HDU 6162 Ch’s gift (线段树+树链剖分)
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题意:给定上一棵树,每个树的结点有一个权值,有 m 个询问,每次询问 s, t , a, b,问你从 s 到 t 这条路上,权值在 a 和 b 之间的和。(闭区间)。
析:很明显的树链剖分,但是要用线段树来维护,首先先离线,然后按询问的 a 排序,每次把小于 a 的权值先更新上,然后再查询,这样就是区间求和了,算完小于a的,再算b的,最答案相减就好了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const LL mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r > 0 && r <= n && c > 0 && c <= m; } struct Val{ int x, id; }; Val a[maxn]; struct Edge{ int to, next; }; Edge edge[maxn<<1]; int head[maxn<<1], cnt; void add(int u, int v){ edge[cnt].to = v; edge[cnt].next = head[u]; head[u] = cnt++; } struct Query{ int s, t, a, b; int id; }; Query q[maxn]; inline bool cmpVal(const Val &lhs, const Val &rhs){ return lhs.x < rhs.x; } inline bool cmpL(const Query &lhs, const Query &rhs){ return lhs.a < rhs.a; } inline bool cmpR(const Query &lhs, const Query &rhs){ return lhs.b < rhs.b; } int fa[maxn], top[maxn], p[maxn]; int pos, son[maxn], num[maxn], dep[maxn]; void init(){ cnt = 0; pos = 0; ms(head, -1); ms(son, -1); } void dfs1(int u, int f, int d){ fa[u] = f; dep[u] = d; num[u] = 1; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == f) continue; dfs1(v, u, d+1); num[u] += num[v]; if(son[u] == -1 || num[son[u]] < num[v]) son[u] = v; } } void dfs2(int u, int sp){ top[u] = sp; p[u] = ++pos; if(son[u] == -1) return ; dfs2(son[u], sp); for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == son[u] || v == fa[u]) continue; dfs2(v, v); } } LL sum[maxn<<2]; void push_up(int rt){ sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void update(int M, int val, int l, int r, int rt){ if(l == r){ sum[rt] = val; return ; } int m = l + r >> 1; if(M <= m) update(M, val, lson); else update(M, val, rson); pu(rt); } LL query(int L, int R, int l, int r, int rt){ if(L <= l && r <= R) return sum[rt]; int m = l + r >> 1; LL ans = 0; if(L <= m) ans = query(L, R, lson); if(R > m) ans += query(L, R, rson); return ans; } LL solve(int u, int v){ int f1 = top[u], f2 = top[v]; LL ans = 0; while(f1 != f2){ if(dep[f1] < dep[f2]){ swap(f1, f2); swap(u, v); } ans += query(p[f1], p[u], all); u = fa[f1]; f1 = top[u]; } if(dep[u] > dep[v]) swap(u, v); return ans += query(p[u], p[v], all); } LL ansL[maxn], ansR[maxn]; int main(){ while(scanf("%d %d", &n, &m) == 2){ init(); for(int i = 1; i <= n; ++i){ scanf("%d", &a[i].x); a[i].id = i; } for(int i = 1; i < n; ++i){ int u, v; scanf("%d %d", &u, &v); add(u, v); add(v, u); } dfs1(1, -1, 0); dfs2(1, 1); for(int i = 0; i < m; ++i){ scanf("%d %d %d %d", &q[i].s, &q[i].t, &q[i].a, &q[i].b); --q[i].a; q[i].id = i; } ms(sum, 0); sort(a + 1, a + n + 1, cmpVal); sort(q, q + m, cmpL); int idx = 1; for(int i = 0; i < m; ++i){ while(idx <= n && a[idx].x <= q[i].a) update(p[a[idx].id], a[idx].x, all), idx++; ansL[q[i].id] = solve(q[i].s, q[i].t); } ms(sum, 0); sort(q, q + m, cmpR); idx = 1; for(int i = 0; i < m; ++i){ while(idx <= n && a[idx].x <= q[i].b) update(p[a[idx].id], a[idx].x, all), idx++; ansR[q[i].id] = solve(q[i].s, q[i].t); } for(int i = 0; i < m; ++i){ if(i) putchar(‘ ‘); printf("%I64d", ansR[i]-ansL[i]); } printf("\n"); } return 0; }
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