bzoj 4515: 游戏 树链剖分+线段树
题目大意:
http://www.lydsy.com/JudgeOnline/problem.php?id=4515
题解:
先让我%一发lych大佬点我去看dalao的题解
讲的很详细.
这里纠正一个地方,lych大佬的式子中有一个\(a*(d[s]-d[x])+b=-a*d[x]+(b-a*d[s])\)
应该是\(a*(d[s]-d[x])+b = -a*d[x] + (b+a*d[s])\)
看来是大爷敲式子的时候敲错了...
除去这个就很完美啦
新get了线段树标记永久化的姿势
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
template<typename T>inline void read(T &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const ll maxn = 210010;
const ll inf = 123456789123456789LL;
struct Edge{
ll to,next;
ll dis;
}G[maxn<<1];
ll head[maxn],cnt;
void add(ll u,ll v,ll d){
G[++cnt].to = v;
G[cnt].next = head[u];
head[u] = cnt;
G[cnt].dis = d;
}
ll son[maxn],siz[maxn],fa[maxn],top[maxn];
ll dfn[maxn],dfs_clock,num[maxn],dep[maxn];
ll d[maxn];
#define v G[i].to
void dfs(ll u){
siz[u] = 1;
for(ll i = head[u];i;i=G[i].next){
if(v == fa[u]) continue;
fa[v] = u;
d[v] = d[u] + G[i].dis;
dep[v] = dep[u] + 1;
dfs(v);
siz[u] += siz[v];
if(siz[son[u]] < siz[v]) son[u] = v;
}
}
void dfs(ll u,ll tp){
top[u] = tp;
dfn[u] = ++dfs_clock;
num[dfs_clock] = u;
if(son[u]) dfs(son[u],tp);
for(ll i = head[u];i;i=G[i].next){
if(v == fa[u] || v == son[u]) continue;
dfs(v,v);
}
}
#undef v
struct Node{
ll k,b,minn;
bool lazy;
Node(){k=lazy=0;b = minn = inf;}
}T[maxn<<2];
ll K,B,L,R;ll n;
ll query(ll rt,ll l,ll r){
if(L <= l && r <= R) return T[rt].minn;
ll ret = inf;
if(T[rt].lazy) ret = min(T[rt].k*d[num[max(L,l)]],T[rt].k*d[num[min(R,r)]]) + T[rt].b;
ll mid = l+r >> 1;
if(L <= mid) ret = min(ret,query(rt<<1,l,mid));
if(R > mid) ret = min(ret,query(rt<<1|1,mid+1,r));
return ret;
}
inline ll query(ll l,ll r){
if(l > r) swap(l,r);
L = l;R = r;
return query(1,1,n);
}
void update(ll rt,ll l,ll r,ll K,ll B){
if(T[rt].lazy == 0){
T[rt].k = K;T[rt].b = B;T[rt].lazy = 1;
if(l < r) T[rt].minn = min(T[rt<<1].minn,T[rt<<1|1].minn);
T[rt].minn = min(T[rt].minn,min(T[rt].k*d[num[l]],T[rt].k*d[num[r]])+T[rt].b);
return;
}
ll mid = l+r >> 1;
ll y1 = K*d[num[r]] + B;
ll y2 = T[rt].k*d[num[r]] + T[rt].b;
ll y3 = K*d[num[l]] + B;
ll y4 = T[rt].k*d[num[l]] + T[rt].b;
if(y3 <= y4 && y1 <= y2) T[rt].k = K,T[rt].b = B;
else if(y3 >= y4 && y1 >= y2) return;
else if(T[rt].k > K){
ll x = (B-T[rt].b)/(T[rt].k-K) + 1;
if(x <= d[num[mid]]){
swap(T[rt].b,B);swap(T[rt].k,K);
update(rt<<1,l,mid,K,B);
}else update(rt<<1|1,mid+1,r,K,B);
}else{
ll x = (T[rt].b - B - 1)/(K - T[rt].k);
if(x > d[num[mid]]){
swap(T[rt].b,B);swap(T[rt].k,K);
update(rt<<1|1,mid+1,r,K,B);
}else update(rt<<1,l,mid,K,B);
}
if(l < r) T[rt].minn = min(T[rt<<1].minn,T[rt<<1|1].minn);
if(T[rt].lazy) T[rt].minn = min(T[rt].minn,
min(T[rt].k*d[num[l]],T[rt].k*d[num[r]])+T[rt].b);
}
void ins(ll rt,ll l,ll r){
if(L <= l && r <= R){
update(rt,l,r,K,B);return;
}
ll mid = l+r >> 1;
if(L <= mid) ins(rt<<1,l,mid);
if(R > mid) ins(rt<<1|1,mid+1,r);
T[rt].minn = min(T[rt<<1].minn,T[rt<<1|1].minn);
if(T[rt].lazy) T[rt].minn = min(T[rt].minn,
min(T[rt].k*d[num[l]],T[rt].k*d[num[r]])+T[rt].b);
}
inline void ins(ll l,ll r){
if(l > r) swap(l,r);
L=l;R=r;ins(1,1,n);
}
inline ll lca(ll u,ll v){
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u,v);
u = fa[top[u]];
}return dep[u] < dep[v] ? u : v;
}
inline void insert(ll u,ll v,ll a,ll b){
ll rt = lca(u,v);
K = -a;B = b+a*d[u];
while(top[u] != top[rt]){
ins(dfn[top[u]],dfn[u]);
u = fa[top[u]];
}ins(dfn[rt],dfn[u]);
K = a;B -= (d[rt]<<1)*a;
while(top[v] != top[rt]){
ins(dfn[top[v]],dfn[v]);
v = fa[top[v]];
}ins(dfn[rt],dfn[v]);
}
inline ll query_all(ll u,ll v){
ll ret = inf;
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u,v);
ret = min(ret,query(dfn[top[u]],dfn[u]));
u = fa[top[u]];
}if(dep[u] < dep[v]) swap(u,v);
ret = min(ret,query(dfn[v],dfn[u]));
return ret;
}
int main(){
ll m;read(n);read(m);
for(ll i=1,u,v;i<n;++i){ll d;
read(u);read(v);read(d);
add(u,v,d);add(v,u,d);
}dfs(1);dfs(1,1);
ll op,s,t;ll a,b;
while(m--){
read(op);read(s);read(t);
if(op == 1){
read(a);read(b);
insert(s,t,a,b);
}else{
printf("%lld\n",query_all(s,t));
}
}
getchar();getchar();
return 0;
}
我能说我从昨天下午调到了今天上午,调了6个小时吗?
人就像命运下的蝼蚁,谁也无法操控自己的人生.