11.26 模拟赛
写的分基本全挂了非常的菜
T1 password
题目大意:
$m$个模式串 求长度为$n$的串中包含所有$m$个串的方案数
$m\le4,len\le50$ $len=字符串总长度$
思路:
可以想到一个$n\times len \times 2^m$的dp
然后将状压部分转换成容斥 使用矩阵加速 矩阵i j 表示节点i - j的转移
(死于全集没写快速幂)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 #include<cstring> 7 #include<vector> 8 #include<queue> 9 #include<map> 10 #define rep(i,s,t) for(register int i=(s);i<=(t);++i) 11 #define dwn(i,s,t) for(register int i=(s);i>=(t);--i) 12 #define ren for(register int i=fst[x];i;i=nxt[i]) 13 #define Fill(x,t) memset(x,t,sizeof(x)) 14 #define ll long long 15 #define inf 2139062143 16 #define MAXN 1400 17 #define MOD 998244353 18 using namespace std; 19 inline int read() 20 { 21 int x=0,f=1;char ch=getchar(); 22 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 23 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 24 return x*f; 25 } 26 inline int add(ll a,ll b) {return a+b<MOD?a+b:a+b-MOD;} 27 char s[5][60]; 28 int n,m,ch[MAXN][10],fail[MAXN],sz,ed[MAXN],cnt; 29 ll tot; 30 void ins(int x) 31 { 32 int len=strlen(s[x]),pos=0; 33 rep(i,0,len-1) 34 { 35 if(!ch[pos][s[x][i]-'0']) ch[pos][s[x][i]-'0']=++sz; 36 pos=ch[pos][s[x][i]-'0']; 37 } 38 cnt++,ed[pos]|=(1<<cnt); 39 } 40 int q[MAXN],l=1,r,vis[MAXN]; 41 void build() 42 { 43 int x; 44 rep(i,0,9) if(ch[0][i]) q[++r]=ch[0][i]; 45 while(l<=r) 46 { 47 x=q[l++]; 48 rep(i,0,9) if(ch[x][i]) fail[ch[x][i]]=ch[fail[x]][i],q[++r]=ch[x][i]; 49 else ch[x][i]=ch[fail[x]][i]; 50 ed[x]|=ed[fail[x]]; 51 } 52 } 53 struct Mat 54 { 55 ll num[55][55]; 56 Mat(){Fill(num,0);} 57 Mat operator * (const Mat &a) const 58 { 59 Mat res; 60 rep(i,0,sz) rep(j,0,sz) rep(k,0,sz) 61 res.num[i][j]=add(res.num[i][j],(num[i][k]*a.num[k][j])%MOD); 62 return res; 63 } 64 }; 65 Mat res; 66 ll calc(int x) 67 { 68 vis[x]=1;ll ans=res.num[0][x]; 69 rep(i,0,9) if(!vis[ch[x][i]]&&!ed[ch[x][i]]) ans=add(ans,calc(ch[x][i])); 70 return ans; 71 } 72 ll q_pow(Mat x,ll t) 73 { 74 Fill(res.num,0);rep(i,0,sz) res.num[i][i]=1; 75 for(;t;t>>=1,x=x*x) 76 if(t&1) res=x*res; 77 //rep(i,0,sz) {rep(j,0,sz) cout<<res.num[i][j]<<" ";puts("");} 78 return calc(0); 79 } 80 ll qaq(ll bas,ll t) 81 { 82 ll res=1; 83 for(;t;t>>=1,(bas*=bas)%=MOD) 84 if(t&1) (res*=bas)%=MOD; 85 return res; 86 } 87 int main() 88 { 89 freopen("password.in","r",stdin); 90 freopen("password.out","w",stdout); 91 m=read(),n=read();int ms=(1<<m)-1;Mat tmp; 92 rep(i,1,m) scanf("%s",s[i]);tot=1; 93 tot=qaq(10LL,n); 94 rep(t,1,ms) 95 { 96 Fill(fail,0);Fill(ch,0);Fill(vis,0);Fill(tmp.num,0);Fill(ed,0);sz=cnt=0; 97 rep(i,1,m) if(1&(t>>(i-1))) ins(i); 98 build();rep(i,0,sz) if(!ed[i]) rep(j,0,9) tmp.num[i][ch[i][j]]++; 99 if(cnt&1) tot=(tot+MOD-q_pow(tmp,n))%MOD;else tot=add(tot,q_pow(tmp,n)); 100 } 101 printf("%lld\n",tot); 102 }
T2 paint
题目大意:
一个数列分成若干段 每段长度属于$[l,r]$ 设该段和为x 每一段的价值为$a*x^2+b*x+c$ 求最终的最大价值
思路:
看上去像是一个斜优 但是由于斜率和x坐标都不单调,还有长度的限制
考虑线段树分治 每个点可以对之后的一段区间造成影响 用线段树记录
对于每个点 我们将包含它的线段树上的节点的凸包搞出来 在每个凸包上二分出答案 并在这$log_n$个答案中取最优
(中间有结果会爆long long 所以特判了一波)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 #include<cstring> 7 #include<vector> 8 #include<queue> 9 #include<map> 10 #define rep(i,s,t) for(register int i=(s);i<=(t);++i) 11 #define dwn(i,s,t) for(register int i=(s);i>=(t);--i) 12 #define ren(x) for(register int i=fst[x];i;i=nxt[i]) 13 #define Fill(x,t) memset(x,t,sizeof(x)) 14 #define ll long long 15 #define inf 1152921504606846976LL 16 #define maxi 2147483647 17 #define MAXN 600100 18 #define MOD 998244353 19 using namespace std; 20 inline ll read() 21 { 22 ll x=0,f=1;char ch=getchar(); 23 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 24 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 25 return x*f; 26 } 27 ll a,b,c,sum[MAXN],dp[MAXN]; 28 int n,fst[MAXN],to[MAXN<<4],nxt[MAXN<<4],cnt,l,r; 29 vector <int> vec[MAXN]; 30 void add(int u,int v) {nxt[++cnt]=fst[u],fst[u]=cnt,to[cnt]=v;} 31 void mdf(int k,int l,int r,int a,int b,int w) 32 { 33 if(l==a&&r==b) {add(k,w);return ;} 34 int mid=l+r>>1; 35 if(b<=mid) mdf(k<<1,l,mid,a,b,w); 36 else if(a>mid) mdf(k<<1|1,mid+1,r,a,b,w); 37 else {mdf(k<<1,l,mid,a,mid,w);mdf(k<<1|1,mid+1,r,mid+1,b,w);} 38 } 39 int top,q[MAXN],tl; 40 struct Point 41 { 42 ll x,y,fi;int id; 43 Point operator - (const Point &a) const {return (Point){x-a.x,y-a.y,fi,id};} 44 bool operator < (const Point &a) const {return x<a.x;} 45 ll operator * (const Point &a) const {return (ll)x*a.y-y*a.x;} 46 }st[MAXN]; 47 ll calc(int x,int y) {if(sum[y]-sum[x]>maxi&&a<0) return -inf;return dp[x]+a*(sum[y]-sum[x])*(sum[y]-sum[x])+c;} 48 void solve(int k,int l,int r) 49 { 50 top=tl=0;int mid; 51 ren(k) st[++top]=(Point){sum[to[i]],dp[to[i]]+a*sum[to[i]]*sum[to[i]]-b*sum[to[i]],dp[to[i]],to[i]}; 52 if(top) 53 { 54 sort(st+1,st+top+1);q[++tl]=1; 55 rep(i,2,top) 56 { 57 if(st[i].fi<=-inf+100000000000000LL) continue; 58 while(tl>1&&(st[q[tl]]-st[q[tl-1]])*(st[i]-st[q[tl]])>=0LL) tl--; 59 q[++tl]=i; 60 } 61 rep(i,1,tl) vec[k].push_back(st[q[i]].id); 62 } 63 if(l==r) 64 { 65 int ml,mr,mid,res; 66 for(int t=k;t;t>>=1) 67 { 68 ml=0,res=mr=vec[t].size()-1;if(mr<0) continue;mr--; 69 while(ml<=mr) {mid=ml+mr>>1;if(calc(vec[t][mid+1],l)<=calc(vec[t][mid],l)) mr=mid-1,res=mid;else ml=mid+1;} 70 dp[l]=max(calc(vec[t][res],l),dp[l]); 71 } 72 return ; 73 } 74 mid=l+r>>1;solve(k<<1,l,mid);solve(k<<1|1,mid+1,r); 75 } 76 int main() 77 { 78 freopen("paint.in","r",stdin); 79 freopen("paint.out","w",stdout); 80 n=read(),a=read(),b=read(),c=read(),l=read(),r=read(); 81 rep(i,1,n) sum[i]=sum[i-1]+read(),dp[i]=-1LL<<60; 82 rep(i,0,n-l) mdf(1,0,n,i+l,min(i+r,n),i); 83 solve(1,0,n);printf("%lld\n",dp[n]+b*sum[n]); 84 }
T3 route
题目大意:
一棵树 一个路径的权值为这个路径上最小边的权值*权值和 求一条路径上最大值与最小值之差<m的路径的权值最大值
思路:
首先可以知道一个结论:
还是线段树分治 按边权排序后每条边对一个区间内边有影响
我们使用带撤销并查集维护这个图的直径 每个点的答案即为当前图的直径$\times$ 这个点的权值(由于从小到大加入边)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 #include<cstring> 7 #include<vector> 8 #include<queue> 9 #include<map> 10 #define rep(i,s,t) for(register int i=(s);i<=(t);++i) 11 #define dwn(i,s,t) for(register int i=(s);i>=(t);--i) 12 #define ren for(register int i=fst[x];i;i=nxt[i]) 13 #define ren0 for(register int i=fst0[k];i;i=nxt0[i]) 14 #define Fill(x,t) memset(x,t,sizeof(x)) 15 #define ll long long 16 #define inf 2139062143 17 #define MAXN 170100 18 using namespace std; 19 inline int read() 20 { 21 int x=0,f=1;char ch=getchar(); 22 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 23 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 24 return x*f; 25 } 26 int n,m,fa[MAXN],in[MAXN<<1],rnk[MAXN],l2[MAXN<<1]; 27 int fst[MAXN],to[MAXN<<1],nxt[MAXN<<1],val[MAXN<<1],cnt,tot; 28 int nxt0[MAXN*20],fst0[MAXN<<2],to0[MAXN*20],top; 29 struct Edge{int u,v,w;}e[MAXN]; 30 struct Path{int x,y;}g[MAXN]; 31 struct Cancel{int x,y,rnk;ll len;Path a;}st[MAXN]; 32 bool operator < (const Edge &a,const Edge &b) {return a.w<b.w;} 33 ll ans,dis[MAXN],f[20][MAXN<<1],dmt; 34 void add(int u,int v,int w) {nxt[++cnt]=fst[u],fst[u]=cnt,to[cnt]=v,val[cnt]=w;} 35 void Add(int u,int v) {nxt0[++cnt]=fst0[u],fst0[u]=cnt,to0[cnt]=v;} 36 int find(int x) {return x==fa[x]?x:find(fa[x]);} 37 void dfs(int x,int fa) 38 { 39 in[x]=tot; 40 ren if(to[i]!=fa) {f[0][++tot]=dis[to[i]]=dis[x]+val[i];dfs(to[i],x);f[0][++tot]=dis[x];} 41 } 42 ll calc(int x,int y) 43 { 44 if(in[x]>in[y]) swap(x,y);int t=l2[in[y]-in[x]+1]; 45 return dis[x]+dis[y]-(min(f[t][in[x]],f[t][in[y]-(1<<t)+1])<<1); 46 } 47 Path merge(int x,int y,int xx,int yy) 48 { 49 int ax=x,ay=y; 50 if(calc(x,xx)>calc(ax,ay)) ax=x,ay=xx; 51 if(calc(x,yy)>calc(ax,ay)) ax=x,ay=yy; 52 if(calc(y,xx)>calc(ax,ay)) ax=y,ay=xx; 53 if(calc(y,yy)>calc(ax,ay)) ax=y,ay=yy; 54 if(calc(xx,yy)>calc(ax,ay)) ax=xx,ay=yy; 55 return (Path){ax,ay}; 56 } 57 void Merge(int x,int y) 58 { 59 int u=find(x),v=find(y); 60 if(rnk[u]<rnk[v]) swap(u,v); 61 fa[v]=u,st[++top]=(Cancel){u,v,rnk[u],dmt,g[u]},g[u]=merge(g[u].x,g[u].y,g[v].x,g[v].y); 62 if(rnk[u]==rnk[v]) rnk[u]++; 63 dmt=max(dmt,calc(g[u].x,g[u].y)); 64 } 65 void dlt() 66 { 67 int x=st[top].x,y=st[top].y; 68 fa[y]=y,rnk[x]=st[top].rnk,g[x]=st[top].a,dmt=st[top--].len; 69 } 70 void mdf(int k,int l,int r,int a,int b,int w) 71 { 72 if(l==a&&r==b) {Add(k,w);return ;} 73 int mid=l+r>>1; 74 if(b<=mid) mdf(k<<1,l,mid,a,b,w); 75 else if(a>mid) mdf(k<<1|1,mid+1,r,a,b,w); 76 else {mdf(k<<1,l,mid,a,mid,w);mdf(k<<1|1,mid+1,r,mid+1,b,w);} 77 } 78 void solve(int k,int l,int r) 79 { 80 int pos=top,mid=l+r>>1; 81 ren0 Merge(e[to0[i]].u,e[to0[i]].v); 82 if(l==r) ans=max(ans,dmt*e[l].w); 83 else {solve(k<<1,l,mid);solve(k<<1|1,mid+1,r);} 84 while(top!=pos) dlt(); 85 } 86 int main() 87 { 88 freopen("route.in","r",stdin); 89 freopen("route.out","w",stdout); 90 n=read(),m=read();int a,b,c; 91 rep(i,1,n-1) {a=read(),b=read(),c=read();add(a,b,c);add(b,a,c);e[i]=(Edge){a,b,c};} 92 dfs(1,0);sort(e+1,e+n);rep(i,1,n) fa[i]=i,g[i]=(Path){i,i}; 93 rep(i,2,tot) l2[i]=l2[i>>1]+1;cnt=0; 94 rep(j,1,19) rep(i,1,tot) {if(i+(1<<j)-1>tot) break;f[j][i]=min(f[j-1][i],f[j-1][i+(1<<j-1)]);} 95 a=n-1; 96 dwn(i,n-1,1) {while(e[i].w-e[a].w<=m&&a) a--;mdf(1,1,n-1,a+1,i,i);} 97 solve(1,1,n-1); 98 printf("%lld\n",ans); 99 }