A.00:45:42(-1) solved by zcz
队友说抄个板子好了，我看了一下板子是扩展中国剩余定理
#include<iostream> #include<cstdio> #include<climits> #include<cstring> #include<map> #include<algorithm> using namespace std; #define LL long long const int maxn=1e5+5; int n; LL exgcd(LL a,LL b,LL &x,LL &y){ if(!b){x=1,y=0;return a;} LL re=exgcd(b,a%b,x,y),tmp=x; x=y,y=tmp-(a/b)*y; return re; } LL m[maxn],a[maxn]; //m为模数集，a为余数集 LL exCRT(){ LL M=m[1],A=a[1],t,d,x,y;int i; for(i=2;i<=n;i++){ d=exgcd(M,m[i],x,y);//解方程 if((a[i]-A)%d)return -1;//无解 x*=(a[i]-A)/d,t=m[i]/d,x=(x%t+t)%t;//求x A=M*x+A,M=M/d*m[i],A%=M;//日常膜一膜（划掉）模一模，防止爆 } A=(A%M+M)%M; return A; } map<long long ,int> isF; int main() { long long f1=1,f2=1; while(f1<=1e16) { isF[f1]=1; long long t=f1; f1=f1+f2; f2=t; } int i,j; while(scanf("%d",&n)!=EOF){ for(i=1;i<=n;i++)scanf("%lld%lld",&m[i],&a[i]); long long t=exCRT(); if(t==-1) { cout<<"Tankernb!"<<endl; continue; } if(isF[t]==0) { cout<<"Zgxnb!"<<endl; } else { cout<<"Lbnb!"<<endl; } } return 0; }
B.00:11:42 solved by hl
范围小的话直接并查集，这题范围比较大，所以要离散化(雾)
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 3e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int fa[maxn]; int Hash[maxn]; struct query{ int op,x; }query[maxn]; int find(int x){ if(x == fa[x]) return x; return fa[x] = find(fa[x]); } void Union(int a,int b){ a = find(a); b = find(b); fa[a] = b; } int main(){ Sca2(N,M); int cnt = 0; for(int i = 1; i <= M ; i ++){ query[i].op = read(); query[i].x = read(); Hash[++cnt] = query[i].x; if(query[i].op == 1){ Hash[++cnt] = query[i].x + 1; } } sort(Hash + 1,Hash + 1 + cnt); cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash; for(int i = 1; i <= cnt + 1; i ++) fa[i] = i; Hash[cnt + 1] = -1; for(int i = 1; i <= M ; i ++){ query[i].x = lower_bound(Hash + 1,Hash + 1 + cnt,query[i].x) - Hash; if(query[i].op == 1){ Union(query[i].x,query[i].x + 1); }else{ Pri(Hash[find(query[i].x)]); } } return 0; }
C.00:12:36(-3) solved by zcz
枚举题意，枚举到和出题人心意相通为止
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<vector> #include<map> using namespace std; int main() { int w; while(cin>>w) { if(w%2==0&&w>2) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } } return 0; }
D.1:05:12 solved by hl
我就比较好奇，出这种题也有钱拿吗？
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; string S,T; int main(){ cin >> T; Sca(N); while(N--){ cin >> S; if(S.length() < T.length()){ if(T.find(S) != S.npos) puts("my child!"); else puts("oh, child!"); }else if(S.length() > T.length()){ if(S.find(T) != S.npos) puts("my teacher!"); else puts("senior!"); }else{ if(S == T) puts("jntm!"); else puts("friend!"); } } return 0; }
E.00:38:48 solved by hl
从后往前枚举，树状数组维护下标最大的值，树状数组下标为数的大小
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K,cnt; int a[maxn],Hash[maxn]; int tree[maxn]; void add(int u,int v){ for(;u > 0; u -= u & -u) tree[u] = max(tree[u],v); } int getmax(int u){ int ans = 0; for(;u <= cnt ; u += u & -u) ans = max(ans,tree[u]); return ans; } int ans[maxn]; int main(){ Sca2(N,M); cnt = 0; for(int i = 1; i <= N ; i ++){ a[i] = Hash[++cnt] = read(); Hash[++cnt] = a[i] + M ; } sort(Hash + 1,Hash + 1 + cnt); cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash - 1; for(int i = N ; i >= 1; i --){ int t = getmax(lower_bound(Hash + 1,Hash + 1 + cnt,a[i] + M) - Hash); // cout << "bug" << i << endl; // cout << lower_bound(Hash + 1,Hash + 1 + cnt,a[i] + M) - Hash << endl; // cout << t << endl; a[i] = lower_bound(Hash + 1,Hash + 1 + cnt,a[i]) - Hash; if(!t) ans[i] = -1; else ans[i] = t - i - 1; // cout << ans[i] << endl; add(a[i],i); } for(int i = 1; i <= N ; i ++){ printf("%d",ans[i]); if(i != N) printf(" "); } return 0; }
F.03:14:29 solved by hl
本场唯一做的一道非签到题，提前感谢一下出题人
题目数据范围给的很奥妙重重，5000个询问和最多为100个k的情况仿佛在暗示什么
果不其然，一个询问点最多往上数100个祖先，枚举每个祖先作为lca对这个点产生的贡献，将会产生一个数的深度的范围，这个点的贡献是他所有子树中深度不超过某个数产生的，同时减去他的祖先对他的影响，每个询问就会变成至多100个询问。
求一个dfs序，题目就转化为了多次询问一个序列中区间所有大于等于p的数的权值和，把所有询问离线出来，用树状数组解决就好了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 2e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL a[maxn]; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int dep[maxn],p[maxn * 2],cnt,fa[maxn]; PII pos[maxn]; void dfs(int u,int la){ pos[u].fi = ++cnt; p[cnt] = u; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; fa[v] = u; dep[v] = dep[u] + 1; dfs(v,u); } pos[u].se = ++cnt; p[cnt] = u; } struct Query{ int l,r,id; int k,flag; Query(int l = 0,int r = 0,int id = 0,int k = 0,int flag = 0):l(l),r(r),id(id),k(k),flag(flag){} }q[maxn]; int Stack[maxn]; bool cmp(Query a,Query b){ return a.k < b.k; } vector<int>Q[maxn]; LL tree[maxn],ans[maxn]; void add(int u,LL v){ for(;u < maxn; u += u & -u) tree[u] +=v ; } LL getsum(int u){ LL sum = 0; for(;u > 0; u -= u & -u) sum += tree[u]; return sum; } int main(){ Sca(N); init(); cnt = 0; for(int i = 1; i <= N ; i ++) a[i] = read(); for(int i = 1; i <= N - 1; i ++){ int u = read(),v = read(); add(u,v); add(v,u); } dep[1] = 0; fa[1] = -1; dfs(1,-1); Sca(M); int o = 0; for(int i = 1; i <= M ; i ++){ int u,w; Sca2(u,w); int j,root; Stack[0] = u; for(j = 0,root = u; j < w && fa[root] != -1;){ root = fa[root]; Stack[++j] = root; } int Mi = -1; for(int k = j ;k >= 0 ; k --){ int v = Stack[k]; int Mx = dep[v] + w - (dep[u] - dep[v]); if(Mi >= Mx) continue; if(Mi >= 0) q[++o] = Query(pos[v].fi,pos[v].se,i,Mi,-1); if(Mx >= 0) q[++o] = Query(pos[v].fi,pos[v].se,i,Mx,1); Mi = max(Mi,Mx); } } sort(q + 1,q + 1 + o,cmp); int f = 1; for(int i = 1; i <= N ; i ++){ Q[dep[i]].pb(i); } for(int i = 0; f <= o ; i ++){ for(int j = 0 ; j < Q[i].size(); j ++){ int v = Q[i][j]; add(pos[v].fi,a[v]); } while(f <= o && q[f].k == i){ ans[q[f].id] += getsum(q[f].l - 1) * -1 * q[f].flag; ans[q[f].id] += getsum(q[f].r) * q[f].flag; f++; } } for(int i = 1; i <= M; i ++){ Prl(ans[i]); } return 0; }
G.00:50:53 solved by hl
回文树，bitset维护一下每个结点字母的数量直接冲就行了
不用bitset,建完树之后dfs统计也行，我寻思bitset可能会快一点，空间复杂度也顶住了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; const int maxn = 3e5 + 10; const int maxc = 26; struct Pal_T{ int next[maxn][maxc]; int fail[maxn],cnt[maxn],num[maxn],len[maxn],S[maxn]; bitset<maxc>P[maxn]; int last,n,tot; int newnode(int l){ for(int i = 0 ; i < maxc; i ++) next[tot][i] = 0; cnt[tot] = num[tot] = 0; len[tot] = l; P[tot].reset(); return tot++; } void init(){ tot = 0; newnode(0); //偶数根节点 newnode(-1); //奇数根节点 last = n = 0; S[n] = -1; fail[0] = 1; } int getfail(int x){ while(S[n - len[x] - 1] != S[n]) x = fail[x]; return x; } void add(int c){ c -= 'a'; S[++n] = c; int cur = getfail(last); if(!next[cur][c]){ int now = newnode(len[cur] + 2); P[now] = P[cur]; P[now][c - 'a'] = 1; fail[now] = next[getfail(fail[cur])][c]; next[cur][c] = now; num[now] = num[fail[now]] + 1; } last = next[cur][c]; cnt[last]++; } LL count(){ LL ans = 0; for(int i = tot - 1; i >= 0; i --){ cnt[fail[i]] += cnt[i]; ans += 1ll * P[i].count() * cnt[i]; } return ans; } }PT; char str[maxn]; int main(){ scanf("%s",str); PT.init(); for(int i = 0;str[i]; i ++) PT.add(str[i]); Prl(PT.count()); return 0; }
I.2:01:19 solved by hl
因为原图是个全排列，可以证明即使将所有有关系的数连边，空间复杂度也是nlogn
这就变成了一个区间查询内部包含多少线段的问题了,简单的二位偏序问题，只要把E题敲完的树状数组复制过来就可以了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int maxm = 2e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn],id[maxn]; struct Line{ int l,r; Line(int l = 0,int r = 0):l(l),r(r){} }line[maxm]; struct Query{ int l,r,id; }q[maxn]; bool cmp(Query a,Query b){ return a.r < b.r; } bool cmp2(Line a,Line b){ return a.r < b.r; } int tree[maxn]; void add(int u,int v){ for(;u > 0; u -= u & -u) tree[u] += v; } int getsum(int u){ int ans = 0; for(;u < N; u += u & -u) ans += tree[u]; return ans; } int ans[maxn]; int main(){ Sca2(N,M); for(int i = 1; i <= N; i ++){ a[i] = read(); id[a[i]] = i; } int cnt = 0; for(int i = 1; i <= N ; i ++){ for(int j = i + i ; j <= N ; j += i){ line[++cnt] = Line(id[i],id[j]); if(line[cnt].l > line[cnt].r) swap(line[cnt].l,line[cnt].r); } } for(int i = 1; i <= M ; i ++){ q[i].l = read(); q[i].r = read(); q[i].id = i; } sort(q + 1,q + 1 + M,cmp); sort(line + 1,line + 1 + cnt,cmp2); int f = 1; for(int i = 1; i <= M ; i ++){ while(f <= cnt && line[f].r <= q[i].r){ add(line[f++].l,1); } ans[q[i].id] = getsum(q[i].l); } for(int i = 1; i <= M ; i ++){ Pri(ans[i]); } return 0; }
J.2:11:54 solved by gbs
队友写的
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; const int maxn = 1e6 + 15; const int mod = 1e9 + 7; vector<int> pn[maxn + 15] ; int th_inv[maxn + 15] ; int high1[maxn+15]; int quick_mi(LL a, LL k) { LL ans = 1; while (k) { if (k & 1) { ans *= a; ans %= mod; } a *= a; a %= mod; k >>= 1; } return ans; } int the_hight; int find_hight(int nowi,int fu) { int nowa = pn[nowi].size(); if (fu != 0) { nowa--; } if (nowa == 0) { return high1[nowi] = 1; } for (unsigned int i=0; i<pn[nowi].size(); i++) { if (pn[nowi][i] == fu) { continue; } high1[nowi] = max(high1[nowi],find_hight(pn[nowi][i],nowi)+1); } return high1[nowi]; } int inva(int a) { if (th_inv[a] == 0) return th_inv[a] = quick_mi(a,mod-2); return th_inv[a]; } int the_ans(int nowi, int fu,int di) { int now_ans = 0; if (di + high1[nowi] != the_hight) { return 1; } int nowa = pn[nowi].size(); if (fu != 0) { nowa--; } if (nowa == 0) { return 0; } for (unsigned int i=0; i<pn[nowi].size(); i++) { if (pn[nowi][i] == fu) { continue; } now_ans = (0LL + now_ans + the_ans(pn[nowi][i],nowi,di+1))%mod; } //cout<<nowi <<' '<<nowa<<' '<<now_ans <<endl; now_ans = (1LL * now_ans *inva(nowa))%mod; now_ans = quick_mi(now_ans ,nowa); //cout<<"*"<<' '<<nowi<<' '<<now_ans <<endl; return now_ans ; } int main() { int n; int a, b; cin >> n; for (int i = 0; i < n-1; i++) { scanf("%d%d", &a, &b); pn[a].push_back(b); pn[b].push_back(a); } find_hight(1,0); the_hight = high1[1]; //printf("%d ",(the_ans(1,0,0)+mod)%mod ); printf("%d ",(1-the_ans(1,0,0)+mod)%mod ); #ifdef VSCode system("pause"); #endif return 0; }
K.00:49:48 solved by gbs
队友写的
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; int xxnn[1005]; int yynn[1005]; struct pointa{ int xi; int yi; pointa(int x1 =0,int y1 =0) { xi =x1; yi = y1; } bool operator <(const pointa & s1)const{ if (xi!= s1.xi) return this->xi <s1.xi; return this->yi <s1.yi; } bool operator ==(const pointa & s1)const{ return (this->xi ==s1.xi)&&(this->yi ==s1.yi); } }; int main() { int n; cin >>n; map<pointa,int>s1; for (int i=0; i<n; i++) { scanf("%d%d",&xxnn[i],&yynn[i]); if (s1[pointa(xxnn[i],yynn[i])] >=1) { n--; i--; continue; } else { s1[pointa(xxnn[i],yynn[i])]=1; xxnn[i] *= 2; yynn[i] *= 2; } } int fin_ans = 0; int llx; int lly; map<pointa,int>map1; for (int i=0; i<n; i++) { for (int j=i; j<n; j++) { llx = (xxnn[i]+xxnn[j])/2; lly = (yynn[i]+yynn[j])/2; if (xxnn[i]==xxnn[j] && yynn[i]==yynn[j]) { map1[pointa(llx,lly)]+=1; } else { map1[pointa(llx,lly)]+=2; } fin_ans = max(fin_ans,map1[pointa(llx,lly)]); } } printf("%d ",n-fin_ans); #ifdef VSCode system("pause"); #endif return 0; }
L.3:36:27(-4) solved by zcz,hl
预处理出任意两点间，红点朝上翻滚到另一点红点朝上需要的步数
就是个旅行商问题了
状压dp入门门槛
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1 << 16; const LL INF = 1e18; const int mod = 1e9 + 7; int N,M,K; LL dis[4005][4005]; void init(){ dis[0][0]=0; int k=3; int cnt=1; for(int i=1;i<=4000;i++) { if(cnt%4==0) { dis[i][0]=dis[0][i]=cnt; } else { dis[i][0]=k; dis[0][i]=k; } dis[i][1]=dis[1][i]=dis[i][0]+1; k++; cnt++; } for(int i=2;i<=4000;i++) { for(int j=2;j<=4000;j++) { dis[i][j]=i+j; } } dis[1][1]=6; dis[1][2]=dis[2][1]=5; dis[1][3]=dis[3][1]=6; dis[2][2]=4; } PII node[20]; LL MAP[20][20]; LL dp[20][maxn]; int main(){ init(); int T = read(); while(T--){ Sca(N); for(int i = 0; i < N ; i ++){ node[i].fi = read(); node[i].se = read(); } for(int i = 0; i < N ; i ++){ for(int j = 0; j < N ; j ++){ MAP[i][j] = dis[abs(node[i].fi - node[j].fi)][abs(node[i].se - node[j].se)]; } MAP[N][i] = MAP[i][N] = dis[abs(node[i].fi)][abs(node[i].se)]; } int ed = (1 << N); for(int i = 0 ; i <= N ; i ++){ for(int j = 0 ; j < ed; j ++) dp[i][j] = INF; } dp[N][0] = 0; for(int i = 0 ; i < ed; i ++){ for(int j = 0 ; j <= N ; j ++){ if(dp[j][i] == INF) continue; for(int k = 0; k < N ; k ++){ if(i & (1 << k)) continue; int v = i | (1 << k); dp[k][v] = min(dp[k][v],dp[j][i] + MAP[j][k]); } } } LL ans = INF; for(int i = 0 ; i < N ; i ++) ans = min(ans,dp[i][ed - 1]); Prl(ans); } return 0; }
M.01:47:31(-4) solved by hl
因为眼睛不好使子序列看成了子串，敲完EXKMP还WA了一发
发现a串只要比b串中的某个位置大，a串后面就都符合，前提是当前位置前面需要找的出子序列和b串位置的前缀一样
树状数组维护前缀最大值，表示a串比这个字符大的时候，前缀最多能匹配多少个数，答案就是max(N - i + 1 + max(0 ~ a[i]))
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d ", x) #define Prl(x) printf("%lld ",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; 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 * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; char a[maxn],b[maxn]; int tree[maxn]; void add(int u,int v){ for(;u <= 27 ; u += u & -u)tree[u] = max(tree[u],v); } int getmax(int u){ int ans = -1; for(;u > 0; u -= u & -u)ans = max(ans,tree[u]); return ans; } int main(){ Sca2(N,M); scanf("%s%s",a,b); int cnt = 0,ans = -1; b[M] = 'a' - 1; for(int i = 0 ; i <= 27; i ++) tree[i] = -1; add(b[0] - 'a' + 2,0); for(int i = 0 ; i < N; i ++){ int v = a[i] - 'a' + 2; int t = getmax(v - 1); if(~t) ans = max(ans,N - i + t); if(cnt < M&& a[i] == b[cnt]){ cnt++; add(b[cnt] - 'a' + 2,cnt); } } Pri(ans); return 0; }