标签：long hid for pac names etc 答案 main 技术
练习赛，评测的时候好像出了些问题，最后我拿自己机子测的212/300，第二题负责评测的写的SPJ就判了第一行的答案，不知道有没出什么问题。
先贴代码
T1.palindrome 92/100
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long using namespace std; #define MN 300000 #define MV 1200000 #define MOD 9875321 char s[MN+5],st[MN*2+5]; int l[MN*2+5],f[MN*2+5],cnt,rr[MV+5],ss[MV+5]; ll hs[MV+5]; struct P{int p,l;}pi[MV+5]; bool cmp(P a,P b){return a.l<b.l;} struct map { int h[MOD],nx[MV*3/2+5],z[MV*3/2+5],en,lz; ll s[MV*3/2+5],ls; int&operator[](ll x) { if(x==ls)return z[lz]; int p=(x%MOD+MOD)%MOD,i; for(i=h[p];i;i=nx[i])if(s[i]==x)return z[lz=i]; nx[!en||z[en]?++en:en]=h[p];s[en]=x;h[p]=en;return z[lz=en]; } }mp; int main() { freopen("palindrome.in","r",stdin); freopen("palindrome.out","w",stdout); int i,r=0,p;ll mx=0; scanf("%s",s); for(st[i=0]=‘(‘;s[i];++i)st[(i<<1)+1]=‘.‘,st[i+1<<1]=s[i];st[(i<<1)+1]=‘.‘;st[i+1<<1]=0; for(i=1;st[i];++i) { if(r>i)l[i]=min(l[(p<<1)-i],r-i),f[i]=(p<<1)-i; else l[i]=1; pi[++cnt]=(P){i,l[i]}; while(st[i-l[i]]==st[i+l[i]])pi[++cnt]=(P){i,++l[i]}; if(i+l[i]>r)r=i+l[i],p=i; } sort(pi+1,pi+cnt+1,cmp); for(i=1;i<=cnt;++i) { if(pi[i].l>1) { for(r=pi[i].p;!mp[(ll)r*(MV+3)+pi[i].l-1];r=f[r]); hs[i]=hs[rr[i]=mp[(ll)r*(MV+3)+pi[i].l-1]]; } hs[i]=hs[i]*31+st[pi[i].p+pi[i].l-1]; mp[(ll)pi[i].p*(MV+3)+pi[i].l]=i; } memset(&mp,0,sizeof(mp)); for(i=cnt;i;--i) { if(pi[i].l==l[pi[i].p])++ss[i]; mx=max(mx,(ll)(pi[i].l-1)*(mp[hs[i]]+=ss[i])); ss[rr[i]]+=ss[i]; } cout<<mx; fclose(stdin);fclose(stdout);return 0; }
T2.sequence 100/100
#include<cstdio> #include<iostream> using namespace std; #define ll long long inline int read() { int x=0;char c; while((c=getchar())<‘0‘||c>‘9‘); for(;c>=‘0‘&&c<=‘9‘;c=getchar())x=x*10+c-‘0‘; return x; } #define MK 200 #define MN 100000 int s[MN+5],d[MK+1][MN+5],q[MN+5],l,r; ll f[2][MN+5]; double cal(int p,int i,int j){return double(f[p][i]-f[p][j])/(s[j]-s[i]-1e-9);} int main() { freopen("sequence.in","r",stdin); freopen("sequence.out","w",stdout); int n,k,i,j,p,pl; n=read();k=read(); for(i=1;i<=n;++i)s[i]=s[i-1]+read(); for(i=p=1,pl=0;i<=k;++i,p^=1,pl^=1) { l=r=0; for(j=1;j<=n;++j) { while(l<r&&cal(pl,q[l+1],q[l])<s[j])++l; d[i][j]=q[l]; f[p][j]=f[pl][q[l]]+(ll)s[j]*s[q[l]]; f[pl][j]-=(ll)s[j]*s[j]; while(l<r&&cal(pl,j,q[r])<cal(pl,q[r],q[r-1]))--r; q[++r]=j; } } cout<<f[pl][n]<<endl; for(i=n,j=k;j;--j)printf("%d ",i=d[j][i]); fclose(stdin);fclose(stdout);return 0; }
T3.beads 20/100
#include<cstdio> #include<algorithm> using namespace std; inline int read() { int x=0;char c; while((c=getchar())<‘0‘||c>‘9‘); for(;c>=‘0‘&&c<=‘9‘;c=getchar())x=x*10+c-‘0‘; return x; } #define MN 200000 #define INF 0x3fffffff struct edge{int nx,t,w;}e[MN*2+5]; int h[MN+5],en,f[MN+5][2]; inline void ins(int x,int y,int w) { e[++en]=(edge){h[x],y,w};h[x]=en; e[++en]=(edge){h[y],x,w};h[y]=en; } void dp(int x,int fa) { int i,p,mx=-INF,mx2=-INF; for(i=h[x];i;i=e[i].nx)if(e[i].t!=fa) { dp(e[i].t,x); f[x][0]+=p=max(f[e[i].t][1]+e[i].w,f[e[i].t][0]); p=f[e[i].t][0]+e[i].w-p; if(p>mx)mx2=mx,mx=p; else if(p>mx2)mx2=p; } f[x][1]=f[x][0]+mx; f[x][0]=max(f[x][0],f[x][0]+mx+mx2); } int main() { freopen("beads.in","r",stdin); freopen("beads.out","w",stdout); int n=read(),i,x,y; for(i=1;i<n;++i)x=read(),y=read(),ins(x,y,read()); dp(1,0); printf("%d",f[1][0]); fclose(stdin);fclose(stdout);return 0; }
标签：long hid for pac names etc 答案 main 技术
原文地址：http://www.cnblogs.com/ditoly/p/6413565.html
