题目链接
树链剖分的学习笔记（更新中）
这道题所给的Hint好有迷惑性，它跟我们说注意士兵的数量可能为负数，我的第一反应是，士兵的数量是不是不能为负数，那么我们是不是要做出些什么调整，然而，语文不好的我看了Discuss才知道说的是：士兵的数量可以为负数。这样也好，题目就变成了一道简单的树链剖分问题了，我们只需要记录每个节点的深度，以及它的最顶节点，然后边更新就是了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 50005;
int N, M, Q, cnt, head[maxN], root[maxN], depth[maxN], size[maxN], W_Son[maxN], top[maxN], w[maxN], new_W[maxN], id[maxN], num;
ll tree[maxN<<2], lazy[maxN<<2];
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN<<1];
void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
void dfs1(int u, int fa, int deep)
{
root[u] = fa;
depth[u] = deep;
size[u] = 1;
int maxxSon = -1;
for(int i=head[u]; i!=-1; i=edge[i].nex)
{
int v = edge[i].to;
if(v == fa) continue;
dfs1(v, u, deep+1);
size[u] += size[v];
if(size[v] > maxxSon)
{
maxxSon = size[v];
W_Son[u] = v;
}
}
}
void dfs2(int x, int topf)
{
top[x] = topf;
id[x] = ++num;
new_W[num] = w[x];
if(!W_Son[x]) return;
dfs2(W_Son[x], topf);
for(int i=head[x]; i!=-1; i=edge[i].nex)
{
int y = edge[i].to;
if(y == root[x] || y == W_Son[x]) continue;
dfs2(y, y);
}
}
void pushup(int rt)
{
tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}
void buildTree(int rt, int l, int r)
{
lazy[rt] = 0;
if(l == r)
{
tree[rt] = new_W[l];
return;
}
int mid = (l + r)>>1;
buildTree(rt<<1, l, mid);
buildTree(rt<<1|1, mid+1, r);
pushup(rt);
}
void pushdown(int rt, int l, int r)
{
if(lazy[rt])
{
int mid = (l + r)>>1;
lazy[rt<<1] += lazy[rt];
lazy[rt<<1|1] += lazy[rt];
tree[rt<<1] += lazy[rt]*(mid - l + 1);
tree[rt<<1|1] += lazy[rt]*(r - mid);
lazy[rt] = 0;
}
}
void update(int rt, int l, int r, int ql, int qr, int val)
{
if(ql<=l && qr>=r)
{
lazy[rt] += val;
tree[rt] += val*(r - l + 1);
return;
}
pushdown(rt, l, r);
int mid = (l + r)>>1;
if(ql>mid) update(rt<<1|1, mid+1, r, ql, qr, val);
else if(qr<=mid) update(rt<<1, l, mid, ql, qr, val);
else
{
update(rt<<1|1, mid+1, r, ql, qr, val);
update(rt<<1, l, mid, ql, qr, val);
}
pushup(rt);
}
ll query(int rt, int l, int r, int ql, int qr)
{
if(ql<=l && qr>=r) return tree[rt];
pushdown(rt, l, r);
int mid = (l + r)>>1;
if(ql>mid) return query(rt<<1|1, mid+1, r, ql, qr);
else if(qr<=mid) return query(rt<<1, l, mid, ql, qr);
else
{
ll ans = query(rt<<1|1, mid+1, r, ql, qr);
ans += query(rt<<1, l, mid, ql, qr);
return ans;
}
}
void update_Range(int x, int y, int val)
{
while(top[x] != top[y])
{
if(depth[top[x]] < depth[top[y]]) swap(x, y);
update(1, 1, N, id[top[x]], id[x], val);
x = root[top[x]];
}
if(depth[x] > depth[y]) swap(x, y);
update(1, 1, N, id[x], id[y], val);
}
ll query_Point(int x)
{
return query(1, 1, N, id[x], id[x]);
}
void init()
{
cnt = num = 0;
memset(head, -1, sizeof(head));
memset(W_Son, 0, sizeof(W_Son));
}
int main()
{
while(scanf("%d%d%d", &N, &M, &Q)!=EOF)
{
init();
for(int i=1; i<=N; i++) scanf("%d", &w[i]);
int e1, e2;
for(int i=1; i<=M; i++)
{
scanf("%d%d", &e1, &e2);
addEddge(e1, e2);
addEddge(e2, e1);
}
dfs1(1, 1, 0);
dfs2(1, 1);
buildTree(1, 1, N);
while(Q--)
{
char s[3];
scanf("%s", s);
if(s[0] == 'I')
{
int L, R, val;
scanf("%d%d%d", &L, &R, &val);
update_Range(L, R, val);
}
else if(s[0] == 'D')
{
int L, R, val;
scanf("%d%d%d", &L, &R, &val);
update_Range(L, R, -val);
}
else
{
int X;
scanf("%d", &X);
printf("%lld\n", query_Point(X));
}
}
}
return 0;
}