【luogu P3950 部落冲突】 题解

题目连接：https://www.luogu.org/problemnew/show/P3950
1.像我这种学数据结构学傻了的
2.边权化点权 所有点权初始化0
3.对于战争 将深度较深的-1，对于和平 将深度较深的+1
4.对于历史 用war记录
5.特别注意 在两个查询的LCA处判断是否为-1并且当前的整条路径上的点权也为-1时 同样是yes
代码：

include

include

include

include

define lson left, mid, rt<<1

define rson mid + 1, right, rt<<1|1

using namespace std;
const int maxn = 300000 + 10;
struct war{
int u, v;
}wa[maxn];
int war_cnt = 0;
struct edge{
int from, next, to;
}e[maxn<<2];
int head[maxn], cnt;
int tree[maxn<<2], lazy[maxn<<2];
int seg[maxn], top[maxn], fa[maxn], son[maxn], size[maxn], deep[maxn], rev[maxn], num;
int node[maxn], n, m, root, res;
void add(int u, int v)
{
e[++cnt].from = u;
e[cnt].next = head[u];
e[cnt].to = v;
head[u] = cnt;
}
//-----segment_tree-----
void PushUP(int rt)
{
tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}
void build(int left, int right, int rt)
{
if(left == right)
{
tree[rt] = rev[left];
return;
}
int mid = (left + right)>>1;
build(lson);
build(rson);
PushUP(rt);
}
void PushDOWN(int left, int right, int rt, int mid)
{
lazy[rt<<1] += lazy[rt];
lazy[rt<<1|1] += lazy[rt];
tree[rt<<1] += lazy[rt](mid - left + 1);
tree[rt<<1|1] += lazy[rt](right - mid);
lazy[rt] = 0;
}
void update(int l, int r, int add, int left, int right, int rt)
{
if(l <= left && right <= r)
{
tree[rt] += (right - left + 1)add;
lazy[rt] += add;
return;
}
int mid = (left + right)>>1;
if(lazy[rt]) PushDOWN(left, right, rt, mid);
if(l <= mid) update(l, r, add, lson);
if(r > mid) update(l, r, add, rson);
PushUP(rt);
}
int query(int l, int r, int left, int right, int rt)
{
int res = 0;
if(l <= left && right <= r)
{
return tree[rt];
}
int mid = (left + right)>>1;
if(lazy[rt]) PushDOWN(left, right, rt, mid);
if(l <= mid) res += query(l, r, lson);
if(r > mid) res += query(l, r, rson);
return res;
}
//-----------
void dfs1(int u, int f, int d)
{
int maxson = -1;
size[u] = 1;
deep[u] = d;
fa[u] = f;
for(int i = head[u]; i != -1; i = e[i].next)
{
int v = e[i].to;
if(f != v)
{
dfs1(v, u, d+1);
size[u] += size[v];
if(size[v] > maxson) son[u] = v, maxson = size[v];
}
}
}
void dfs2(int u, int t)
{
seg[u] = ++num;
rev[num] = node[u];
top[u] = t;
if(!son[u]) return;
dfs2(son[u], t);
for(int i = head[u]; i != -1; i = e[i].next)
{
int v = e[i].to;
if(v == son[u] || v == fa[u]) continue;
dfs2(v, v);
}
}
int LCA(int x, int y)
{
while(top[x] != top[y])
{
if(deep[top[x]] < deep[top[y]]) swap(x, y);
x = fa[top[x]];
}
return deep[x] < deep[y] ? x : y;
}
int qRange(int x, int y)
{
int ans = 0;
while(top[x] != top[y])
{
if(deep[top[x]] < deep[top[y]]) swap(x, y);
res = 0;
res = query(seg[top[x]], seg[x], 1, n, 1);
ans = (ans + res);
x = fa[top[x]];
}
if(deep[x] > deep[y]) swap(x, y);
res = 0;
res = query(seg[x], seg[y], 1, n, 1);
ans = (ans + res);
return ans;
}
void updRange(int x, int y, int k)
{
while(top[x] != top[y])
{
if(deep[top[x]] < deep[top[y]]) swap(x, y);
update(seg[top[x]], seg[x], k, 1, n, 1);
x = fa[top[x]];
}
if(deep[x] > deep[y]) swap(x, y);
update(seg[x], seg[y], k, 1, n, 1);
}
int main()
{
//freopen("out.txt","w",stdout);
memset(head, -1, sizeof(head));
memset(node, 0, sizeof(node));
scanf("%d%d",&n,&m);
root = 1;
for(int i = 1; i < n; i++)
{
int u, v;
scanf("%d%d",&u,&v);
add(u, v); add(v, u);
}
dfs1(root,0,1);
dfs2(root,root);
build(1,n,1);
for(int i = 1; i <= m; i++)
{
char opt;
int u, v, x;
cin>>opt;
if(opt == 'Q')
{
scanf("%d%d",&u,&v);
int p = LCA(u, v);
int now = qRange(u, v);
if(now == 0)
{
printf("Yes\n");
continue;
}
int pp = qRange(p, p);
if(now == -1 && pp == -1)
{
printf("Yes\n");
continue;
}
if(now != 0)
{
printf("No\n");
continue;
}
}
if(opt == 'C')
{
scanf("%d%d",&u,&v);
war_cnt++;
wa[war_cnt].u = u;
wa[war_cnt].v = v;
if(deep[u] > deep[v])
updRange(u, u, -1);
else
updRange(v, v, -1);
}
if(opt == 'U')
{
scanf("%d",&x);
if(deep[wa[x].u] > deep[wa[x].v])
updRange(wa[x].u, wa[x].u, 1);
else
updRange(wa[x].v, wa[x].v, 1);
}
}
/for(int i = 1; i <= n; i++)
cout<<qRange(i,i)<<" ";/
return 0;
}
/
7 9
1 2
1 3
3 4
5 3
7 4
4 6
C 3 1
C 3 4
Q 1 2
Q 1 4
Q 4 5
Q 6 7
U 2
Q 5 4
Q 1 5

Yes
No
No
Yes
Yes
No
*/