A:不会
B:暴力搜索,枚举每一次操作的情况,稍微剪枝,避免重复记录即可
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn = 7;
int pz[maxn][maxn];
int lst[maxn];
ll res;
int ex,ey;
unordered_set<ll> win;
int check() {
int f1 = 0, f2 = 0;
for(int i = 1; i <= 4; i++) {
for(int j = 1; j <= 2; j++) {
if(pz[i][j] == 1 && pz[i][j+1] == 1 && pz[i][j+2] == 1) f1++;
if(pz[i][j] == 2 && pz[i][j+1] == 2 && pz[i][j+2] == 2) f2++;
}
}
for(int j = 1; j <= 4; j++)
for(int i = 1; i <= 2; i++) {
if(pz[i][j] == 1 && pz[i+1][j] == 1 && pz[i+2][j] == 1) f1++;
if(pz[i][j] == 2 && pz[i+1][j] == 2 && pz[i+2][j] == 2) f2++;
}
for(int i = 1; i <= 2; i++) {
for(int j = 1; j <= 2; j++) {
if(pz[i][j] == 1 && pz[i+1][j+1] == 1 && pz[i+2][j+2] == 1) f1++;
if(pz[i][j] == 2 && pz[i+1][j+1] == 2 && pz[i+2][j+2] == 2) f2++;
}
}
for(int i = 1; i <= 2; i++) {
for(int j = 3; j <= 4; j++) {
if(pz[i][j] == 1 && pz[i+1][j-1] == 1 && pz[i+2][j-2] == 1) f1++;
if(pz[i][j] == 2 && pz[i+1][j-1] == 2 && pz[i+2][j-2] == 2) f2++;
}
}
if(f1 != 0) return 1;
return f2 == 0 ? 0 : 2;
}
int all;
void dfs(int p,int x,int y){
if(pz[ex][ey]==1) return ;
int tmp=check();
if(tmp==1) return ;
if(pz[ex][ey]==2&&p==0&&ex==x&&ey==y){
if(tmp==2){
ll st=3;
rep(i,1,4)rep(j,1,4){
st*=3;
st+=pz[i][j];
}
win.insert(st);
return ;
}
}
if(tmp==2) return ;
if(all==0) return ;
rep(i,1,4){
if(lst[i]==0) continue;
lst[i]--;
all--;
int pos=1;
while(pz[pos][i]!=0) pos++;
pz[pos][i]=1+p;
dfs(1-p,pos,i);
pz[pos][i]=0;
lst[i]++;
all++;
}
}
int ans[6][6][6];
int main() {
memset(pz,0,sizeof pz);
int a,b,c;cin>>a>>b>>c;
rep(_i,1,4)lst[_i]=4;
pz[1][a]=1;lst[a]=3;
ex=b,ey=c;
all=15;
dfs(1,1,a);
cout<<win.size();
}
C:重定向边,记忆化搜索求DAG上最长路即可
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=3e5+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
int d[maxn],dis[maxn],vis[maxn];
pii e[maxn];
int ans;
vector<int> g[maxn];
int dfs(int now){
if(dis[now]) return dis[now];
int res=1;
for(int to:g[now]){
res=max(res,dfs(to)+1);
}
return dis[now]=res;
}
int main(){
cin>>n>>m;
rep(i,1,m){
int a,b;cin>>a>>b;
e[i]=make_pair(a,b);
d[a]++,d[b]++;
}
rep(i,1,m){
int a=e[i].fi,b=e[i].se;
if(d[a]==d[b]) continue;
if(d[a]>d[b]) swap(a,b);
g[b].push_back(a);
// vis[a]=1;
}
rep(i,1,n) if(!vis[i])dfs(i);
rep(i,1,n) k=max(dis[i],k);
// showa(dis,1,n);
cout<<k<<endl;
return 0;
}
D:暴力
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e3+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
bool vis[maxn];
int f(ll x) {
int res = 0;
while(x) {
int a = x%10;
res += a*a;
x /= 10;
}
return res;
}
int main(){
ll n;
cin >> n;
if(n == 1) {
cout << "HAPPY\n";
return 0;
}
memset(vis, false, sizeof vis);
int now = f(n);
vis[now] = true;
while(now != 1) {
now = f(now);
if(vis[now]) {
cout << "UNHAPPY\n";
return 0;
}
vis[now] = 1;
}
cout << "HAPPY\n";
return 0;
}
E:做法有点像枚举边求最小环
枚举每一条边,把所有边权比它小的边放入图中,然后以这两个边为源点和汇点跑最小割即可
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
template<typename T>class mxf{public:
struct node{int to,next;T cap;}e[maxm<<1];
int cur[maxn],head[maxn],dis[maxn],gap[maxn];
int nume=1,s,t,tot;
void init(int n){
rep(i,0,n) head[i]=gap[i]=dis[i]=0;
nume=1;
}
void add(int a,int b,T c){
e[++nume]={b,head[a],c};head[a]=nume;
e[++nume]={a,head[b],0};head[b]=nume;
}
T dfs(int now,T flow=INF){
if (now==t||!flow) return flow;
T use=0,tmp;
int d=dis[now]-1,to;
for (int &i=cur[now];i;i=e[i].next) {
if(dis[to=e[i].to]==d&&(tmp=e[i].cap)){
e[i].cap-=(tmp=dfs(to,min(flow-use,tmp)));
e[i^1].cap+=tmp;
if((use+=tmp)==flow) return use;
}
}
if (!--gap[dis[now]]) dis[s]=tot+1;
++gap[++dis[now]];
cur[now]=head[now];
return use;
}
T getflow(int ss,int tt,int n,T ans=0){
tot=n;s=ss;t=tt;gap[0]=tot;
memcpy(cur,head,(tot+1)<<2);
while(dis[s]<=tot) ans+=dfs(s);
return ans;
}
};
mxf<int> net;
struct node{int a,b,c;}e[maxn];
bool cmp(node x,node y){
return x.c<y.c;
}
int main() {
cin>>n>>m;
rep(i,1,m){
cin>>e[i].a>>e[i].b>>e[i].c;
}
sort(e+1,e+1+m,cmp);
ll ans=0;
rep(i,1,m){
net.init(n);
int s=e[i].a,t=e[i].b;
rep(j,1,m){
if(j==i) continue;
if(e[j].c>=e[i].c) break;
net.add(e[j].a,e[j].b,1);
net.add(e[j].b,e[j].a,1);
}
int tmp=net.getflow(s,t,n);
ans+=tmp;
}
cout<<ans<<endl;
return 0;
}
// auto _start=chrono::high_resolution_clock::now();
// auto _end=chrono::high_resolution_clock::now();
// cerr<<"elapsed time: "<<chrono::duration<double,milli>(_end-_start).count()<<" ms\n";
F:递归分解,找规律
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
pair<ll, ll> f(ll k, ll step) {
ll len = (k), Size = len*len/4;
if(k == 2) {
if(step == 0) return make_pair(1, 1);
else if(step == 1) return make_pair(1, 2);
else if(step == 2) return make_pair(2, 2);
else return make_pair(2, 1);
}
int num = step/Size;
step = step%Size;
if(num == 0) {
pair<ll, ll> tmp = f(k/2, step);
swap(tmp.fi, tmp.se);
return tmp;
}
else if(num == 1) {
pair<ll, ll> tmp = f(k/2, step);
tmp.se += len/2;
return tmp;
}
else if(num == 2) {
pair<ll, ll> tmp = f(k/2, step);
tmp.fi += len/2; tmp.se += len/2;
return tmp;
}
else {
pair<ll, ll> tmp = f(k/2, step);
swap(tmp.fi, tmp.se);
ll a = len/2-tmp.fi+1, b = len/2-tmp.se+1;
return make_pair(a+len/2, b);
}
}
int main() {
std::ios::sync_with_stdio(false);
ll k; ll step;
cin >> k >> step;
step--;
pair<ll, ll> t = f(k, step);
cout << t.fi << " " << t.se << "\n";
return 0;
}
G:不会
H:一开始想了假算法,卡了很久,后来才写的fft
#include<bits/stdc++.h>
#define ll long long
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define IO std::ios::sync_with_stdio(false)
using namespace std;
const int maxn=8e6+20,maxm=8e3+10;
const double pi=acos(-1.0);
struct cp{double x,y;}a[maxn],b[maxn];
cp operator*(cp a,cp b){return {a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
cp operator+(cp a,cp b){return {a.x+b.x,a.y+b.y};}
cp operator-(cp a,cp b){return {a.x-b.x,a.y-b.y};}
class fourier{public:
int rev[maxn],len,pw;
void init(int n){
len=1,pw=0;
while(len<=n) len<<=1,pw++;
rep(i,0,len-1) rev[i]=0;
rep(i,0,len-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<(pw-1));
}
void transform(cp*a,int flag){
rep(i,0,len-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
for(int mid=1;mid<len;mid<<=1){
cp wn={cos(pi/mid),flag*sin(pi/mid)};
for(int r=mid<<1,j=0;j<len;j+=r){
cp t={1,0};
for(int k=0;k<mid;k++,t=t*wn){
cp x=a[j+k],y=t*a[mid+j+k];
a[j+k]=x+y,a[j+k+mid]=x-y;
}
}
}
if(flag==-1) rep(i,0,len-1) a[i].x/=len;
}
}fft;
char st[] = {'R', 'S', 'P'};
int n, m; string s, t;
int ans[maxn];
void solve(char ch, char ch1) {
for(int i = 0; i <= fft.len; i++)
a[i].x = b[i].x = a[i].y = b[i].y = 0;
for(int i = 1; i <= m; i++)
if(t[m-i+1] == ch) b[i].x = 1;
for(int i = 1; i <= n; i++)
if(s[i] == ch1) a[i].x = 1;
fft.transform(a, 1); fft.transform(b, 1);
for(int i = 0; i < fft.len; i++) {
a[i] = a[i]*b[i];
}
fft.transform(a, -1);
for(int l = 0; l < n; l++) {
ans[l] += (int)(a[l+m+1].x+0.5);
}
}
int main() {
std::ios::sync_with_stdio(false);
cin >> n >> m >> s >> t;
s = " " + s; t = " " + t;
fft.init(n+m);
solve('R', 'S'); solve('S', 'P'); solve('P', 'R');
int res = 0;
for(int i = 0; i < n; i++)
res = max(res, ans[i]);
cout << res << "\n";
return 0;
}
/**
12 4
RSPPSSSRRPPR
RRRR
12 3
RRRRRRRRRRRR
SSS
12 4
PPPRRRRRRRRR
RSSS
12 4
RRRRRRRRRSSS
RRRS
**/
I:KMP,枚举K即可
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
class prefix{public:
int p[maxn],lens;
int *s;
void init(int *_s,int _lens){
s=_s,lens=_lens;
rep(i,0,lens-1) p[i]=0;
p[0]=-1;
int now=0,pos=-1;
while(now<lens)
if(pos==-1||s[now]==s[pos]) p[++now]=++pos;
else pos=p[pos];
}
vector<int> find(int *t,int lent){
int now,pos=0;
vector<int> ans;
while(now<lent) {
if(pos==-1||t[now]==s[pos]) pos++,now++;
else pos=p[pos];
if(pos==lens) pos=p[pos],ans.push_back(now-lens);
}
return ans;
}
}kmp;
int a[maxn];
int main() {IO;
cin>>n;
per(i,0,n-1){
cin>>a[i];
}
kmp.init(a,n);
int ans1=INF,ans2=INF;
rep(i,1,n){
int t=i-kmp.p[i];
if(ans1+ans2==t+n-i){
if(t<ans2) {
ans1=n-i;ans2=t;
}
}else if(ans1+ans2>t+n-i){
ans1=n-i;ans2=t;
}
}
cout<<ans1<<' '<<ans2<<endl;
}
J:不会
K:构造
#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn = 1e4+10;
int a[maxn];
int main() {
std::ios::sync_with_stdio(false);
int n;
cin >> n;
for(int i = 1; i <= n; i++) {
int x;
cin >> x >> a[i];
}
int nowx = 0, nowy = 0, l = 0, r = 0, u = 0, d = 0;
int vx = 1, vy = 0;
cout << "1 "; r = 1; nowx = 1;
for(int i = 1; i < n; i++) {
if(a[i] == 1) {
if(vx == 1 && vy == 0) {
vx = 0, vy = 1;
cout << u+1-nowy << " ";
nowy = u+1; u++;
}
else if(vx == 0 && vy == 1) {
vx = -1, vy = 0;
cout << nowx-(l-1) << " ";
nowx = l-1; l = l-1;
}
else if(vx == -1 && vy == 0) {
vx = 0, vy = -1;
cout << nowy-(d-1) << " ";
nowy = d-1; d--;
}
else {
vx = 1, vy = 0;
cout << r+1-nowx << " ";
nowx = r+1; r++;
}
}
else if(a[i] == -1){
if(vx == 1 && vy == 0) {
vx = 0, vy = -1;
cout << nowy-(d-1) << " ";
nowy = d-1; d--;
}
else if(vx == 0 && vy == 1) {
vx = 1, vy = 0;
cout << r+1-nowx << " ";
nowx = r+1; r++;
}
else if(vx == -1 && vy == 0) {
vx = 0, vy = 1;
cout << u+1-nowy << " ";
nowy = u+1; u++;
}
else {
vx = -1, vy = 0;
cout << nowx-(l-1) << " ";
nowx = l-1; l--;
}
}
}
}
L:赛后补的,枚举天数进行dp,上限大概是n^3天
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 55;
ll dp[3][maxn][maxn*maxn*maxn], a[3][maxn];
vector<pair<int, int> > E[3][maxn];
int n[3], m[3], goal[3];
int main() {
std::ios::sync_with_stdio(false);
int p, all = 1;
cin >> p;
for(int k = 0; k < p; k++) {
cin >> n[k] >> m[k];
all *= n[k];
for(int i = 1; i <= n[k]; i++)
cin >> a[k][i];
for(int i = 1; i <= m[k]; i++) {
int x, y, c;
cin >> x >> y >> c;
//E[k][x].push_back(make_pair(y, c));
E[k][y].push_back(make_pair(x, c)); //对每个点要知道谁能到他 反向存图
}
cin >> goal[k];
}
for(int i = 0; i < p; i++) {
for(int j = 0; j <= n[i]; j++) {
for(int k = 0; k <= all; k++) {
dp[i][j][k] = 1e18;
}
}
}
#define fi first
#define se second
for(int k = 0; k < p; k++) {
// cout << "@@\n";
dp[k][1][0] = 0;
for(int j = 1; j <= all; j++) {
for(int i = 1; i <= n[k]; i++) {
dp[k][i][j] = min(dp[k][i][j], dp[k][i][j-1]+a[k][i]);
for(auto it : E[k][i]) {
int v = it.fi, c = it.se;
dp[k][i][j] = min(dp[k][i][j], dp[k][v][j-1]+c);
}
}
}
}
ll ans = 1e18;
for(int i = 1; i <= all; i++) {
ll tmp = 0;
// cout << "---i = " << i << endl;
for(int k = 0; k < p; k++) {
// cout << "k = " << k << " dp = " << dp[k][goal[k]][i] << endl;
tmp += dp[k][goal[k]][i];
}
ans = min(ans, tmp);
}
cout << ans << "\n";
}