PAT1129 Recommendation System
需要稍加思考，每次都sort会TLE。
#include<iostream>
#include<algorithm>
#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
const int MAXN = 50010;
int arr[20], cnt[MAXN];
//按照次数排序，若相同则按照大小排序
bool cmp(int a, int b) {
if (cnt[a] != cnt[b])
return cnt[a] > cnt[b];
else
return a < b;
}
int main() {
ac
int n, m, k = 0, a;
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> a;
if (i) {
cout << a << ":";
for (int j = 0; j < k; j++)
cout << " " << arr[j];
cout << endl;
}
//次数+1
cnt[a]++;
bool op = false;
for (int j = 0; j < k; j++)
if (arr[j] == a) {
op = true;
break;
}
//没找到，可能有k之外的元素进入前k
if (!op)
arr[k++] = a;
//找到了，那前k个元素肯定还是前k大，只需要内部排序
sort(arr, arr + k, cmp);
//k不能超过m
if (k > m)
k = m;
}
return 0;
}
PAT1121 Damn Single
签到题，可以用一下unordered_set来优化一下。
#include<iostream>
#include<unordered_set>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN = 100010;
//记录着对应couple的编号
int map[MAXN], ans[MAXN];
unordered_set<int> set;
int main() {
int n, a, b, pos = 0;
//因为编号从0开始，故有可能出错，赋初值超过100000即可
memset(map, 0x3f, sizeof(map));
cin >> n;
while (n--) {
cin >> a >> b;
map[a] = b;
map[b] = a;
}
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a;
set.insert(a);
}
for (int i:set)
//找不到自己的情侣
if (!set.count(map[i]))
ans[pos++] = i;
sort(ans, ans + pos);
cout << pos << endl;
for (int i = 0; i < pos; i++) {
printf("%05d", ans[i]);
if (i != pos - 1)
cout << " ";
}
return 0;
}
PAT1055 The World’s Richest
使用邻接表和优先队列，有一个样例TLE。
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
struct node {
string name;
int age, wealth;
node(string a, int b, int c) : name(a), age(b), wealth(c) {
};
node() {
};
//注意优先队列的重载运算符是反的
bool operator<(const node &a) const {
if (wealth != a.wealth)
return wealth < a.wealth;
else if (age != a.age)
return age > a.age;
else
return name > a.name;
}
};
vector<node> G[210];
int N, K;
bool cmp(const node &a, const node &b) {
return a.age < b.age;
}
int main() {
ac
string s;
int a, b, lim;
cin >> N >> K;
for (int i = 0; i < N; i++) {
cin >> s >> a >> b;
G[a].push_back(node(s, a, b));
}
node tmp;
for (int i = 1; i <= K; i++) {
cin >> lim >> a >> b;
priority_queue<node> Q;
//年龄之间的节点全部加入到优先队列中，这里可以优化
for (int i = a; i <= b; i++)
for (node &j:G[i])
Q.push(j);
cout << "Case #" << i << ":" << endl;
if (Q.empty())
cout << "None" << endl;
while (lim-- && !Q.empty()) {
tmp = Q.top();
Q.pop();
cout << tmp.name << " " << tmp.age << " " << tmp.wealth << endl;
}
}
return 0;
}
PAT1051 Pop Sequence
很经典的一题，推荐做一下，提高对栈的理解。
#include<iostream>
#include<stack>
#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
const int MAXN = 100010;
int arr[MAXN];
int m, n, k;
int main() {
ac
cin >> m >> n >> k;
while (k--) {
stack<int> ST;
for (int i = 1; i <= n; i++)
cin >> arr[i];
int pos = 1;
bool op = true;
for (int i = 1; i <= n && op; i++) {
//栈为空或者不等于目标元素
while (ST.empty() || ST.top() != arr[i]) {
ST.push(pos++);
//超过最大容量
if (ST.size() > m) {
op = false;
break;
}
}
ST.pop();
}
if (op)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
PAT1091 Acute Stroke
Blood Fill裸题，可以bfs也可dfs，拓展到了三维，只需要稍稍改动fx数组。
#include<iostream>
#include<queue>
#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
const int M = 1300;
const int N = 130;
const int L = 70;
bool G[L][M][N];
int m, n, l, t;
struct node {
int l, m, n;
node(int _x, int _y, int _z) : l(_x), m(_y), n(_z) {
};
};
//三维
int fx[][3] = {
{
0, 0, 1},
{
0, 0, -1},
{
0, 1, 0},
{
0, -1, 0},
{
1, 0, 0},
{
-1, 0, 0},
};
bool check(const node &a) {
return a.l >= 0 && a.m >= 0 && a.n >= 0 && a.l < L && a.n < N && a.m < M;
}
int bfs(int l, int m, int n) {
int ans = 0;
node tmp(l, m, n);
G[l][m][n] = false;
queue<node> Q;
Q.push(tmp);
while (!Q.empty()) {
node cur(Q.front());
Q.pop();
++ans;
for (int i = 0; i < 6; i++) {
node nxt(cur);
//三维方向拓展
nxt.l += fx[i][0];
nxt.m += fx[i][1];
nxt.n += fx[i][2];
if (check(nxt) && G[nxt.l][nxt.m][nxt.n]) {
Q.push(nxt);
//要变为false
G[nxt.l][nxt.m][nxt.n] = false;
}
}
}
return ans;
}
int main() {
ac
int ans = 0;
cin >> m >> n >> l >> t;
for (int i = 0; i < l; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < n; k++)
cin >> G[i][j][k];
for (int i = 0; i < l; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < n; k++)
if (G[i][j][k]) {
int cnt = bfs(i, j, k);
if (cnt >= t)
ans += cnt;
}
cout << ans;
return 0;
}
官网的题快刷完了，考试前一定刷完，奥里给><
