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;
}
原文:http://www.cnblogs.com/dwtfukgv/p/7426596.html