Solutions - NOISG 2020 重现赛
T1
主观难度：【0】
不会做的出门左转，一机房有初一小朋友正在学前缀和。
T2
主观难度：【1】
容易看出方程式 \(dp_i = dp_j + (n-j) \max_{k-j+1}^{i} a_k\)。这个式子是很容易想到斜率优化的。但是合并凸壳不太好做。
观察 Subtask 4，我们发现这个 Subtask 的答案显然等于 \(na_1\)。仔细思考后我们发现对于一个连续的不增子序列，设子序列开头为 \(i\)，中间任一元素为 \(j\)，如果你把 \(j\) 选为一段的开头那么在付出 \(a_i(n-i)\) 的代价之外你还需要付出额外的 \(a_j(n-j)\) 的代价，但是如果 \(i\) 的段包含了 \(j\) 那么只需要付出 \(a_i(n-i)\) 的代价。故对于一个连续的不增子序列，只选开头最优。
于是我们把这个序列变成了不降的，发现方程式变为了 \(dp_i = dp_j + (n-j)a_i\)，斜率优化搞搞即可。
#include <bits/stdc++.h>
#define llong long long
#define N 1000006
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n;
llong a[N], dp[N];
struct Seg{
llong k, b;
};
#define gety(s,x) (s.k*x+s.b)
Seg val[N<<6]; int ls[N<<6], rs[N<<6], tsiz, root;
#define mid ((l+r)>>1)
inline int newnode(){
return ++tsiz;
}
inline void modify(Seg k, int& x = root, int l = 1, int r = 1e9){
if(!x) return val[x = newnode()] = k, void();
Seg k1 = val[x], k2 = k;
if(gety(k1, mid) > gety(k2, mid)) swap(k1, k2);
val[x] = k1;
if(l == r) return;
if(gety(k1, l) > gety(k2, l)) modify(k2, ls[x], l, mid );
if(gety(k1, r) > gety(k2, r)) modify(k2, rs[x], mid+1, r);
return;
}
inline llong query(int pos, int& x = root, int l = 1, int r = 1e9){
if(!x) return 1e18+3;
llong res = gety(val[x], pos);
if(pos <= mid) res = min(res, query(pos, ls[x], l, mid ));
else res = min(res, query(pos, rs[x], mid+1, r));
return res;
}
int main(){
// freopen("in.txt", "r", stdin);
read(n);
for(int i = 1; i <= n; ++i) read(a[i]);
for(int i = 1; i <= n; ++i) a[i] = max(a[i], a[i-1]);
modify({n, 0});
for(int i = 1; i <= n; ++i){
dp[i] = query(a[i]);
modify({n-i, dp[i]});
}
printf("%lld", dp[n]);
return 0;
}
T3
主观难度：【0】
花了 2.5h 吃了一坨大的，非常美味。
容易想到维护斜率。但是碍于区间赋值操作，区间的两端不好维护。于是我们开两个线段树，一个维护实际值，一个维护斜率。就做完了。
时间复杂度 \(\mathrm O(n \log n)\)。
代码长度 6.5k。好大一坨。
#include <bits/stdc++.h>
#define llong long long
#define N 300005
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, q;
int a[N];
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define mid ((l+r)>>1)
class SegT1{
private:
struct Seg{llong k, b; bool vis;};
Seg val[N<<2], tag1[N<<2], tag2[N<<2];
inline void addtag1(int x, Seg k){
tag2[x] = {0, 0, false};
val[x] = tag1[x] = k;
return;
}
inline void addtag2(int x, Seg k){
val[x].k += k.k, tag2[x].k += k.k;
val[x].b += k.b, tag2[x].b += k.b;
tag2[x].vis = true;
return;
}
inline void pushdown(int x){
if(tag1[x].vis){
addtag1(ls(x), tag1[x]);
addtag1(rs(x), tag1[x]);
tag1[x] = {0, 0, false};
}
if(tag2[x].vis){
addtag2(ls(x), tag2[x]);
addtag2(rs(x), tag2[x]);
tag2[x] = {0, 0, false};
}
return;
}
public:
inline void build(int x = 1, int l = 1, int r = n){
tag1[x] = tag2[x] = {0, 0, false};
if(l == r) return addtag1(x, {0, a[l], true});
build(ls(x), l, mid), build(rs(x), mid+1, r);
return;
}
inline void modify1(int L, int R, Seg k, int x = 1, int l = 1, int r = n){
if(L <= l && R >= r) return addtag1(x, k);
pushdown(x);
if(L <= mid) modify1(L, R, k, ls(x), l, mid );
if(R > mid) modify1(L, R, k, rs(x), mid+1, r);
return;
}
inline void modify2(int L, int R, Seg k, int x = 1, int l = 1, int r = n){
if(L <= l && R >= r) return addtag2(x, k);
pushdown(x);
if(L <= mid) modify2(L, R, k, ls(x), l, mid );
if(R > mid) modify2(L, R, k, rs(x), mid+1, r);
return;
}
inline llong query(int pos, int x = 1, int l = 1, int r = n){
if(l == r) return val[x].k*pos+val[x].b;
pushdown(x);
if(pos <= mid) return query(pos, ls(x), l, mid );
else return query(pos, rs(x), mid+1, r);
}
inline llong operator[](int pos){
return query(pos);
}
} val;
class SegT2{
private:
struct Node{
int l, r; int val, len1, len2;
llong val1, val2;
bool vis;
};
Node val[N<<2]; llong tag1[N<<2], tag2[N<<2];
const llong NOTAG = (llong)1e18+3;
Node merge(Node a, Node b){
if(!a.vis) return b;
if(!b.vis) return a;
Node res; res.vis = true;
res.l = a.l, res.r = b.r;
res.val = max(a.val, b.val);
if(a.val2 == b.val1) res.val = max(res.val, a.len2+b.len1);
res.val1 = a.val1, res.val2 = b.val2;
if(a.len1 == a.r-a.l+1 && a.val1 == b.val1) res.len1 = a.len1+b.len1;
else res.len1 = a.len1;
if(b.len2 == b.r-b.l+1 && a.val2 == b.val2) res.len2 = a.len2+b.len2;
else res.len2 = b.len2;
return res;
}
inline void pushup(int x){
val[x] = merge(val[ls(x)], val[rs(x)]);
return;
}
inline void addtag1(int x, llong k){
tag1[x] = k, tag2[x] = NOTAG;
Node res;
res.l = val[x].l, res.r = val[x].r;
res.val1 = res.val2 = k;
res.val = res.len1 = res.len2 = val[x].r-val[x].l+1;
res.vis = true;
val[x] = res;
return;
}
inline void addtag2(int x, llong k){
if(tag2[x] == NOTAG) tag2[x] = 0;
val[x].val1 += k, val[x].val2 += k;
tag2[x] += k;
return;
}
inline void pushdown(int x){
if(tag1[x] != NOTAG){
addtag1(ls(x), tag1[x]);
addtag1(rs(x), tag1[x]);
tag1[x] = NOTAG;
}
if(tag2[x] != NOTAG){
addtag2(ls(x), tag2[x]);
addtag2(rs(x), tag2[x]);
tag2[x] = NOTAG;
}
}
public:
inline void build(int x = 1, int l = 1, int r = n-1){
tag1[x] = tag2[x] = NOTAG;
if(l == r) return val[x] = {l, r, 1, 1, 1, a[l+1]-a[l], a[l+1]-a[l], true}, void();
build(ls(x), l, mid), build(rs(x), mid+1, r);
pushup(x);
}
inline void modify1(int L, int R, llong k, int x = 1, int l = 1, int r = n-1){
if(L <= l && R >= r) return addtag1(x, k);
pushdown(x);
if(L <= mid) modify1(L, R, k, ls(x), l, mid );
if(R > mid) modify1(L, R, k, rs(x), mid+1, r);
pushup(x);
return;
}
inline void modify2(int L, int R, llong k, int x = 1, int l = 1, int r = n-1){
if(L <= l && R >= r) return addtag2(x, k);
pushdown(x);
if(L <= mid) modify2(L, R, k, ls(x), l, mid );
if(R > mid) modify2(L, R, k, rs(x), mid+1, r);
pushup(x);
return;
}
inline Node query(int L, int R, int x = 1, int l = 1, int r = n-1){
if(L <= l && R >= r) return val[x];
pushdown(x);
Node res; res.vis = false;
if(L <= mid) res = merge(res, query(L, R, ls(x), l, mid ));
if(R > mid) res = merge(res, query(L, R, rs(x), mid+1, r));
return res;
}
inline int evaluate(int L, int R){
return query(L, R-1).val+1;
}
} seg;
// In all SegT options, 1 = cover, 2 = add
inline int solve1(){
while(q--){
int op, l, r, k, b;
read(op, l, r);
if(op == 1 || op == 2) read(k, b);
if(op == 3) puts("1");
}
return 0;
}
int main(){
// freopen("in.txt", "r", stdin);
read(n, q);
for(int i = 1; i <= n; ++i) read(a[i]);
if(n == 1) return solve1();
val.build(), seg.build();
while(q--){
int op, l, r; read(op, l, r);
if(op == 1){ // Add
llong k, b; read(b, k);
b -= l*k;
val.modify2(l, r, {k, b, true});
if(l != r) seg.modify2(l, r-1, k);
if(l != 1) seg.modify1(l-1, l-1, val[l]-val[l-1]);
if(r != n) seg.modify1(r, r, val[r+1]-val[r]);
}
if(op == 2){ // Cover
llong k, b; read(b, k);
b -= l*k;
val.modify1(l, r, {k, b, true});
if(l != r) seg.modify1(l, r-1, k);
if(l != 1) seg.modify1(l-1, l-1, val[l]-val[l-1]);
if(r != n) seg.modify1(r, r, val[r+1]-val[r]);
}
if(op == 3){ // Evaluate
if(l == r) puts("1");
else printf("%d\n", seg.evaluate(l, r));
}
}
return 0;
}
T4
主观难度：【1+】
模拟赛场上糖了。
我们把每个点位能到达的区域画在图上，发现是从原点开始向左下和右下无限延伸的两条射线组成的图形。这不太好看，于是我们令 \(x_i = t_i+a_i, y_i = t_i-a_i\)，坐标系被我们旋转了 \(45 \degree\)（顺便还放大到了 \(\sqrt{2}\)，但这不重要）。于是发现一个点可达另一个点当且仅当 \(x_i \le x_j \land y_i \le y_j\)。这下方便多了。
然而 Hootime 场上做到这里不会做了。
其实接下来要求的就是最小链覆盖。我们通过 Dilworth 定理将其转化为最长反链，于是这题就变成了导弹拦截，用你喜欢的方式求出 LIS 即可。
就做完了。
#include <bits/stdc++.h>
#define llong long long
#define N 500005
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, m;
struct Node{int x, y;};
Node a[N];
int tmp[N], cnt;
int main(){
freopen("in.txt", "r", stdin);
read(m, n);
for(int i = 1; i <= n; ++i) read(a[i].x);
for(int i = 1; i <= n; ++i) read(a[i].y);
for(int i = 1; i <= n; ++i) a[i] = {a[i].x+a[i].y, a[i].x-a[i].y};
sort(a+1, a+n+1, [&](Node o1, Node o2){return o1.x>o2.x || (o1.x==o2.x&&o1.y>o2.y);});
for(int i = 1; i <= n; ++i){
int pos = lower_bound(tmp+1, tmp+cnt+1, a[i].y)-tmp;
tmp[pos] = a[i].y;
if(pos > cnt) ++cnt;
}
printf("%d", cnt);
return 0;
}
T5
主观难度：【2】
脑子抽了导致没看懂题解。
我们建出最短路树。然后答案肯定只有走最短路加权和不走最短路。
我们枚举每条非树边 \((u, v)\)，容易发现 \((u, v)\) 可以替换最短路上的一段。我们记其在最短路上的祖先与最短路第一次相交的节点为 \(b_u, b_v\)，容易发现路径 \(1 \to u \to v \to n\) 与最短路不同的段只有 \(b_u \to u \to v \to b_v\)。于是我们在 \(b_u\) 标记一下有一条距离为距离为 \(dis_{1,u}+len_{(u, v)}+dis_{v, n}\) 的路径可以替换 \(b_u\) 到以后某点的路径，在 \(b_v\) 删除标记。
然后我们枚举最短路上的边，该边被伸长后的最短路即为 \(\min\{ dis_{1, n}+w_i, \min_{k \in s_i} k \}\)，其中 \(k\) 为可替换该边的路径长度，\(w_i\) 为可对该边加权的边的最大权。
然后对所有东西取 \(\max\) 即可。
复杂度 \(\mathrm O(n \log n)\)。
Tip：其实 multiset 可以当可删堆用。
#include <bits/stdc++.h>
#define llong long long
#define N 300005
#define M 300005
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, m, l;
int to[M<<1], nxt[M<<1], id[M<<1], tag[M<<1], head[N], gsiz; llong val[M<<1];
#define mkarc(u,v,w1,w2) (++gsiz, to[gsiz]=v, val[gsiz]=w1, id[gsiz]=w2, nxt[gsiz]=head[u], head[u]=gsiz)
int fa[N], bel[N], vis[N], sta[N], nid[N], eid[N], top;
llong eval[N], dis1[N], dis2[N], ans;
typedef pair<llong, int> Node;
priority_queue<Node, vector<Node>, greater<Node> > pq;
vector<llong> tmp1[N], tmp2[N];
multiset<llong> tmp;
int main(){
freopen("in.txt", "r", stdin);
read(n, m);
for(int i = 1; i <= m; ++i){
int u, v; llong w; read(u, v, w);
mkarc(u, v, w, i), mkarc(v, u, w, i);
eval[i] = w;
}
for(int i = m; i >= 1; --i) eval[i] = max(eval[i], eval[i+1]);
for(int i = 1; i <= n; ++i) dis1[i] = dis2[i] = (llong)1e18+3;
dis1[1] = 0, pq.emplace(0, 1);
while(pq.size()){
int u = pq.top().second; pq.pop();
if(vis[u]) continue;
vis[u] = true;
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(dis1[u]+val[i] < dis1[v]){
dis1[v] = dis1[u]+val[i], fa[v] = u;
pq.emplace(dis1[v], v);
}
}
}
for(int i = 1; i <= n; ++i) vis[i] = false;
dis2[n] = 0, pq.emplace(0, n);
while(pq.size()){
int u = pq.top().second; pq.pop();
if(vis[u]) continue;
vis[u] = true;
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(dis2[u]+val[i] < dis2[v]){
dis2[v] = dis2[u]+val[i];
pq.emplace(dis2[v], v);
}
}
}
for(int i = 1; i <= n; ++i) vis[i] = false;
int now = n;
while(now != 0) bel[now] = now, vis[now] = true, now = fa[now];
for(int i = 2; i <= n; ++i){
int j = i;
while(!bel[j]) sta[++top] = j, j = fa[j];
while(top) bel[sta[top]] = bel[j], --top;
}
now = n;
while(now != 0) sta[++l] = now, now = fa[now];
reverse(sta+1, sta+l+1);
nid[n] = l;
for(int u = n; u != 1; u = fa[u]){
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(v != fa[u]) continue;
eid[v] = id[i], nid[v] = nid[u]-1;
break;
}
}
for(int u = 1; u <= n; ++u){
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(fa[u] == v || fa[v] == u) continue;
if(bel[u] == bel[v]) continue;
if(nid[bel[u]] > nid[bel[v]]) continue;
tmp1[nid[bel[u]]].push_back(dis1[u]+val[i]+dis2[v]);
tmp2[nid[bel[v]]].push_back(dis1[u]+val[i]+dis2[v]);
}
}
for(int i = 1; i < l; ++i){
for(llong val : tmp1[i]) tmp.insert(val);
for(llong val : tmp2[i]) tmp.erase(tmp.lower_bound(val));
ans = max(ans, min(dis1[n]+eval[eid[sta[i]]+1], tmp.size() ? *tmp.begin() : (llong)1e18+3));
}
printf("%lld", ans);
return 0;
}
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