[CLYZ2017]day8
异色弧
solution
70分
\(Let's\;orz\) BearChild!
朗格拉姆计数
solution
60分
将环复制两遍,预处理出前\(i\)个数中比\(j\)大的数的个数 ,枚举\(a,b\),答案加上\(c\)的个数.
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 5001
#define M 10001
using namespace std;
typedef long long ll;
//s[i][j]表示前i个数中比j大的数的个数
ll ans;
int s[M][N],a[M],n;
inline int read(){
int ret=0;char c=getchar();
while(!isdigit(c))
c=getchar();
while(isdigit(c)){
ret=(ret<<1)+(ret<<3)+c-'0';
c=getchar();
}
return ret;
}
inline void Aireen(){
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
for(int i=1;i<=n;++i)
a[i+n]=a[i];
for(int j=1;j<=n;++j)
for(int i=1;i<=(n<<1);++i){
s[i][j]=s[i-1][j];
if(a[i]>j) ++s[i][j];
}
for(int i=1;i<=n;++i)
for(int j=i;j<i+n;++j)
if(a[j]>a[i]){
ans+=(ll)(s[i+n-1][a[j]]-s[j][a[j]]);
}
printf("%lld\n",ans);
}
int main(){
freopen("counter.in","r",stdin);
freopen("counter.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}
100分
三元组的形式为\(123,231,312\).
\(231=XX1-321\).
\(312=3XX-321\).
树状数组或线段树维护即可.
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 200005
using namespace std;
typedef long long ll;
ll ans;
int a[N],s[N],g[N],fr[N],be[N],n;
inline int read(){
int ret=0;char c=getchar();
while(!isdigit(c))
c=getchar();
while(isdigit(c)){
ret=(ret<<1)+(ret<<3)+c-'0';
c=getchar();
}
return ret;
}
inline int lowbit(int x){
return x&(-x);
}
inline void add(int k){
for(int i=k;i<=n;i+=lowbit(i))
++s[i];
}
inline int ask(int k){
int ret=0;
for(int i=k;i;i-=lowbit(i))
ret+=s[i];
return ret;
}
inline ll c(int k){
return 1ll*k*(k-1)>>1ll;
}
inline void Aireen(){
n=read();
for(int i=1;i<=n;++i){
a[i]=read();g[a[i]]=i;
}
for(int i=1;i<=n;++i){
fr[i]=ask(g[i]-1);
be[i]=ask(n)-ask(g[i]);
add(g[i]);
}
memset(s,0,sizeof(s));
for(int i=n,f,b;i;--i){
f=ask(g[i]-1);
b=ask(n)-ask(g[i]);
add(g[i]);
ans+=1ll*fr[i]*b;//123
ans+=c(be[i])-1ll*f*be[i];//312
ans+=c(f)-1ll*f*be[i];//231
}
printf("%lld\n",ans);
}
int main(){
freopen("counter.in","r",stdin);
freopen("counter.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}
修路
solution
斯坦纳树
\(f[i][j]\)表示以\(i\)为根的树中,连通的节点的状态为\(j\)的最小/大权值.
\(f[i][j]=min\{f[i][l]+f[i][l^j]\}\),\(l\)为\(j\)的子集.
\(f[i][j]=min\{f[k][j]+g[k][i]\}\),这可以用\(spfa\)或\(dijkstra\)求解.
100分
用斯坦纳树求出\(f[\;][\;]\)数组.
如果在状态\(j\)中,\(i\)与\(n-i+1\)在\(j\)中或都不在\(j\)中,\(g[j]=min\{f[i][j]\}\)(保证\(i\)与\(n-i+1\)连通).
\(g[i]=min\{g[j]+g[i^j]\}j\)为\(i\)的子集\()\).
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define K 8
#define N 10005
#define M 20005
#define INF 10000000
#define min(a,b) a<b?a:b
using namespace std;
int nxt[M],to[M],w[M];
int f[N][1<<K],ff[1<<K],g[N],q[INF],h,t,n,m,k,d,cnt;
bool inq[N];
inline int read(){
int ret=0;char c=getchar();
while(!isdigit(c))
c=getchar();
while(isdigit(c)){
ret=(ret<<1)+(ret<<3)+c-'0';
c=getchar();
}
return ret;
}
inline void addedge(int x,int y,int z){
nxt[++cnt]=g[x];g[x]=cnt;to[cnt]=y;w[cnt]=z;
}
inline void spfa(int k){
int u;
while(h<=t){
u=q[h++];inq[u]=false;
for(int i=g[u];i;i=nxt[i])
if(f[u][k]+w[i]<f[to[i]][k]){
f[to[i]][k]=f[u][k]+w[i];
if(!inq[to[i]]){
inq[to[i]]=true;q[++t]=to[i];
}
}
}
}
inline bool chk(int s){
for(int i=1;i<=d;++i)
if((bool)(s&(1<<i-1))^(bool)(s&(1<<k-i)))
return false;
return true;
}
inline void Aireen(){
n=read();m=read();d=read();
for(int i=1,x,y,w;i<=m;++i){
x=read();y=read();w=read();
addedge(x,y,w);addedge(y,x,w);
}
k=d<<1;
for(int i=1;i<=n;++i)
for(int j=0;j<(1<<k);++j)
f[i][j]=INF;
for(int i=1;i<=d;++i)
f[i][1<<i-1]=f[n-i+1][1<<k-i]=0;
for(int j=0;j<(1<<k);++j){
h=1;t=0;
for(int i=1;i<=n;++i){
for(int l=j;l;l=(l-1)&j)
f[i][j]=min(f[i][j],f[i][l]+f[i][j^l]);
if(f[i][j]<INF){
inq[i]=true;q[++t]=i;
}
}
spfa(j);
}
for(int j=0;j<(1<<k);++j){
ff[j]=INF;
if(chk(j)){
for(int i=1;i<=n;++i)
ff[j]=min(ff[j],f[i][j]);
}
}
for(int j=0;j<(1<<k);++j)
for(int l=j;l;l=(l-1)&j)
ff[j]=min(ff[j],ff[j^l]+ff[l]);
if(ff[(1<<k)-1]==INF) puts("-1");
else printf("%d\n",ff[(1<<k)-1]);
}
int main(){
freopen("road.in","r",stdin);
freopen("road.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}
