SMU Summer 2024 Contest Round 5
A Robot Takahashi
思路：
将所有数排序，枚举孩子成人的分解点X，同时根据s的标识维护正真的孩子成人的个数
void solve() {
int n;
cin >> n;
string s;
cin >> s;
int sum = 0;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '1') sum ++;
}
vector<PII> w(n);
for (int i = 0; i < n; ++i) {
cin >> w[i].first;
w[i].second = i;
}
sort(w.begin(), w.end());
int a1 = sum, a0 = 0, ans = a1;
for (int i = 0; i < w.size(); ++i) {
int j = i;
while (j < w.size() && w[j].first == w[i].first) {
if (s[w[j ++].second] == '1') a1 --;
else a0 ++;
}
i = j - 1;
ans = max(ans, a1 + a0);
}
cout << ans;
}B Connect 6
思路：
暴力枚举，注意对角线有两种\和/
void solve() {
int n;
cin >> n;
vector<string> ve(n + 1);
for (int i = 1; i <= n; ++i) {
cin >> ve[i];
ve[i] = ' ' + ve[i];
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
int a = 0, b = 0, c = 0, d = 0;
for (int k = 0; k < 6; ++k) {
if (i >= 6 && j >= 6 && ve[i - k][j - k] == '#') a++;
if (i >= 6 && j + 5 <= n && ve[i - k][j + k] == '#') d++;
if (j >= 6 && ve[i][j - k] == '#') b++;
if (i >= 6 && ve[i - k][j] == '#') c++;
}
if (a >= 4 || b >= 4 || c >= 4 || d >= 4) {
cout << "Yes";
return ;
}
}
}
cout << "No";
return ;
}C Strange Balls
思路：
用栈维护连续相同的数字及其个数，若满足数字等于个数就取出，同时记录总个数即为答案
void solve() {
int n;
cin >> n;
vector<int> cnt (n + 1);
vector<PII> ve;
int ans = 0;
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
if (ve.size() && ve.back().first == x) {
ve.back().second ++, ans ++;
} else {
ve.push_back({x, 1}), ans ++;
}
if (ve.size() && ve.back().first == ve.back().second) {
ans -= ve.back().second;
ve.pop_back();
}
cout << ans << '\n';
}
}D Linear Probing
思路：
用类似路径压缩的方式找到下一个空的位置
数组不是很大，也可以直接用set存所有空的位置，每次赋值时二分找位置，找到后从set中删掉该位置，注意的是可能位置太靠后以至于后面没有空位置，那就手动把位置调到0即可
void solve() {
int q;
cin >> q;
int n = pow(2, 20);
vector<int> val(n, -1);
set<int> se;
for (int i = 0; i < n; ++i) {
se.insert(i);
}
while (q --) {
int t, x;
cin >> t >> x;
int s = x;
x %= n;
if (t == 2) {
cout << val[x] << '\n';
} else {
auto p = se.lower_bound(x);
if (p == se.end()) {
x = 0;
p = se.lower_bound(x);
}
val[*p] = s;
se.erase(p);
}
}
}F Stronger Takahashi
思路：01bfs，优先做0步的操作，再做1步的操作，用双端队列维护
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
int dxx[20] = {-2, -2, -2, -1, -1,
-1, -1, -1, 0, 0,
0, 0, 1, 1, 1,
1 ,1, 2, 2, 2};
int dyy[20] = {-1, 0, 1, -2, -1,
0, 1, 2, -2, -1,
1, 2, -2, -1, 0,
1, 2, -1, 0, 1};
void solve() {
int h, w;
cin >> h >> w;
vector<string> ve(h + 1);
for (int i = 1; i <= h; ++i) {
cin >> ve[i];
ve[i] = ' ' + ve[i];
}
deque<pair<PII, int> > q;
vector<vector<int> > dis(h + 1, vector<int> (w + 1, -1));
q.push_back({{1, 1}, 0});
while (q.size()) {
auto [u, ww] = q.front();
q.pop_front();
int x = u.first, y = u.second;
if (dis[x][y] != -1) continue;
dis[x][y] = ww;
for (int i = 0; i < 4; ++i) {
int xx = x + dx[i], yy = y + dy[i];
if (xx < 1 || xx > h || yy < 1 || yy > w || ve[xx][yy] != '.') continue;
q.push_front({{xx, yy}, ww});
}
for (int i = 0; i < 20; ++i) {
int xx = x + dxx[i], yy = y + dyy[i];
if (xx < 1 || xx > h || yy < 1 || yy > w) continue;
q.push_back({{xx, yy}, ww + 1});
}
}
cout << dis[h][w];
}E Red Polyomino
思路：暴力dfs+剪枝，枚举下一步改颜色的位置
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
void solve() {
int n, k;
cin >> n >> k;
vector<string> s(n);
for (int i = 0; i < n; ++i) cin >> s[i];
map<vector<string>, int> st;
int ans = 0;
auto dfs = [&] (int cnt, auto dfs) -> void {
if (st.count(s)) return ;
st[s] = 1;
if (cnt == 0) {
ans ++;
return ;
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i][j] == '.') {
for (int k = 0; k < 4; ++k) {
int xx = i + dx[k], yy = j + dy[k];
if (xx < 0 || xx >= n || yy < 0 || yy >= n) continue;
if (s[xx][yy] == 'j') {
s[i][j] = 'j';
dfs(cnt - 1, dfs);
s[i][j] = '.';
break;
}
}
}
}
}
};
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i][j] == '.') {
s[i][j] = 'j';
dfs(k - 1, dfs);
s[i][j] = '.';
}
}
}
cout << ans;
}
G Predilection
思路：
递推，相邻的数才进行合并，用f[i]表示前i个数的答案
第i个数没合并前，与f[i - 1]个序列构成一种序列，贡献为f[i - 1]
第i个数合并后，与f[i - 1]个序列的最后一个数进行合并，贡献为f[i - 1]
但是存在0的话会有重复的情况，就需要消去加入a[i]后出现的0带来的贡献，即找到后缀和为0的最大下标j，消去的贡献为f[j - 1]
为啥只消最大的下标j呢，因为之前的已经f[j]消掉了
void solve() {
int n;
cin >> n;
vector<int> a(n + 1), r(n + 5);
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
for (int i = n; i >= 1; --i) {
r[i] = a[i] + r[i + 1];
}
map<int, int> pre;
vector<int> f(n + 1);
f[1] = 1;
for (int i = 2; i <= n; ++i) {
f[i] = f[i - 1] * 2 % mod;
f[i] = (f[i] - pre[r[i]] + mod) % mod;
pre[r[i]] = f[i - 1];
}
cout << f[n];
}
