目录
最长回文子串
又一道好题
信号塔
网格树
异色边2
| M | 最长回文子串 |
构造+马拉车
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
//#define f first
//#define s second
const int inf = 0x3f3f3f3f3f3f3f3f, N = 1e5 + 5, mod = 1e9 + 7;
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);int T;cin >> T;while (T--){int n;cin >> n;vector<int>a(n + 1);for (int i = 1; i <= n; i++) cin >> a[i];vector<int>f(n + 1);f[1] = f[2] = 1;for (int i = 2; i < n; i++) {if (a[i] == 1) {f[i + 1] = 1 - f[i - 1];}else {f[i + 1] = f[i - 1];}}vector<int>tmp(n * 2 + 5);int cnt = 0;tmp[++cnt]=-2, tmp[++cnt] = -1;for (int i = 1; i <= n; i++) {tmp[++cnt] = f[i];tmp[++cnt] = -1;}tmp[++cnt] = -3;int mr = 0, mid = 0, ans = 0;vector<int>p(n * 2 + 5);for (int i = 2; i <= cnt - 1; i++) {//mid-(i-mid)if (i < mr)p[i] = min(p[mid - (i - mid)], mr - i + 1);else p[i] = 1;while (tmp[i - p[i]] == tmp[i + p[i]]) ++p[i];if (i + p[i] > mr) mr = i + p[i] - 1, mid = i;}bool ok = true;for (int i = 3; i <= cnt - 2; i += 2) {if ((p[i] / 2-1)*2+1 != a[i/2])ok = false;}if (ok) {for (int i = 1; i <= n; i++) {cout << f[i];}cout << "\n";}else {cout << -1 << '\n';}}
}
| K | 又一道好题 |
启发式合并
int g[N], idx[N], val[N], m;
unordered_set<int>st[N];
int getid(int v) {if (g[v] == -1) {g[v] = ++m;val[g[v]] = v;}return g[v];
}
void join(int x, int y, int v) {val[x] = v;g[v] = x;for (int w : st[y]) {st[x].insert(w);idx[w] = x;}st[y].clear();g[v - 1] = -1;
}
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);int T;cin >> T;while (T--){int n, Q;cin >> n >> Q;for (int i = 0; i <= Q; ++i) g[i] = -1;for (int i = 1; i <= n; ++i) {st[0].insert(i);idx[i] = 0;}g[0] = 0;val[0] = 0;m = 0;int QQ=Q;while (Q--){int op, x;cin >> op >> x;if (op == 1) {int p = idx[x];int v = val[p] + 1;st[p].erase(x);int w = getid(v);st[w].insert(x);idx[x] = w;}else {if (g[x] == -1) continue;int x1 = g[x];int x2 = getid(x + 1);if (st[x1].size() < st[x2].size()) {join(x2, x1, x + 1);}else {join(x1, x2, x + 1);}}}for (int i = 1; i <= n; i++) {cout << val[idx[i]] << ' ';}cout << "\n";for (int i = 0; i <= QQ; i++) {st[i].clear();}}
}
| H | 信号塔 |
换根dp
dp 0表示当前没有信号塔子树最少需要信号塔数，1 当前有信号塔子树需要的信号塔数
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
//#define f first
//#define s second
const int inf = 0x3f3f3f3f3f3f3f3f, N = 2e5 + 5, mod = 1e9 + 7;
vector<int>q[N];
PII dp[N];//0 no 1 yes
int sz[N];
int f[N];
void dfs(int x, int fa)
{f[x] = fa;sz[x] = 1;dp[x] = { 0,1 };for (auto w : q[x]) {if (w == fa) continue;dfs(w, x);sz[x] += sz[w];dp[x].first += dp[w].second;dp[x].second += min(dp[w].first, dp[w].second);}
}
map<PII, int>mp;
int res[N];
void dfs2(int x, int fa) {for (auto w : q[x]) {if (w == fa) continue;dp[x].first -= dp[w].second;dp[x].second -= min(dp[w].first, dp[w].second);dp[x].first += dp[fa].second;dp[x].second += min(dp[fa].first, dp[fa].second);res[mp[{min(x, w), max(x, w)}]] = min(dp[x].first, dp[x].second) + min(dp[w].first, dp[w].second);dfs2(w, x);dp[x].second-= min(dp[fa].first, dp[fa].second);dp[x].first-= dp[fa].second;dp[x].first += dp[w].second;dp[x].second += min(dp[w].first, dp[w].second);}
}
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);int T;cin >> T;while (T--){int n;cin >> n;for (int i = 1; i <= n; i++) q[i].clear();for (int i = 1; i < n; i++){int u, v;cin >> u >> v;mp[{min(u, v), max(u, v)}] = i;q[u].push_back(v);q[v].push_back(u);}dfs(1, 0);dfs2(1, 0);for (int i = 1; i < n; i++) {cout << n - res[i] << ' ';}cout << "\n";mp.clear();}
}| G | 网格树 |
暴力dfs+剪枝
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
const int inf = 0x3f3f3f3f3f3f3f3f, N = 1e5 + 5, mod = 1e9 + 7;
int i, j, k, n, m, t, tot, cur, li, xx, yy;
vector<pair<int, int> > d = { {1,0},{-1,0},{0,1},{0,-1} };
string s[66];
int res, f[33][33], op[33][33], vis[33][33];
bool dfs2(int x, int y, int fx, int fy) {//用来检查当前放置方法是否符合if (vis[x][y]) return 0;vis[x][y] = 1; tot++;for (auto w : d) {int i = w.first, j = w.second;i += x, j += y;if (i<1 || i>n || j<1 || j>m || f[i][j]==0) continue;if (i == fx && j == fy) continue;if (dfs2(i, j, x, y)==0) return 0;}return 1;
}
void dfs(int x, int y)
{memset(vis, 0, sizeof vis); tot = 0;if (y > m) { x++; y = 1; }int sb = n * m - (x - 1) * m - (y - 1);if (sb + cur <= res)return;if (x > n) {//终止条件if (cur <= res || dfs2(xx, yy, -1, -1)==0) return;if (tot != cur) return;memcpy(op, f, sizeof f); res = cur;//op当前最优的放置方法return;}f[x][y] = 1; cur++;//f=1表示当前点位'.' 0表示'#'if (dfs2(x, y, -1, -1)==1&&tot==cur)dfs(x, y + 1);f[x][y] = 0; cur--;if (s[x][y] != '+')dfs(x, y + 1);
}
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);cin >> n >> m;for (int i = 1; i <= n; i++) {s[i] = string(m, '.');s[i] = "$" + s[i];}xx = yy = 1; s[1][1] = '+';//表示以1 1为跟的树dfs(1, 1);for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {if (op[i][j]) cout << ".";else cout << "W";}cout << "\n";}
}
| J | 异色边2 |
就是特殊处理一下环的元素，找到环后其他就都是树，贪心一下就行了,有点像基环树，找环的时候简单用一下tarjan
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
const int inf = 0x3f3f3f3f3f3f3f3f, N = 3e5 + 5, mod = 1e9 + 7;
vector<int>v[N];
int f[N];
int cnt, top;
int k;
int sk[N];
int h[N];
int col[N];
void dfs(int x,int fa)
{sk[++top]=x;f[x]=1;for(int y:v[x]){if(y==fa) continue;if(f[y]==1){if(cnt!=0){continue;}for(int i=top;sk[i]!=y&&i>=1;i--)//找到唯一的环 h[++cnt]=sk[i];h[++cnt]=y;}else dfs(y,x);}top--;
}
void dfs1(int x,int fa)
{for(int y:v[x]){if(y==fa||!f[y]) continue;if(k>0){k--;col[y]=col[x]^1;}else col[y]=col[x];dfs1(y,x);}
}
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);int T;cin >> T;while (T--){int n;cin >> n >> k;for (int i = 1; i <= n; i++) v[i].clear(), f[i] = 0;for (int i = 1; i <= n; i++) {int a, b;cin >> a >> b;v[a].push_back(b);v[b].push_back(a);}cnt = top = 0;dfs(1,0);int t=cnt/2*2;//如果时是奇数的话，有一个是没用的 if(t<=k){for(int i=1;i<=cnt;i++){col[h[i]]=i&1;}k-=t;}else{t=k/2*2;for(int i=1;i<=t;i++){col[h[i]]=i&1;}for(int i=t+1;i<=cnt;i++){col[h[i]]=1;}k%=2;}for(int i=1;i<=cnt;i++){f[h[i]]=0;}for(int i=1;i<=cnt;i++){dfs1(h[i],0);}if(k>0){cout<<"-1\n";}else{for(int i=1;i<=n;i++){cout<<col[i];}cout<<"\n";}}
}
)
