POJ.2186 Popular Cows

1.Problem
2186 -- Popular Cows (poj.org)

2.Idea
SCC+Topological Sort

3.Source

 const int N = 100005;  
 //////////////////////////////  
 int V;  
 vector<int> G[N];  
 vector<int> rG[N];  
 vector<int> vs;  
 bool used[N];  
 int cmp[N];  
 void add_edge(int from, int to)  
 {  
      G[from].push_back(to);  
      rG[to].push_back(from);  
 }  
 void dfs(int v)  
 {  
      used[v] = true;  
      for (int i = 0; i < G[v].size(); i++) {  
           if (!used[G[v][i]]) dfs(G[v][i]);  
      }  
      vs.push_back(v);  
 }  
 void rdfs(int v, int k)  
 {  
      used[v] = true;  
      cmp[v] = k;  
      for (int i = 0; i < rG[v].size(); i++) {  
           if (!used[rG[v][i]]) rdfs(rG[v][i], k);  
      }  
 }  
 int scc()  
 {  
      memset(used, 0, sizeof used);  
      vs.clear();  
      for (int v = 0; v < V; v++) {  
           if (!used[v]) dfs(v);  
      }  
      memset(used, 0, sizeof used);  
      int k = 0;  
      for (int i = vs.size() - 1; i >= 0; i--) {  
           if (!used[vs[i]]) rdfs(vs[i], k++);  
      }  
      return k;  
 }  
 int n, m;  
 int a[N], b[N];  
 void solve()  
 {  
      scanf("%d%d", &n, &m);  
      REP(i, m) scanf("%d%d", &a[i], &b[i]);  
      V = n;  
      for (int i = 0; i < m; i++) {  
           add_edge(a[i] - 1, b[i] - 1);  
      }  
      int nn = scc();  
      int u = 0, num = 0;  
      for (int v = 0; v < V; v++) {  
           if (cmp[v] == nn - 1) {  
                u = v;  
                num++;  
           }  
      }  
      memset(used, 0, sizeof(used));  
      rdfs(u, 0);  
      for (int v = 0; v < V; v++) {  
           if (!used[v]) {  
                num = 0;  
                break;  
           }  
      }  
      printf("%d\n", num);  
 }  

Atcoder.abc185_f F - Range Xor Query

1.Problem
F - Range Xor Query (atcoder.jp)

2.Idea
XOR segtree

3.Source

 Int seg[1 << 20];  
 void set_value(Int pos, Int val) {  
      pos += 1 << 19;  
      seg[pos] = seg[pos] ^ val;  
      while ((pos /= 2) > 0) {  
           seg[pos] = seg[pos * 2] ^ seg[pos * 2 + 1];  
      }  
 }  
 Int get_xor(int ql, int qr, int sl = 0, int sr = 1 << 19, int pos = 1) {  
      //no overlap  
      if (qr <= sl || sr <= ql) return 0;  
      //fit in the segment  
      if (ql <= sl && sr <= qr) return seg[pos];  
      //partially overlap  
      Int sm = (sl + sr) / 2;  
      Int lxor = get_xor(ql, qr, sl, sm, pos * 2);  
      Int rxor = get_xor(ql, qr, sm, sr, pos * 2 + 1);  
      return lxor ^ rxor;  
 }  
 void solve()  
 {  
      Int n, q, t, x, y;  
      cin >> n >> q;  
      for (int i = 0; i < n; i++) {  
           int x; cin >> x;  
           set_value(i, x);  
      }  
      for (int i = 0; i < q; i++) {  
           cin >> t >> x >> y;  
           if (t == 1) {  
                set_value(x - 1, y);  
           }  
           else {  
                cout << get_xor(x - 1, y) << endl;  
           }  
      }  
 }  

LeetCode.699 Falling Squares

1.Problem
Falling Squares - LeetCode

2.Idea
Using range max query on segtree

3.Source

 class Solution {  
 public:  
      map<int, int> mp;  
      int tree[8000];  
      void update(int t, int low, int high, int i, int j, int h) {  
           if (i > j) return;  
           if (low == high) {  
                tree[t] = h;  
                return;  
           }  
           int mid = (low + high) / 2;  
           update(2 * t, low, mid, i, min(mid, j), h);  
           update(2 * t + 1, mid + 1, high, max(mid + 1, i), j, h);  
           tree[t] = max(tree[2 * t], tree[2 * t + 1]);  
      }  
      int query(int t, int low, int high, int i, int j) {  
           if (i > j) return -2e9;  
           if (low == i && high == j) return tree[t];  
           int mid = (low + high) / 2;  
           return max(  
                query(2 * t, low, mid, i, min(mid, j)),  
                query(2 * t + 1, mid + 1, high, max(mid + 1, i), j)  
           );  
      }  
      vector<int> fallingSquares(vector<vector<int>>& positions) {  
           set<int> s;  
           memset(tree, 0, sizeof(tree));  
           for (auto it : positions) {  
                s.insert(it[0]);  
                s.insert(it[0] + it[1] - 1);  
           }  
           int compressed = 1, n = positions.size();  
           vector<int> ans(n);  
           for (auto it : s)  
                mp[it] = compressed++;  
           for(int i=0; i<n; i++){  
       int start=positions[i][0], end=positions[i][1]+start-1, h=positions[i][1];  
       int curr=query(1, 1, 2*n, mp[start], mp[end]), ncurr=curr+h;  
       update(1, 1, 2*n, mp[start], mp[end], ncurr);  
       ans[i]=tree[1];  
     }  
           return ans;  
      }  
 };