《Codeforces Round #661 (Div. 3)》
阿新 • • 發佈：2020-08-13
A：讀錯題導致難度飆升。
這裡題意是任意兩個下標滿足條件都能換，那就是個水題了
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 1e5+5; const int M = 1e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e8 #defineView CodeINM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}return x*f; } int a[55],vis[105]; int main() { int ca;ca = read(); while(ca--) { int n;n = read(); memset(vis,0,sizeof(vis)); for(int i = 1;i <= n;++i) { a[i] = read(); vis[a[i]]++; } sort(a+1,a+n+1); int f = 0; for(int i = 1;i < n;++i) { if(vis[a[i]+1] != 0) continue; if(vis[a[i]] > 1){vis[a[i]]--;continue;} f = 1; } if(f) printf("NO\n"); else printf("YES\n"); } // system("pause"); return 0; }
B：水題
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 1e3+5; const int M = 1e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e18 #define INM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline LL read() { LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } LL a[55],b[55]; int main() { int ca;ca = read(); while(ca--) { int n;n = read(); LL mi1 = INF,mi2 = INF; for(int i = 1;i <= n;++i) a[i] = read(),mi1 = min(mi1,a[i]); for(int i = 1;i <= n;++i) b[i] = read(),mi2 = min(mi2,b[i]); LL ans = 0; for(int i = 1;i <= n;++i) { int dis = a[i]-mi1; ans += dis; int ma = b[i]-mi2; if(ma > dis) ans += ma-dis; // dbg(ans); } printf("%lld\n",ans); } // system("pause"); return 0; }View Code
C：一開始想到二分，但發現不滿足二分性質。
但是這也為解題提供了一個思路，可以發現組合的權值最多隻有[1,100]的可能性，那麼取列舉這個權值即可
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 1e3+5; const int M = 1e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e18 #define INM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } int w[55],n,vis[105],num[105]; int check(int x) { for(int i = 0;i <= 100;++i) num[i] = vis[i]; int ans = 0; for(int i = 1;i <= n;++i) { if(w[i] > x) continue; if(x-w[i] == w[i]) { if(num[w[i]] > 1) num[w[i]] -= 2,ans++; } else if(num[x-w[i]] > 0 && num[w[i]] > 0) { num[x-w[i]]--; num[w[i]]--; ans++; } } return ans; } int main() { int ca;ca = read(); while(ca--) { n = read(); memset(vis,0,sizeof(vis)); for(int i = 1;i <= n;++i) w[i] = read(),vis[w[i]]++; int ans = 0; for(int i = 1;i <= 100;++i) ans = max(ans,check(i)); printf("%d\n",ans); } // system("pause"); return 0; }View Code
D：感覺我想的稍微麻煩了。。能AC就是好題解（逃
可以發現對於每個1，我們要將後面的沒有被接入的第一個0接入當前序列。（0也是同理）
這樣顯然是可以滿足最小的子序列數的。
那麼對於這個後面的最小位置，我們可以用線段樹維護這個位置。
然後一個位置被接入，就去更新線段樹。因為是單點更新，複雜度nlogn，也挺快了。
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 2e5+5; const int M = 1e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e8 #define INM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline int read() { int x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } int ans[N];//0,1 char s[N]; struct Node{int L,r,mi;}node[2][N<<2]; void build(int L,int r,int idx,int id)//id-0,id-1 { node[id][idx].L = L,node[id][idx].r = r; if(L == r) { if(s[L] == '0' && id == 0 || s[L] == '1' && id == 1) node[id][idx].mi = L; else node[id][idx].mi = INF; return ; } int mid = (L+r)>>1; build(L,mid,idx<<1,id); build(mid+1,r,idx<<1|1,id); node[id][idx].mi = min(node[id][idx<<1].mi,node[id][idx<<1|1].mi); } void update(int x,int idx,int id) { if(node[id][idx].L == node[id][idx].r) { node[id][idx].mi = INF; return ; } int mid = (node[id][idx].L + node[id][idx].r)>>1; if(mid >= x) update(x,idx<<1,id); else update(x,idx<<1|1,id); node[id][idx].mi = min(node[id][idx<<1].mi,node[id][idx<<1|1].mi); } int query_mi(int L,int r,int idx,int id) { if(node[id][idx].L >= L && node[id][idx].r <= r) return node[id][idx].mi; int mid = (node[id][idx].L+node[id][idx].r)>>1,mi = INF; if(mid >= L) mi = min(mi,query_mi(L,r,idx<<1,id)); if(mid < r) mi = min(mi,query_mi(L,r,idx<<1|1,id)); return mi; } int main() { int ca;ca = read(); while(ca--) { int n;n = read(); scanf("%s",s+1); build(1,n,1,0); build(1,n,1,1); int ma = 0; memset(ans,0,sizeof(ans)); for(int i = 1;i <= n;++i) { if(ans[i] == 0) ans[i] = ++ma; int tmp; if(s[i] == '0') tmp = query_mi(i+1,n,1,1); else tmp = query_mi(i+1,n,1,0); if(tmp == INF) continue; ans[tmp] = ans[i]; if(s[tmp] == '0') update(tmp,1,0); else update(tmp,1,1); } printf("%d\n",ma); for(int i = 1;i <= n;++i) printf("%d%c",ans[i],i == n ? '\n' : ' '); } //system("pause"); return 0; }View Code
E1：讀錯題導致難度飆升（梅開二度，老笨蛋了）
注意的是，這裡是要使總(root-leaves)路徑權值和小於s。
那麼，我們可以去記錄路徑被經過的次數，然後乘上這個路徑的w-w/2，這就是讓它折半一次的能對總權值減少的貢獻。
那麼顯然我們每次都選最大的，就能在最少的次數內滿足。
這裡這個被經過次數很顯然就是子節點數。堆來維護最大值即可
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 1e5+5; const int M = 1e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e18 #define INM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline LL read() { LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } struct Node{int to;LL dis;}; vector<Node> G[N]; int ssize[N]; LL sum = 0; struct Point{ LL cost,w,cnt; bool operator < (const Point &a)const{ return cost < a.cost; } }; priority_queue<Point> Q; LL cal(LL w,int cnt) { return 1LL*cnt*(w-w/2); } void dfs(int u,int fa,LL dis) { if(G[u].size() == 1) ssize[u] = 1; for(auto v : G[u]) { if(v.to == fa) continue; dfs(v.to,u,v.dis); ssize[u] += ssize[v.to]; } if(dis != 0) { //printf("dis is %lld u is %d ssize is %d\n",dis,u,ssize[u]); sum += 1LL*ssize[u]*dis; Q.push(Point{cal(dis,ssize[u]),dis,ssize[u]}); } } int main() { int ca;ca = read(); while(ca--) { int n;LL s; n = read(),s = read(); for(rg int i = 1;i <= n;++i) G[i].clear(),ssize[i] = 0; for(rg int i = 1;i < n;++i) { int u,v;LL w; u = read(),v = read(),w = read(); G[u].push_back(Node{v,w}); G[v].push_back(Node{u,w}); } while(!Q.empty()) Q.pop(); sum = 0; dfs(1,0,0); int ans = 0; while(!Q.empty() && sum > s) { ans++; Point q = Q.top(); Q.pop(); sum -= q.cost; q.w /= 2; q.cost = cal(q.w,q.cnt); Q.push(q); } printf("%d\n",ans); } // system("pause"); return 0; }View Code
E2：最小化花費了，可以用二分來。
因為c只有1,2，那麼我們可以去維護兩個堆。
一個維護1，一個維護2。開兩個字首和陣列
用陣列1來維護花費代價為i時能減少的最大權值。
陣列2來維護花費代價為2*i時能減少的最大權值。
我們列舉1的花費，然後去二分最小的2代價使得sum-兩個字首和陣列和 <= s
memset初始化陣列T了，其實沒必要，只要讓第0個為0即可（改了之後快的飛起~~蕪湖）
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double ld; typedef pair<LL,int> pii; const int N = 1e5+5; const int M = 2e6+5; const LL Mod = 998244353; #define rg register #define pi acos(-1) #define INF 1e18 #define INM INT_MIN #define dbg(ax) cout << "now this num is " << ax << endl; inline LL read() { LL x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();} return x*f; } struct Node{int to;LL dis,cost;}; vector<Node> G[N]; struct Point{ int cnt;LL w,cost; bool operator < (const Point &a)const{ return cost < a.cost; } }; LL cal(LL w,int cnt) { return 1LL*cnt*(w-w/2); } priority_queue<Point> Q1,Q2; int ssize[N]; LL sum = 0,pre1[M],pre2[M],s; void dfs(int u,int fa,LL dis,int c) { if(G[u].size() == 1) ssize[u] = 1; for(auto v : G[u]) { if(v.to == fa) continue; dfs(v.to,u,v.dis,v.cost); ssize[u] += ssize[v.to]; } if(dis != 0) { sum += 1LL*ssize[u]*dis; if(c == 1) Q1.push(Point{ssize[u],dis,cal(dis,ssize[u])}); else Q2.push(Point{ssize[u],dis,cal(dis,ssize[u])}); } } bool check(LL tmp,int x) { return (tmp-pre2[x]) <= s; } int main() { int ca;ca = read(); while(ca--) { int n; n = read(),s = read(); for(int i = 1;i <= n;++i) G[i].clear(),ssize[i] = 0; for(rg int i = 1;i < n;++i) { int u,v,w,c; u = read(),v = read(),w = read(),c = read(); G[u].push_back(Node{v,w,c}); G[v].push_back(Node{u,w,c}); } while(!Q1.empty()) Q1.pop(); while(!Q2.empty()) Q2.pop(); pre1[0] = pre2[0] = sum = 0; dfs(1,0,0,0); int cnt1 = 0,cnt2 = 0; while(!Q1.empty() && sum-pre1[cnt1] > s) { Point q = Q1.top();Q1.pop(); ++cnt1; pre1[cnt1] = pre1[cnt1-1]+q.cost; q.w /= 2; q.cost = cal(q.w,q.cnt); if(q.cost != 0) Q1.push(q); } while(!Q2.empty() && sum-pre2[cnt2] > s) { Point q = Q2.top();Q2.pop(); ++cnt2; pre2[cnt2] = pre2[cnt2-1]+q.cost; q.w /= 2; q.cost = cal(q.w,q.cnt); if(q.cost != 0) Q2.push(q); } LL ans = INF; for(int i = 0;i <= cnt1;++i) { LL ma = sum-pre1[i]; int L = 0,r = cnt2,ta = 1e8; while(L <= r) { int mid = (L+r)>>1; if(check(ma,mid)) { ta = mid; r = mid-1; } else L = mid+1; } ans = min(ans,1LL*i+ta*2); } printf("%lld\n",ans); } //system("pause"); return 0; }View Code
F：待補