10.19-10.20 test
2016 10.19-10.20 两天 题目by mzx
Day1:
T1:loverfinding
题解：hash
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<string> 6 #include<algorithm> 7 #include<ctime> 8 #include<cmath> 9 #include<queue> 10 using namespace std; 11 #define FILE "loverfinding" 12 #define pii pair<long long,long long> 13 #define LL long long 14 #define up(i,j,n) for(int i=j;i<=n;i++) 15 #define down(i,n,j) for(int i=n;i>=j;i--) 16 #define max(x,y) ((x)>(y)?(x):(y)) 17 #define min(x,y) ((x)<(y)?(x):(y)) 18 #define abs(x) ((x)<0?(-(x)):(x)) 19 template<typename T>inline bool chkmax(T &a,T b){return a<b?a=b,true : false;} 20 template<typename T>inline bool chkmin(T &a,T b){return a>b?a=b,true : false;} 21 int read(){ 22 int x=0;char ch=getchar();bool flag=0; 23 while(ch<'0'||ch>'9'){if(ch=='-')flag=1;ch=getchar();} 24 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 25 return flag?-x:x; 26 } 27 namespace OI{ 28 void print(int x){printf("%d\n",x);exit(0);} 29 int n,x0,y0,xt,yt,cnt=1,nx,ny,dx,dy; 30 int hx[10000000],hy[10000000],step=7,mod=9999777; 31 bool vis[10000000]; 32 bool hash(int x,int y){ 33 int pos=abs((LL)(x+15214)*97%mod*(y+147)%mod*143%mod*71373)%mod; 34 while(vis[pos]&&!(hx[pos]==x&&hy[pos]==y)){ 35 pos+=step; 36 if(pos>mod)pos-=mod; 37 } 38 if(!vis[pos]){ 39 vis[pos]=1;hx[pos]=x;hy[pos]=y;return 1; 40 } 41 return 0; 42 } 43 void slove(){ 44 n=read(),nx=x0=read(),ny=y0=read(),xt=read(),yt=read(); 45 up(i,1,n){ 46 if(nx==xt&&ny==yt)print(cnt); 47 if(hash(nx,ny))cnt++; 48 dx=read(),dy=read(); 49 nx+=dx,ny+=dy; 50 } 51 printf("SingleDogMZX\n"); 52 } 53 } 54 int main(){ 55 freopen(FILE".in","r",stdin); 56 freopen(FILE".out","w",stdout); 57 using namespace OI; 58 slove(); 59 return 0; 60 }
T2:snackstore
题解：floyed
#include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define FILE "snackstore" #define up(i,j,n) for(int i=j;i<=n;i++) template<typename T>inline bool chkmax(T &a,T b){return a<b?a=b,true : false;} template<typename T>inline bool chkmin(T &a,T b){return a>b?a=b,true : false;} namespace IO{ char buf[1<<15],*fs,*ft; inline char gc(){return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;} inline int read(){ int x=0,ch=gc(); while(ch<'0'||ch>'9')ch=gc(); while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=gc(); return x; } }using namespace IO; namespace OI{ const int maxn=110; const int inf=1000000001LL; int n,m,Q; struct node{ int id,v; bool operator<(const node& b)const{return v<b.v;} }v[maxn]; int A[maxn],pow[maxn]; int f[maxn][maxn][maxn]; void init(){ n=read(),m=read(),Q=read(); up(i,1,n)v[i].v=read(),v[i].id=i; int x,y,vv; for(int c=0;c<=n;c++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) f[c][i][j]=inf; for(int i=1;i<=n;i++)f[0][i][i]=0; up(i,1,m){ x=read(),y=read(),vv=read(); if(vv<f[0][y][x])f[0][x][y]=f[0][y][x]=vv; } sort(v+1,v+n+1); for(int i=1;i<=n;i++)A[i]=v[i].id; for(int i=1;i<=n;i++)pow[i]=v[i].v; for(int c=1;c<=n;c++){ memcpy(f[c],f[c-1],sizeof(f[c-1])); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(f[c][i][A[c]]+f[c][A[c]][j]<f[c][i][j]) f[c][i][j]=f[c][i][A[c]]+f[c][A[c]][j]; } for(int c=0;c<=n;c++) for(int i=1;i<=n;i++) sort(f[c][i]+1,f[c][i]+n+1); } int left,right,mid,ans; inline int find(int c){ left=0,right=n; while(left<=right){ mid=(left+right)>>1; if(c>=pow[mid])ans=mid,left=mid+1; else right=mid-1; } return ans; } inline int Find(int *a,int d){ left=1,right=n; while(left<=right){ mid=(left+right)>>1; if(a[mid]>d)right=mid-1; else left=mid+1,ans=mid; } return ans; } void slove(){ init(); int s,c,d,pos; up(i,1,Q){ s=read(),c=read(),d=read(); printf("%d\n",Find(f[find(c)][s],d)-1); } return; } } int main(){ freopen(FILE".in","r",stdin); freopen(FILE".out","w",stdout); using namespace OI; slove(); return 0; }
T3：有点恶心，暂时搁置
Day2:
T1:three_squirrels
题解：拓扑排序
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<string> #include<algorithm> #include<ctime> #include<cmath> #include<queue> using namespace std; #define FILE "three_squirrels" #define pii pair<long long,long long> #define LL long long #define up(i,j,n) for(int i=j;i<=n;i++) #define down(i,n,j) for(int i=n;i>=j;i--) #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define abs(x) ((x)<0?(-(x)):(x)) template<typename T>inline bool chkmax(T &a,T b){return a<b?a=b,true : false;} template<typename T>inline bool chkmin(T &a,T b){return a>b?a=b,true : false;} int read(){ int x=0;char ch=getchar();bool flag=0; while(ch<'0'||ch>'9'){if(ch=='-')flag=1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return flag?-x:x; } namespace OI{ const int maxn=101000,mod=1000000007; int f[maxn]; int n; struct node{ int y,next; }e[2010000]; int len=0,linkk[maxn]; void insert(int x,int y){ e[++len].y=y; e[len].next=linkk[x]; linkk[x]=len; } int dfs(int x){ if(f[x]!=-1)return f[x]; f[x]=0; for(int i=linkk[x];i;i=e[i].next)f[x]=(f[x]+dfs(e[i].y))%mod; return f[x]; } void slove(){ memset(f,-1,sizeof(f)); f[0]=1; n=read();int k,x; up(i,1,n){ k=read(); up(j,1,k){ x=read(); insert(i,x); } } printf("%d\n",dfs(n)); } } int main(){ freopen(FILE".in","r",stdin); freopen(FILE".out","w",stdout); using namespace OI; slove(); return 0; }
T2:savemzx
题解：树形动规
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<string> #include<algorithm> #include<ctime> #include<cmath> #include<queue> using namespace std; #define FILE "savemzx" #define pii pair<long long,long long> #define LL long long #define up(i,j,n) for(int i=j;i<=n;i++) #define down(i,n,j) for(int i=n;i>=j;i--) #define max(x,y) ((x)>(y)?(x):(y)) #define min(x,y) ((x)<(y)?(x):(y)) #define abs(x) ((x)<0?(-(x)):(x)) template<typename T>inline bool chkmax(T &a,T b){return a<b?a=b,true : false;} template<typename T>inline bool chkmin(T &a,T b){return a>b?a=b,true : false;} int read(){ int x=0;char ch=getchar();bool flag=0; while(ch<'0'||ch>'9'){if(ch=='-')flag=1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return flag?-x:x; } namespace OI{ const int maxn=1010000; const LL inf=10000000000000000LL; char s[maxn]; int next[maxn],n; bool vis[maxn]; inline LL squ(int x){return (LL)x*x;} struct node{ int y,next; LL v; }e[maxn<<1]; int linkk[maxn],len=0; void insert(int x,int y,LL v){ e[++len].y=y; e[len].next=linkk[x]; linkk[x]=len; e[len].v=v; } LL w[maxn],maxx[maxn],maxx2[maxn],child[maxn]; int fa[maxn]; LL ans=0; void dfs(int x){ for(int i=linkk[x];i;i=e[i].next){ if(e[i].y==fa[x])continue; fa[e[i].y]=x; dfs(e[i].y); chkmax(ans,e[i].v); w[e[i].y]+=e[i].v; if(w[e[i].y]>maxx[x])maxx2[x]=maxx[x],maxx[x]=w[e[i].y]; else if(w[e[i].y]>maxx2[x])maxx2[x]=w[e[i].y]; } chkmax(ans,maxx[x]+maxx2[x]); w[x]=maxx[x]; } void slove(){ scanf("%s",s+1);n=strlen(s+1); int j=0;next[1]=0; insert(1,0,1);insert(0,1,1); for(int i=2;i<=n;i++){ while(s[i]!=s[j+1]&&j)j=next[j]; if(s[i]==s[j+1])j++; next[i]=j; insert(j,i,squ(j-i)); insert(i,j,squ(i-j)); } dfs(0); printf("%lld\n",ans); return; } } int main(){ freopen(FILE".in","r",stdin); freopen(FILE".out","w",stdout); using namespace OI; slove(); return 0; }
T3:chocolatebox
题解：请自行推导公式，数学题
总的来说，这套题，两个T3（实现了让人不想看的目标），T2都是图论题，也需要一些分析和转化，总体不错；
但这套题很恶心的地方是，卡常卡的太厉害（你能想象day1T1被卡的全T吗？），day1T2写完了正解后，交上去发现一直T，直到我抄了std的输入之后......
桑心过度；
祝初赛考出好成绩,啦啦啦......
