}else {
s.pop();
if (! s.empty())
idx = fun(s.top());
else
idx = 0;
}
}
}
return 0;
}
愚人节的礼物
#include <iostream>
using namespace std;
int main(){
char a[1010];
while (cin>>a) {
int i=0;
int num=0;
while (a[i]!='\0') {
if (a[i]=='(') {
num++;
}else if(a[i]==')'){
num--;
}else{
cout<<num<<endl;
break;
}
i++;
}
}
return 0;
}
迷宫
#include<string.h>
#include<ctype.h>
#include<malloc.h>
#include<limits.h>
#include<stdio.h>
#include<stdlib.h>
#define TRUE 1
#define FALSE 0
#define OK 1
#define ERROR 0
#define INFEASIBLE -1
#define OVERFLOW 0
typedef int Status;
typedef int Boolean;
#define MAXLENGTH 25
#define STACK_INIT_SIZE 10
#define STACKINCREMENT 2
typedef struct
{
int r, c;
} PosType;
typedef struct
{
int ord;
PosType seat;
int di;
} SElemType;
typedef struct
{
SElemType * base;
SElemType * top;
int stacksize;
} SqStack;
typedef struct
{
char arr[10][11];
} MazeType;
Status Pass(MazeType MyMaze, PosType CurPos)
{
if (MyMaze.arr[CurPos.r][CurPos.c]==' ' || MyMaze.arr[CurPos.r][CurPos.c]=='S' || MyMaze.arr[CurPos.r][CurPos.c]=='E')
return 1;
else
return 0;
}
void FootPrint(MazeType &MyMaze, PosType CurPos)
{
MyMaze.arr[CurPos.r][CurPos.c] = '*';
}
PosType NextPos(PosType CurPos, int Dir)
{
PosType ReturnPos;
switch (Dir)
{
case 1:
ReturnPos.r = CurPos.r;
ReturnPos.c = CurPos.c + 1;
break;
case 2:
ReturnPos.r = CurPos.r + 1;
ReturnPos.c = CurPos.c;
break;
case 3:
ReturnPos.r = CurPos.r;
ReturnPos.c = CurPos.c - 1;
break;
case 4:
ReturnPos.r = CurPos.r - 1;
ReturnPos.c = CurPos.c;
break;
}
return ReturnPos;
}
void MarkPrint(MazeType &MyMaze, PosType CurPos)
{
MyMaze.arr[CurPos.r][CurPos.c] = '!';
}
Status InitStack(SqStack *S)
{
(*S).base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));
if(!(*S).base)
exit(OVERFLOW);
(*S).top=(*S).base;
(*S).stacksize=STACK_INIT_SIZE;
return OK;
}
Status Push(SqStack *S,SElemType e)
{
if((*S).top-(*S).base>=(*S).stacksize)
{
(*S).base=(SElemType *)realloc((*S).base,((*S).stacksize+STACKINCREMENT)*sizeof(SElemType));
if(!(*S).base)
exit(OVERFLOW);
(*S).top=(*S).base+(*S).stacksize;
(*S).stacksize+=STACKINCREMENT;
}
*((*S).top)++=e;
return OK;
}
Status StackEmpty(SqStack S)
{
if(S.top==S.base)
return TRUE;
else
return FALSE;
}
Status Pop(SqStack *S,SElemType *e)
{
if((*S).top==(*S).base)
return ERROR;
*e=*--(*S).top;
return OK;
}
Status MazePath(MazeType &maze, PosType start, PosType end)
{
SqStack S;
PosType curpos;
int curstep;
SElemType e;
InitStack(&S);
curpos = start;
curstep = 1;
do
{
if (Pass(maze, curpos))
{
FootPrint(maze, curpos);
e.di = 1;
e.ord = curstep;
e.seat = curpos;
Push(&S, e);
if (curpos.r == end.r && curpos.c == end.c)
return (TRUE);
curpos = NextPos(curpos, 1);
curstep++;
}
else
{
if (!StackEmpty(S))
{
Pop(&S, &e);
while (e.di == 4 && !StackEmpty(S))
{
MarkPrint(maze, e.seat);
Pop(&S, &e);
}
if (e.di < 4)
{
e.di++;
Push(&S, e);
curpos = NextPos(e.seat, e.di);
}
}
}
}
while (!StackEmpty(S));
return FALSE;
}
int main()
{
int i, j;
PosType start, end;
MazeType maze;
memset(maze.arr, 0, sizeof(maze.arr));
for(i=0; i<10; i++)
{
gets(maze.arr[i]);
for(j=0; j<10; j++)
{
if(maze.arr[i][j] == 'S')
{
start.r = i;
start.c = j;
}
else if(maze.arr[i][j] == 'E')
{
end.r = i;
end.c = j;
}
}
}
MazePath(maze, start, end);
for(i=0; i<10; i++)
{
puts(maze.arr[i]);
}
return 0;
}
表达式求值
#include <iostream>
#include <stack>
#include <queue>
using namespace std;
int isp(char x) {
if (x == '#')
return 0;
if (x == '(')
return 1;
if (x == '*' || x == '/' || x == '%')
return 5;
if (x == '+' || x == '-')
return 3;
if (x == ')')
return 6;
}
int icp(char x) {
if (x == '#') {
return 0;
}
if (x == '(')
return 6;
if (x == '*' || x == '/' || x == '%')
return 4;
if (x == '+' || x == '-')
return 2;
if (x == ')')
return 1;
}
double fun(double a, double b, char x) {
if (x == '+')
return a + b;
if (x == '-')
return a - b;
if (x == '*')
return a * b;
if (x == '/')
return a / b;
}
int main () {
char x;
int n = 0;
stack<char> op;
stack<double> num;
op.push('#');
double a, b;
queue<int> q;
while (cin >> x) {
if (x >= '0' && x <= '9') {
n = n * 10 + x - '0';
cin >> x;
if (! (x >= '0' && x <= '9')) {
num.push(n);
n = 0;
}
cin.putback(x);
continue;
}
if (icp(x) < isp(op.top())) {
while (icp(x) < isp(op.top())) {
b = num.top();
num.pop();
a = num.top();
num.pop();
num.push(fun(a, b, op.top()));
op.pop();
}
}
if (icp(x) > isp(op.top())) {
op.push(x);
}
if (x == ')') {
op.pop();
}
if (x == '#') {
cout << num.top() << endl;
while (! num.empty()) {
num.pop();
}
while (! op.empty()) {
op.pop();
}
op.push('#');
n = 0;
}
}
return 0;
}
n阶Hanoi塔问题
#include <iostream>
#include <iomanip>
using namespace std;
char s1[] = {'Y', 'Z'};
char s2[] = {'X', 'Y'};
void hanoi(int n, char from, char to, char xx, int &i) {
if (n == 1) {
cout << setw(2) << i << ". Move disk " << n << " from " << from << " to " << to << endl;
i++;
return;
}
hanoi(n - 1, from, xx, to, i);
cout << setw(2) << i << ". Move disk " << n <<" from " << from << " to " << to << endl;
i++;
hanoi(n - 1, xx, to, from, i);
}
int main() {
int n;
while (cin >> n) {
int i = 1;
hanoi(n, 'X', 'Z', 'Y', i);
cout << endl;
}
return 0;
}
定位子串
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char str1[150], str2[150];
char *s;
while (cin >> str1) {
cin >> str2;
s = strstr(str1, str2);
if (s == NULL)
cout << 0 << endl;
else
cout << (s - str1) / sizeof(char) + 1 << endl;
}
return 0;
}
字符串插入
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char s1[200], s2[200];
int n;
while (cin >> s1 >> s2 >> n) {
int l1 = strlen(s1);
int l2 = strlen(s2);
for (int i = 0; i < n - 1; i++) {
cout << s1[i];
}
for (int i = 0; i < l2; i++) {
cout << s2[i];
}
for (int i = n - 1; i < l1; i++) {
cout << s1[i];
}
cout << endl;
}
return 0;
}
求子串位置的定位函数
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char s1[1000],s2[1000];
while (cin >> s1 >> s2) {
int i = 0;
for (; i < strlen(s1); i++) {
cout << s1[i];
if (s1[i] == s2[0]) {
int j = 1;
for (; j < strlen(s2); j++) {
if (i + j < strlen(s1))
cout << s1[i + j];
if (i + j >= strlen(s1) || s1[i + j] != s2[j])
break;
}
if (j == strlen(s2)) {
cout << endl << i + 1 << endl;
break;
}
}
}
if (i == strlen(s1)) {
cout << endl << 0 << endl;
}
}
return 0;
}
KMP算法中的模式串移动数组
#include <iostream>
#include <cstring>
using namespace std;
void get_next(char *t, int *next) {
int i = 1;
next[1] = 0;
int j = 0;
while (i < t[0]) {
if (j == 0 || t[i] == t[j]) {
++i;
++j;
next[i] = j;
}else
j = next[j];
}
}
int main() {
char s[105],t[105];
while (cin>>s){
t[0]=strlen(s);
strcpy(t+1,s);
int next[1000];
get_next(t,next);
for (int i=1;i<=t[0];i++) {
cout<<next[i]<< " ";
}
cout<<endl;
}
return 0;
}
KMP字符串模式匹配算法实现
#include <iostream>
#include <cstring>
using namespace std;
class SString{
private:
char *s;
char len;
public:
SString(char *ss) {
set(ss);
}
void set(char *ss) {
s = ss;
len = strlen(s);
}
char& operator[](int index) {
if (index == 0) {
return len;
}
return s[index - 1];
}
};
void get_next(SString t, int *next) {
int i = 1;
next[1] = 0;
int j = 0;
while (i < t[0]) {
if (j == 0 || t[i] == t[j]) {
++i;
++j;
next[i] = j;
}else
j = next[j];
}
}
int Index_KMP(SString S, SString T, int pos) {
int next[255];
int i = pos;
int j = 1;
get_next(T, next);
while (i <= S[0] && j <= T[0]) {
if (j == 0 || S[i] == T[j]) {
++i;
++j;
}else {
j = next[j];
}
}
if (j > T[0])
return i - T[0];
else
return 0;
}
int main() {
char s1[105], s2[105];
while (cin >> s1 >> s2) {
cout << Index_KMP(SString(s1), SString(s2), 1) << endl;
}
return 0;
}
稀疏矩阵转置
#include <iostream>
#include <iomanip>
using namespace std;
template<class T>
struct Trituple
{
int row;
int col;
T val;
};
template<class T>
class SparseMatrix
{
public:
SparseMatrix(int maxt=100);
~SparseMatrix();
bool TransposeTo(SparseMatrix &);
bool AddTo(const SparseMatrix&);
void Input();
void Output();
private:
Trituple<T>* data;
int rows,cols,terms;
int maxterms;
};
template<class T>
SparseMatrix<T>::SparseMatrix(int maxt)
{
maxterms=maxt;
data=new Trituple<T>[maxterms];
terms=rows=cols=0;
}
template<class T>
SparseMatrix<T>::~SparseMatrix()
{
if (data!=NULL)
{
delete[] data;
}
}
template<class T>
bool SparseMatrix<T>::TransposeTo(SparseMatrix &B)
{
if (terms>B.maxterms)
{
return false;
}
B.rows=cols;
B.cols=rows;
B.terms=terms;
if (terms>0)
{
int p=0;
for (int j=1;j<=cols;j++)
{
for (int k=0;k<terms;k++)
{
if (data[k].col==j)
{
B.data[p].row=j;
B.data[p].col=data[k].row;
B.data[p].val=data[k].val;
p++;
}
}
}
}
return true;
}
template<class T>
void SparseMatrix<T>::Input()
{
int row1,col1,number;
cin>>row1>>col1;
rows=row1;
cols=col1;
for (int i=0;i<rows;i++)
{
for (int j=0;j<cols;j++)
{
cin>>number;
if (number!=0)
{
data[terms].row=i+1;
data[terms].col=j+1;
data[terms].val=number;
terms++;
}
}
}
}
template<class T>
void SparseMatrix<T>::Output()
{
T **tempArray=new T*[rows];
for (int i1=0;i1<rows;i1++)
{
tempArray[i1]=new int[cols];
}
for (int j=0;j<rows;j++)
{
for (int k=0;k<cols;k++)
{
tempArray[j][k]=0;
}
}
for (int i=0;i<terms;i++)
{
tempArray[data[i].row-1][data[i].col-1]=data[i].val;
}
for (int j=0;j<rows;j++)
{
for (int k=0;k<cols;k++)
{
cout<<tempArray[j][k]<<" ";
}
cout<<endl;
}
for (int l=0;l<rows;l++)
{
delete[] tempArray[l];
}
delete tempArray;
cout<<endl;
}
template<class T>
bool SparseMatrix<T>::AddTo(const SparseMatrix& B)
{
if (rows!=B.rows||cols!=B.cols)
{
return false;
}
bool flag=false;
int tempTerms=terms;
for (int i=0;i<B.terms;i++)
{
flag=false;
for (int j=0;j<tempTerms;j++)
{
if (data[j].col==B.data[i].col&&data[j].row==B.data[i].row)
{
data[j].val+=B.data[i].val;
flag=true;
}
}
if (flag==false)
{
data[++terms-1].col=B.data[i].col;
data[terms-1].row=B.data[i].row;
data[terms-1].val=B.data[i].val;
}
}
return true;
}
int main()
{
SparseMatrix<int> sm(50);
SparseMatrix<int> sm1(50);
SparseMatrix<int> sm2(50);
sm.Input();
sm.TransposeTo(sm1);
sm1.Output();
return 0;
}
稀疏矩阵快速转置
这道题我提交了十次才AC。。。。
#include<iostream>
using namespace std;
#define MAXROW 250
#define MAXSIZE 12500
typedef int ElemType;
typedef struct
{
int i, j;
ElemType e;
}Triple;
typedef struct
{
Triple data[MAXSIZE + 1];
int n, m, t;
}Matrix;
void Inst(Matrix &a)
{
int i, j;
ElemType tmp;
cin>>a.n>>a.m;
a.t = 0;
for (i = 1; i <= a.n; i++)
{
for (j = 1; j <= a.m; j++)
{
cin>>tmp;
if (tmp)
{
a.t++;
a.data[a.t].i = i;
a.data[a.t].j = j;
a.data[a.t].e = tmp;
}
}
}
}
void Trans(Matrix a, Matrix &b)
{
int num[MAXROW + 1] = { 0 }, pos[MAXROW + 1] = { 0 };
int i, j, p;
b.n = a.m; b.m = a.n; b.t = a.t;
for (i = 1; i <= a.t; i++) num[a.data[i].j]++;
for (i = 1; i <= a.m; i++) pos[i] = pos[i - 1] + num[i];
for (i = a.t; i >= 1; i--)
{
p = pos[a.data[i].j]--;
b.data[p].i = a.data[i].j;
b.data[p].j = a.data[i].i;
b.data[p].e = a.data[i].e;
}
}
void Out(Matrix a)
{
int i, j;
int u = 1;
for (i = 1; i <= a.n; i++)
{
for (j = 1; j <= a.m; j++)
if (i == a.data[u].i && j == a.data[u].j)
cout<<a.data[u++].e<<" ";
else
cout<<"0"<<" ";
cout<<endl;
}
}
int main()
{
Matrix a, b;
Inst(a);
Trans(a, b);
Out(b);
return 0;
}
