数据结构专题
地址 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=73154#overview
密码 acmore
基础数据结构的专题， 单调队列，单调栈，优先队列，堆，并查集，RMQ，线段树，树状数组
不错的单调队列的题目 题解
1 //#pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 100000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 int a[MAXN], L, H, n; 41 int q1[MAXN], q2[MAXN]; 42 43 int find_ans() { 44 int f1, f2, t1, t2, l1 = -1, l2 = -1, ans = 0; 45 f1 = f2 = t1 = t2 = 0; 46 rep (i, 0, n - 1) { 47 while(f1 < t1 && a[q1[t1 - 1]] <= a[i]) t1 --;//维护最大值队列（递减） 48 q1[t1++] = i; 49 while(f2 < t2 && a[q2[t2 - 1]] >= a[i]) t2 --;//维护最小值队列(递增) 50 q2[t2++] = i; 51 while(a[q1[f1]] - a[q2[f2]] > H) {//差值过大 52 q1[f1] < q2[f2] ? l1 = q1[f1 ++] : l2 = q2[f2 ++]; 53 } 54 if(a[q1[f1]] - a[q2[f2]] >= L) {//差值满足条件 55 ans = max(ans, i - max(l1, l2)); 56 } 57 } 58 return ans; 59 } 60 61 int main() 62 { 63 while(~scanf("%d %d %d", &n, &L, &H)) { 64 rep (i, 0, n - 1) scanf("%d", a + i); 65 printf("%d\n", find_ans()); 66 } 67 return 0; 68 }
好像不是优先队列，只是一个搜索
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 110 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 const int dx[4] = {0, 1, 0, -1}; 40 const int dy[4] = {1, 0, -1, 0}; 41 42 template <class T> 43 class MyQueue 44 { 45 private: 46 T* que; 47 int si, fr, re; 48 void setValue(int _size) { 49 fr = 0; re = 0; 50 si = _size; 51 que = (T*)malloc(si * sizeof(T)); 52 } 53 public: 54 MyQueue() { 55 this->setValue(100005); 56 } 57 MyQueue(int _size) { 58 this->setValue(_size); 59 } 60 T front() { 61 if(fr != re) 62 return que[fr]; 63 return NULL; 64 } 65 void pop() { 66 if(fr != re) 67 fr = (fr + 1) % si; 68 } 69 void push(T e) { 70 if((re + 1) % si == fr) return ; 71 que[re] = e; 72 re = (re + 1) % si; 73 } 74 bool empty() { 75 if(fr == re) return 1; 76 return 0; 77 } 78 void clear() { 79 fr = 0; 80 re = 0; 81 } 82 }; 83 84 struct Node { 85 int x, y, s; 86 Node(int _x = 0, int _y = 0, int _s = 0) { 87 x = _x; y = _y; s = _s; 88 } 89 }; 90 bool vis[MAXN][MAXN]; 91 MyQueue<Node> q(41000); 92 int m, n; 93 char ma[MAXN][MAXN]; 94 95 bool inside(int x, int N) { 96 return x >= 0 && x < N; 97 } 98 99 int bfs(int x, int y) { 100 mem0(vis); 101 q.clear(); 102 vis[x][y] = 1; 103 q.push(Node(x, y, 0)); 104 while(!q.empty()) { 105 Node fr = q.front(); q.pop(); 106 int px = fr.x, py = fr.y, ps = fr.s; 107 rep (i, 0, 3) { 108 int nx = px + dx[i], ny = py + dy[i]; 109 if(inside(nx, n) && inside(ny, m) 110 && ma[nx][ny] != '#' && !vis[nx][ny]) { 111 int ns = ps + 1 + (ma[nx][ny] == 'x'); 112 if(ma[nx][ny] == 'a') 113 return ns; 114 vis[nx][ny] = 1; 115 q.push(Node(nx, ny, ns)); 116 } 117 } 118 } 119 return -1; 120 } 121 122 int main() 123 { 124 while(~scanf("%d %d%*c", &n, &m)) { 125 int st_x = -1, st_y; 126 rep (i, 0, n - 1) { 127 scanf("%s", ma[i]); 128 if(st_x == -1) rep (j, 0, m - 1) { 129 if(ma[i][j] == 'r') st_x = i, st_y = j; 130 } 131 } 132 int ans = bfs(st_x, st_y); 133 if(ans == -1) { 134 puts("Poor ANGEL has to stay in the prison all his life."); 135 } 136 else cout << ans << endl; 137 } 138 return 0; 139 }
一个不错的RMQ的应用 题解
#pragma comment(linker, "/STACK:1677721600") #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <vector> #include <cstdio> #include <cctype> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; #define INF 0x3f3f3f3f #define inf (-((LL)1<<40)) #define lson k<<1, L, (L + R)>>1 #define rson k<<1|1, ((L + R)>>1) + 1, R #define mem0(a) memset(a,0,sizeof(a)) #define mem1(a) memset(a,-1,sizeof(a)) #define mem(a, b) memset(a, b, sizeof(a)) #define FIN freopen("in.txt", "r", stdin) #define FOUT freopen("out.txt", "w", stdout) #define rep(i, a, b) for(int i = a; i <= b; i ++) #define dec(i, a, b) for(int i = a; i >= b; i --) template<class T> T MAX(T a, T b) { return a > b ? a : b; } template<class T> T MIN(T a, T b) { return a < b ? a : b; } template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } //typedef __int64 LL; typedef long long LL; const int MAXN = 50000 + 100; const int MAXM = 110000; const double eps = 1e-8; LL MOD = 1000000007; struct Node { int v, s; Node(int _v = 0, int _s = 0) { v = _v; s = _s; } }; vector<Node>G[MAXN]; int mx[MAXN], se[MAXN], d[MAXN], idx[MAXN]; int ma[MAXN][20], mi[MAXN][20]; int n, u, v, w; int m, Q; void init(int n) { rep (i, 0, n) G[i].clear(); mem0(mx); mem0(se); } void dfs1(int u, int fa) { mx[u] = se[u] = 0; int sz = G[u].size(); rep (i, 0, sz - 1) { int v = G[u][i].v; if(v == fa) continue; dfs1(v, u); int len = G[u][i].s + mx[v]; if(len > mx[u]) se[u] = mx[u], mx[u] = len; else if(len > se[u]) se[u] = len; } } void dfs2(int u, int fa, int dis) { d[u] = max(dis, max(mx[u], se[u])); int sz = G[u].size(); rep (i, 0, sz - 1) { int v = G[u][i].v, len = G[u][i].s; if(v == fa) continue; if(len + mx[v] == mx[u]) dfs2(v, u, max(dis, se[u]) + len); else dfs2(v, u, max(dis, mx[u]) + len); } } void rmq_init(int n) { rep (i, 1, n) ma[i][0] = mi[i][0] = d[i]; for(int j = 1; (1<<j) <= n; j ++) { for(int i = 1; i + (1<<j) - 1 <= n; i ++) { ma[i][j] = max(ma[i][j - 1], ma[i + (1 << (j - 1))][j - 1]); mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]); } } rep (len, 1, n) { idx[len] = 0; while((1 << (idx[len] + 1)) <= len) idx[len] ++; } } int rmq(int l, int r) { int k = idx[r - l + 1]; return max(ma[l][k], ma[r - (1 << k) + 1][k]) - min(mi[l][k], mi[r - (1 << k) + 1][k]); } int main() { // FIN; while(~scanf("%d %d", &n, &m) && n) { init(n); rep (i, 1, n - 1) { scanf("%d %d %d", &u, &v, &w); G[u].push_back(Node(v, w)); G[v].push_back(Node(u, w)); } //计算每个点到叶子节点的最远距离 dfs1(1, -1); dfs2(1, -1, 0); //计算答案 rmq_init(n); while(m --) { scanf("%d", &Q); int l = 1, ans = 0; rep (i, 1, n) { while(l < i && rmq(l, i) > Q) l ++; ans = max(ans, i - l + 1); } cout << ans << endl; } } return 0; }
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 50000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 int m, n; 41 int mx[302][9][302][9]; 42 int idx[302], q, lx, ly, rx, ry; 43 44 void rmq_init(int m, int n) { 45 for(int i = 0; (1<<i) <= m; i ++) { 46 for(int j = 0; (1<<j) <= n; j ++) { 47 if(i == 0 && j == 0) continue; 48 int len2 = (1 << j), len1 = (1 << i); 49 for(int x = 1; x + len1 - 1 <= m; x ++) { 50 for(int y = 1; y + len2 - 1 <= n; y ++) { 51 if(i == 0) mx[x][i][y][j] = max(mx[x][i][y][j - 1], mx[x][i][y + (len2 >> 1)][j - 1]); 52 else mx[x][i][y][j] = max(mx[x][i - 1][y][j], mx[x + (len1 >> 1)][i - 1][y][j]); 53 } 54 } 55 } 56 } 57 for(int i = 1; i <= m || i <= n; i ++) { 58 idx[i] = 0; 59 while((1 << (idx[i] + 1)) <= i) idx[i] ++; 60 } 61 } 62 63 int rmq(int lx, int rx, int ly, int ry) { 64 int a = idx[rx - lx + 1], la = (1 << a); 65 int b = idx[ry - ly + 1], lb = (1 << b); 66 return max(max(max(mx[lx][a][ly][b], 67 mx[rx - la + 1][a][ly][b]), 68 mx[lx][a][ry - lb + 1][b]), 69 mx[rx - la + 1][a][ry - lb + 1][b]); 70 } 71 72 int main() 73 { 74 //FIN; 75 while(~scanf("%d %d", &m, &n)) { 76 mem0(mx); 77 rep (i, 1, m) rep (j, 1, n) 78 scanf("%d", &mx[i][0][j][0]); 79 rmq_init(m, n); 80 scanf("%d", &q); 81 while(q--) { 82 scanf("%d %d %d %d", &lx, &ly, &rx, &ry); 83 int ma = rmq(lx, rx, ly, ry); 84 printf("%d %s\n", ma, ma == mx[lx][0][ly][0] || ma == mx[lx][0][ry][0] 85 || ma == mx[rx][0][ly][0] || ma == mx[rx][0][ry][0] 86 ? "yes" : "no"); 87 } 88 } 89 return 0; 90 }
做了好多遍了
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 100000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 int n, a[MAXN], ans_l[MAXN], ans_r[MAXN]; 41 42 void find_ans(int *ans, int *a, int n) { 43 stack<int>s; 44 rep (i, 1, n) { 45 while(!s.empty() && a[s.top()] >= a[i]) { 46 s.pop(); 47 } 48 if(s.empty()) ans[i] = i; 49 else ans[i] = i - s.top(); 50 s.push(i); 51 } 52 } 53 54 void reverse_arr(int *arr, int n) { 55 for(int i = 1; i + i <= n; i ++) swap(a[i], a[n + 1 - i]); 56 } 57 58 int main() 59 { 60 // FIN; 61 while(~scanf("%d", &n) && n) { 62 rep (i, 1, n) scanf("%d", a + i); 63 find_ans(ans_l, a, n); 64 reverse_arr(a, n); 65 find_ans(ans_r, a, n); 66 LL ans = 0; 67 rep (i, 1, n) { 68 ans = max(ans, (LL)a[i] * (ans_l[n + 1 - i] + ans_r[i] - 1)); 69 } 70 cout << ans << endl; 71 } 72 return 0; 73 }
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 200000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 const double PI = 4.0 * atan(1.0); 40 41 int T, N, Q; 42 int x, h; 43 struct Node { 44 int x, h; 45 Node(int _x = 0, int _h = -1) { 46 x = _x; h = _h; 47 } 48 }node[MAXN]; 49 int sta[MAXN >> 1]; 50 double ans_l[MAXN >> 1], ans_r[MAXN >> 1]; 51 52 int cmp_up(Node A, Node B) { return A.x < B.x; } 53 54 int cmp_down(Node A, Node B) { return A.x > B.x; } 55 56 double pre(int r, int l) { 57 return (double)(node[r].h - node[l].h) / fabs(1.0 * node[r].x - node[l].x); 58 } 59 60 double pre(Node a, int l) { 61 return (double)(a.h - node[l].h) / fabs(1.0 * a.x - node[l].x); 62 } 63 64 void record_ans(double *angle) { 65 int tp = 0; 66 rep (i, 1, N + Q) { 67 if(node[i].h < 0) { 68 while(tp >= 2 && pre(Node(node[i].x, 0), sta[tp - 1]) - pre(sta[tp - 1], sta[tp - 2]) > eps) tp --; 69 angle[-node[i].h] = atan(fabs(1.0 * node[i].x - node[sta[tp - 1]].x) / node[sta[tp - 1]].h); 70 } 71 else { 72 while(tp >= 2 && pre(i, sta[tp - 1]) - pre(sta[tp - 1], sta[tp - 2]) > eps) tp --; 73 sta[tp++] = i; 74 } 75 } 76 } 77 78 int main() 79 { 80 // FIN; 81 cin >> T; 82 rep (cas, 1, T) { 83 scanf("%d", &N); 84 rep (i, 1, N) { 85 scanf("%d %d", &x, &h); 86 node[i] = Node(x, h); 87 } 88 scanf("%d", &Q); 89 rep (i, 1, Q) { 90 scanf("%d", &x); 91 node[N + i] = Node(x, -i); 92 } 93 sort(node + 1, node + 1 + N + Q, cmp_up); 94 record_ans(ans_l); 95 sort(node + 1, node + 1 + N + Q, cmp_down); 96 record_ans(ans_r); 97 cout << "Case #" << cas <<":" << endl; 98 rep (i, 1, Q) { 99 printf("%.10f\n", (ans_l[i] + ans_r[i]) * 180 / PI); 100 } 101 } 102 return 0; 103 }
简单线段树
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 100000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 const double PI = 4.0 * atan(1.0); 40 41 int T, n, m; 42 char ch[3]; 43 int A, B, x; 44 45 struct Node { int l, r, mx; }; 46 int a[MAXN]; 47 48 struct SegTree { 49 Node node[MAXN<<2]; 50 51 void push_up(int k, int L, int R) { 52 int mid = (L + R) >> 1; 53 node[k].l = node[k << 1].l; 54 node[k].r = node[k << 1 | 1].r; 55 node[k].mx = max(node[k << 1].mx, node[k<<1 | 1].mx); 56 if(a[mid] < a[mid + 1]) { 57 if(node[k << 1].l == mid - L + 1) node[k].l += node[k << 1 | 1].l; 58 if(node[k << 1 | 1].r == R - mid) node[k].r += node[k << 1].r; 59 node[k].mx = max(node[k].mx, node[k << 1].r + node[k << 1 | 1].l); 60 } 61 } 62 63 void update(int k, int L, int R, int p, int v) { 64 if(L == R) { 65 a[L] = v; 66 node[k].l = node[k].r = node[k].mx = 1; 67 return ; 68 } 69 int mid = (L + R) >> 1; 70 if(p <= mid) update(lson, p, v); 71 else update(rson, p, v); 72 push_up(k, L, R); 73 } 74 75 int query(int k, int L, int R, int l, int r) { 76 if(L > r || R < l) return 0; 77 if(l <= L && R <= r) return node[k].mx; 78 int mid = (L + R) >> 1; 79 int ans = max(query(lson, l, r), query(rson, l, r)); 80 if(a[mid] < a[mid + 1] && l <= mid && mid <= r) { 81 ans = max(ans, min(node[k << 1].r, mid - l + 1) + min(node[k << 1 | 1].l, r - mid)); 82 } 83 return ans; 84 } 85 86 }segTree; 87 88 int main() 89 { 90 // FIN; 91 cin >> T; 92 while(T--) { 93 cin >> n >> m; 94 rep (i, 1, n) { 95 scanf("%d", &x); 96 segTree.update(1, 1, n, i, x); 97 } 98 while(m --) { 99 scanf("%s %d %d", &ch, &A, &B); 100 if(ch[0] == 'Q') { 101 printf("%d\n", segTree.query(1, 1, n, ++A, ++B)); 102 } 103 else { 104 segTree.update(1, 1, n, ++A, B); 105 //rep (i, 1, n) printf("%d%c", a[i], i == n ? '\n' : ' '); 106 } 107 } 108 } 109 return 0; 110 }
简单并查集
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 100000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 const double PI = 4.0 * atan(1.0); 40 41 int fa[100002], a, b; 42 bool vis[100002]; 43 44 int find_fa(int u) { 45 return u == fa[u] ? u : fa[u] = find_fa(fa[u]); 46 } 47 48 int main() 49 { 50 // FIN; 51 while(1) { 52 mem0(vis); 53 int ok = 1, cnt_v = 0, cnt_e = -1; 54 while(scanf("%d %d", &a, &b) && a) { 55 if(a == -1) return 0; 56 if(!ok) continue; 57 if(!vis[a]) fa[a] = a, vis[a] = 1, cnt_v ++; 58 if(!vis[b]) fa[b] = b, vis[b] = 1, cnt_v ++; 59 int x = find_fa(a), y = find_fa(b); 60 if(x == y) ok = 0; 61 if(cnt_e == -1) cnt_e++; 62 fa[y] = x; 63 cnt_e ++; 64 } 65 puts(ok && cnt_e + 1 == cnt_v ? "Yes" : "No"); 66 } 67 return 0; 68 }
J,K比较简单，不放上来了。I不会= =！
