队友写的。 好像水水的。
//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include<bits/stdc++.h> #define fi first #define se second #define db double #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 998244353 #define ld long double //#define C 0.5772156649 //#define ls l,m,rt<<1 //#define rs m+1,r,rt<<1|1 #define pll pair<ll,ll> #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> #define ull unsigned long long //#define base 1000000000000000000 #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;} inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const ull ba=233; const db eps=1e-5; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=1000000+10,maxn=1000000+10,inf=0x3f3f3f3f; ll dp[200][2]; int main() { int l,k;scanf("%d%d",&l,&k); dp[1][1]=dp[k][1]=1; for(int i=1;i<=l;i++) { dp[i+1][0]+=dp[i][1]; dp[i+1][1]+=dp[i][0]; dp[i+k][1]+=dp[i][0]; } ll ans=0; for(int i=1;i<=l;i++)ans+=dp[i][1]; printf("%lld ",ans); return 0; } /******************** ********************/
队友写的。
//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include<bits/stdc++.h> #define fi first #define se second #define db double #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 998244353 #define ld long double //#define C 0.5772156649 //#define ls l,m,rt<<1 //#define rs m+1,r,rt<<1|1 #define pll pair<ll,ll> #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> #define ull unsigned long long //#define base 1000000000000000000 #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;} inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const ull ba=233; const db eps=1e-5; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f; ll a[N]; int main() { ll n,t;scanf("%lld%lld",&n,&t); for(int i=1;i<=n;i++)scanf("%lld",&a[i]); ll ma=0,sum=0; for(int i=1;i<=n;i++) { ma=max(ma,a[i]); sum=sum+a[i]; if(t<sum)printf("%d ",1); else { ll te=(t-sum)/ma+2; printf("%lld ",te); } } return 0; } /******************** ********************/
dfs暴力枚举两两组合的情况， 枚举选第一个没被选的和枚举的组合， 这样能把复杂度降到最低。
然后答案一遍加入一遍更新， 好像我常数扣的有点厉害。 复杂度是15 * 13 * 11 * ... * 3
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 16 + 7; const int M = 2000 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); struct Point { int x, y; } p[N]; int X[16][16], Y[16][16], all; int num[4007][4007]; int n, cnt[1 << 16], who[1 << 16]; int ans, tmp; vector<int> go[1 << 16]; inline void change(int x, int y, int op) { x += 2000, y += 2000; if(op == 1) tmp += num[x][y]; else tmp -= num[x][y] - 1; num[x][y] += op; } void dfs(int S) { if(cnt[S] + 2 > n) { ans = max(ans, tmp); return; } for(int i = 1; i < SZ(go[S]); i++) { change(X[go[S][0]][go[S][i]], Y[go[S][0]][go[S][i]], 1); dfs(S | (1 << go[S][0]) | (1 << go[S][i])); change(X[go[S][0]][go[S][i]], Y[go[S][0]][go[S][i]], -1); } } int main() { for(int i = 1; i < (1 << 16); i++) cnt[i] = cnt[i - (i & -i)] + 1; scanf("%d", &n); all = n / 2; for(int i = 0; i < n; i++) { scanf("%d%d", &p[i].x, &p[i].y); } for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { X[i][j] = p[i].x - p[j].x; Y[i][j] = p[i].y - p[j].y; int gcd = __gcd(X[i][j], Y[i][j]); X[i][j] /= gcd; Y[i][j] /= gcd; if(X[i][j] < 0) X[i][j] = -X[i][j], Y[i][j] = -Y[i][j]; // printf("%d %d: %d %d ", i, j, X[i][j], Y[i][j]); } } for(int i = 0; i < (1 << n); i++) for(int k = 0; k < n; k++) if(!(i >> k & 1)) go[i].push_back(k); dfs(0); printf("%d ", ans); return 0; } /* */
I - Starting a Scenic Railroad Service
队友写的。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> #include<iostream> #include<stack> #include<string> #include<map> using namespace std; #define pb push_back #define pf push_front #define lb lower_bound #define ub upper_bound #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define mst(x,a) memset(x,a,sizeof(x)) #define all(x) (x).begin(),(x).end() #define CLOSE ios::sync_with_stdio(false) typedef pair<int,int> pii; typedef long long ll; typedef vector<int> vi; #define fi first #define se second #define sz(x) ((int)x.size()) #define cl(x) x.clear() const int mod = 1000000007; const int N = 200010; const int INF=0x3f3f3f3f; void MOD(ll &a){if(a>=mod) a-=mod;} void MOD(ll &a,ll c){if(a>=c) a-=c;} void ADD(ll &a,ll b){ a+=b; MOD(a);} void ADD(ll &a,ll b,ll c){a+=b;MOD(a,c);} ll qpow(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod;b/=2;}return ans;} ll qpow(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c;b/=2;}return ans;} int n; int x[N],y[N],B[N]; struct seg{ int l,r; }a[N],b[N]; bool cmp1(seg a,seg b){ if(a.l==b.l) return a.r<b.r; return a.l<b.l; } bool cmp2(seg a,seg b){ if(a.r==b.r) return a.l<b.l; return a.r<b.r; } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d%d",&x[i],&y[i]); y[i]--; B[x[i]]++; B[y[i]+1]--; a[i].l=x[i]; a[i].r=y[i]; b[i].l=x[i]; b[i].r=y[i]; } sort(a+1,a+1+n,cmp1); sort(b+1,b+1+n,cmp2); int ans1=0,ans2=0; for(int i=1;i<=100001;i++){ B[i]+=B[i-1]; ans2=max(ans2,B[i]); } //a sort by l b sort by r for(int i=1;i<=n;i++){ int tmp=n; int l=1,r=n+1; while(l+1<r){ int m=l+r>>1; if(a[m].l>y[i]) r=m; else l=m; } tmp-=n-r+1; l=0,r=n; while(l+1<r){ int m=l+r>>1; if(b[m].r<x[i]) l=m; else r=m; } tmp-=l; ans1=max(ans1,tmp); } printf("%d %d ",ans1,ans2); } /* */
感觉是比较常规的套路题。
把边(u, v, w)反转以后如果 dis[ 1 ][ v ] + dis[ 2 ][ u ] + w < 最短路则是happy， 这里有个问题就是反转之后1到不了v， 但是反转前到到的了，
那么dis[ 1 ][ v ] 就不是INF， 但是通过分析我们能发现这不影响结果。
然后就是 sad 和 soso， 这个对 1 到 2 之间的最短路图求个桥就好了，看看反转的是不是桥。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); int n, m, a[N], b[N], c[N]; LL dis[2][N]; bool flag[N]; struct node { LL d; int to, id; bool operator < (const node& rhs) const { return d > rhs.d; } }; vector<node> G[N], rG[N], sG[N]; priority_queue<node> que; void Dij(int S, LL dis[N]) { dis[S] = 0; que.push(node{0, S, 0}); while(!que.empty()) { int u = que.top().to; LL d = que.top().d; que.pop(); if(d > dis[u]) continue; for(auto& t : G[u]) { if(d + t.d < dis[t.to]) { dis[t.to] = d + t.d; que.push(node{dis[t.to], t.to, 0}); } } } } void Dij2(int S, LL dis[N]) { dis[S] = 0; que.push(node{0, S, 0}); while(!que.empty()) { int u = que.top().to; LL d = que.top().d; que.pop(); if(d > dis[u]) continue; for(auto& t : rG[u]) { if(d + t.d < dis[t.to]) { dis[t.to] = d + t.d; que.push(node{dis[t.to], t.to, 0}); } } } } int dfn[N], low[N], idx; bool in[N]; void tarjan(int u, int fa) { in[u] = true; ++idx; dfn[u] = low[u] = idx; for(node& t : sG[u]) { int v = t.to; if(v == fa) continue; if(!dfn[v]) { tarjan(v, u); low[u] = min(low[u], low[v]); if(dfn[u] < low[v]) flag[t.id] = true; } else if(in[v]) { low[u] = min(low[u], dfn[v]); } } in[u] = false; } int main() { memset(dis, INF, sizeof(dis)); scanf("%d%d", &n, &m); for(int i = 1; i <= m; i++) { scanf("%d%d%d", &a[i], &b[i], &c[i]); G[a[i]].push_back(node{c[i], b[i], i}); rG[b[i]].push_back(node{c[i], a[i], i}); } Dij(1, dis[0]); Dij2(2, dis[1]); LL mn = dis[0][2]; if(mn != INF) { for(int i = 1; i <= n; i++) { for(node t : G[i]) { if(t.d + dis[0][i] + dis[1][t.to] == mn) { sG[i].push_back(t); sG[t.to].push_back(node{t.d, i, t.id}); } } } tarjan(1, 0); for(int i = 1; i <= m; i++) { if(dis[0][b[i]] + dis[1][a[i]] + c[i] < mn) { puts("HAPPY"); } else { if(flag[i]) puts("SAD"); else puts("SOSO"); } } } else { for(int i = 1; i <= m; i++) { if(dis[0][b[i]] + dis[1][a[i]] + c[i] < mn) { puts("HAPPY"); } else { puts("SAD"); } } } return 0; } /* */
赛后补题**********************************************************************************************************************************************
G - Rendezvous on a Tetrahedron
训练结束半个多小时才码出来。。。 把四面体展开， 然后暴力的取找到它最后的位置在哪里， 好像有更方便的方法， 我写得好麻烦啊啊。。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); struct Point { double x, y; void print() { printf("%.3f %.3f ^^ ", x, y); } Point(double x = 0, double y = 0) : x(x), y(y) { } }; typedef Point Vector; int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } Point operator + (Vector A, Vector B) {return Point(A.x + B.x, A.y + B.y);} Point operator - (Vector A, Vector B) {return Point(A.x - B.x, A.y - B.y);} Point operator * (Vector A, double p) {return Point(A.x * p, A.y * p);} Point operator / (Vector A, double p) {return Point(A.x / p, A.y / p);} bool operator < (const Vector &A, const Vector &B) {return A.y < B.y || (A.y == B.y && A.x < B.x);} bool operator == (const Vector &A, const Point &B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0;} double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} double Length(Vector A) {return sqrt(Dot(A, A));} double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} Vector Rotate(Vector A, double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad)); } double GetLineIntersectionTime(Point P, Vector v, Point Q, Vector w) { Vector u = P - Q; double t = Cross(w, u) / Cross(v, w); return t; } double dist(const Point& a, const Point &b) { return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); } bool isPointOnSegment(const Point &p, const Point &a1, const Point &a2) { if(dcmp(Cross(a1-p,a2-p))) return 0; else if(dcmp(p.x-min(a1.x,a2.x))>=0&&dcmp(p.x-max(a1.x,a2.x))<=0 &&dcmp(p.y-min(a1.y,a2.y))>=0&&dcmp(p.y-max(a1.y,a2.y))<=0) return 1; else return 0; } int isPointInPolygon(Point p, Point *poly, int n) { int wn = 0; for(int i = 0; i < n; i++) { if(isPointOnSegment(p, poly[i], poly[(i+1)%n])) return -1; //在边界上 int k = dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i])); int d1 = dcmp(poly[i].y-p.y); int d2 = dcmp(poly[(i+1)%n].y-p.y); if(k>0 && d1<=0 && d2>0) wn++; if(k<0 && d2<=0 && d1>0) wn--; } if(wn != 0) return 1; //内部 return 0; //外部 } Point A2 = Point(0, 0); Point A3 = Point(2, 0); Point A1 = Point(1, sqrt(3.0)); Point D = (A1 + A2) / 2; Point B = (A1 + A3) / 2; Point C = (A2 + A3) / 2; Point poly1[] = {A1, D, B}; Point poly2[] = {D, C, B}; Point poly3[] = {D, A2, C}; Point poly4[] = {B, C, A3}; bool isA12(Point x) { return isPointOnSegment(x, A1, A2); } bool isA23(Point x) { return isPointOnSegment(x, A2, A3); } bool isA31(Point x) { return isPointOnSegment(x, A3, A1); } int belong(Point x) { if(isPointInPolygon(x, poly1, 3) == 1) return 1; if(isPointInPolygon(x, poly2, 3) == 1) return 2; if(isPointInPolygon(x, poly3, 3) == 1) return 3; if(isPointInPolygon(x, poly4, 3) == 1) return 4; } Point dfs(Point p, Vector v, double len) { if(isA12(p)) { double t1 = GetLineIntersectionTime(p, v, A2, A3 - A2); double t2 = GetLineIntersectionTime(p, v, A1, A3 - A1); Point nxtp; if(dcmp(t1) > 0 && (dcmp(t2) <= 0 || dcmp(t1) > 0 && t1 < t2)) { nxtp = p + (v * t1); double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = C * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } else { nxtp = p + v * t2; double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = B * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } } else if(isA23(p)) { double t1 = GetLineIntersectionTime(p, v, A1, A1 - A2); double t2 = GetLineIntersectionTime(p, v, A1, A1 - A3); Point nxtp; if(dcmp(t1) > 0 && (dcmp(t2) <= 0 || dcmp(t1) > 0 && t1 < t2)) { nxtp = p + v * t1; double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = D * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } else { nxtp = p + v * t2; double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = B * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } } else if(isA31(p)) { double t1 = GetLineIntersectionTime(p, v, A2, A2 - A1); double t2 = GetLineIntersectionTime(p, v, A2, A2 - A3); Point nxtp; if(dcmp(t1) > 0 && (dcmp(t2) <= 0 || dcmp(t1) > 0 && t1 < t2)) { nxtp = p + v * t1; double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = D * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } else { nxtp = p + v * t2; double gg = dist(p, nxtp); if(dcmp(len - gg) < 0) { return p + (v / Length(v)) * len; } else { nxtp = C * 2.0 - nxtp; v = Rotate(v, PI); return dfs(nxtp, v, len - gg); } } } } Point read() { char s[10]; double d, l; scanf("%s%lf%lf", s, &d, &l); if(s[0] == 'C' && s[1] == 'D') { Point startp = A2; Vector startv = Rotate(C - A2, d / 180 * PI); return dfs(startp, startv, l); } else if(s[0] == 'D' && s[1] == 'B') { Point startp = A1; Vector startv = Rotate(D - A1, d / 180 * PI); return dfs(startp, startv, l); } else { Point startp = A3; Vector startv = Rotate(B - A3, d / 180 * PI); return dfs(startp, startv, l); } } int main() { Point Point1 = read(); Point Point2 = read(); if(belong(Point1) == belong(Point2)) puts("YES"); else puts("NO"); return 0; } /* */
不会写， 但是看了题解觉得好有道理啊啊啊， 我怎么没想到呢。
我们把t[ i ] == s[ i ]的位置的颜色保留， 把颜色不同的当成没有颜色就好了。
我们定义dp[ i ] 表示把 1 - i 全部染成对应颜色所需要的次数。
我们把 t 中连续相同的一段看成一个整体。变成 WBWBWB的形式， 如果长度为 m 则答案为 (m / 2) + 1
如果t[ i ] == s[ i ] dp[ i ] = dp[ i - 1]
否则 dp[ i ] = min(dp[ j ] + (sum[ i ] - sum[ j + 1 ]) / 2 + 1)
很显然能发现能用单调队列优化。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); int n, k, head, rear, sum[N], que[N]; int dp[N]; char s[N], t[N]; int main() { scanf("%d%d", &n, &k); scanf("%s%s", s + 1, t + 1); for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (t[i] != t[i - 1]); head = 1, rear = 0; dp[0] = 0; que[++rear] = 0; for(int i = 1; i <= n; i++) { if(t[i] == s[i]) { dp[i] = dp[i - 1]; } else { while(head <= rear && i - que[head] > k) head++; dp[i] = dp[que[head]] + (sum[i] - sum[que[head] + 1] + 1) / 2 + 1; } while(head <= rear && 2*dp[i] - sum[i + 1] <= 2*dp[que[rear]] - sum[que[rear] + 1]) rear--; que[++rear] = i; } printf("%d ", dp[n]); return 0; } /* */