HDOJ 3966 树链剖分
链接：
http://acm.hdu.edu.cn/showproblem.php?pid=3966
代码：
1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <stack> 6 #include <cstdio> 7 #include <string> 8 #include <vector> 9 #include <cstdlib> 10 #include <cstring> 11 #include <sstream> 12 #include <iostream> 13 #include <algorithm> 14 #include <functional> 15 using namespace std; 16 #define rep(i,a,n) for (int i=a;i<n;i++) 17 #define per(i,a,n) for (int i=n-1;i>=a;i--) 18 #define all(x) (x).begin(),(x).end() 19 #define pb push_back 20 #define mp make_pair 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 typedef long long ll; 24 typedef vector<int> VI; 25 typedef pair<int, int> PII; 26 const ll MOD = 1e9 + 7; 27 const int INF = 0x3f3f3f3f; 28 const int MAXN = 5e4 + 7; 29 // head 30 31 int n, m, p; 32 int a[MAXN]; 33 VI G[MAXN]; 34 35 int tim; 36 int siz[MAXN]; //保存以x为根的子树节点个数 37 int top[MAXN]; //保存当前节点的所在链的顶端节点 38 int son[MAXN]; //保存重儿子 39 int dep[MAXN]; //保存当前节点的深度 40 int par[MAXN]; //保存当前节点的父亲 41 int tid[MAXN]; //保存树中每个节点剖分后的新编号 42 int ran[MAXN]; //保存当前节点在线段树中的位置 43 44 void init() { 45 rep(i, 0, MAXN) { 46 G[i].clear(); 47 tim = 0; 48 siz[i] = top[i] = son[i] = dep[i] = par[i] = tid[i] = ran[i] = 0; 49 } 50 } 51 52 void dfs1(int u, int pre){ 53 siz[u] = 1; 54 par[u] = pre; 55 dep[u] = dep[pre] + 1; 56 rep(i, 0, G[u].size()) { 57 int v = G[u][i]; 58 if (v != par[u]){ 59 dfs1(v, u); 60 siz[u] += siz[v]; 61 if (son[u] == 0 || siz[v] > siz[son[u]]) 62 son[u] = v; 63 } 64 } 65 } 66 67 void dfs2(int u, int tp) { 68 top[u] = tp; 69 tid[u] = ++tim; 70 ran[tid[u]] = u; 71 if (son[u]) dfs2(son[u], tp); 72 rep(i, 0, G[u].size()) { 73 int v = G[u][i]; 74 if (v != son[u] && v != par[u]) 75 dfs2(v, v); 76 } 77 } 78 79 int tree[MAXN << 2], lazy[MAXN << 2]; 80 81 void pushdown(int rt, int m) { 82 if (lazy[rt] != 0) { 83 tree[rt << 1] += lazy[rt] * (m - (m >> 1)); 84 tree[rt << 1 | 1] += lazy[rt] * (m >> 1); 85 lazy[rt << 1] += lazy[rt]; 86 lazy[rt << 1 | 1] += lazy[rt]; 87 lazy[rt] = 0; 88 } 89 } 90 91 void build(int l, int r, int rt) { 92 lazy[rt] = 0; 93 if (l == r) { 94 tree[rt] = a[ran[l]]; 95 return; 96 } 97 int m = (l + r) >> 1; 98 build(lson); 99 build(rson); 100 } 101 102 int query(int k, int l, int r, int rt) { 103 if (l == r) return tree[rt]; 104 pushdown(rt, r - l + 1); 105 int m = (l + r) >> 1; 106 if (k <= m) return query(k, lson); 107 else return query(k, rson); 108 } 109 110 void update(int L, int R, int x, int l, int r, int rt) { 111 if (L <= l && r <= R) { 112 tree[rt] += (r - l + 1) *x; 113 lazy[rt] += x; 114 return; 115 } 116 pushdown(rt, r - l + 1); 117 int m = (l + r) >> 1; 118 if (L <= m) update(L, R, x, lson); 119 if (R > m) update(L, R, x, rson); 120 } 121 122 void change(int x, int y, int val) 123 { 124 while (top[x] != top[y]){ 125 if (dep[top[x]] < dep[top[y]]) swap(x, y); 126 update(tid[top[x]], tid[x], val, 1, n, 1); 127 x = par[top[x]]; 128 } 129 if (dep[x] > dep[y]) swap(x, y); 130 update(tid[x], tid[y], val, 1, n, 1); 131 } 132 133 int main() { 134 while (cin >> n >> m >> p) { 135 init(); 136 rep(i, 1, n + 1) scanf("%d", a + i); 137 while (m--) { 138 int u, v; 139 scanf("%d%d", &u, &v); 140 G[u].pb(v); 141 G[v].pb(u); 142 } 143 dfs1(1, 0); 144 dfs2(1, 1); 145 build(1, n, 1); 146 while (p--) { 147 char s[2]; 148 scanf("%s", s); 149 if (s[0] == 'Q') { 150 int c; 151 scanf("%d", &c); 152 printf("%d\n", query(tid[c], 1, n, 1)); 153 } 154 else { 155 int c1, c2, k; 156 scanf("%d%d%d", &c1, &c2, &k); 157 if (s[0] == 'I') change(c1, c2, k); 158 else change(c1, c2, -k); 159 } 160 } 161 } 162 return 0; 163 }
