【AtCoder】CODE FESTIVAL 2017 Final
A - AKIBA
模拟即可
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
string s,tar = "AKIHABARA";
bool Solve() {
cin >> s;
if(s.length() > 9) {
return false;
}
for(int i = 0 ; i < 9 ; ++i) {
if(s.length() <= i) s += "A";
if(s[i] != tar[i]) {
if(tar[i] == 'A') {
s.insert(i,1,'A');
}
else return false;
}
}
if(s != tar) return false;
return true;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
if(Solve()) puts("YES");
else puts("NO");
return 0;
}
B - Palindrome-phobia
题解
abc出现次数的最大值和最小值相差不超过1
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
char s[MAXN];
int cnt[4];
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
scanf("%s",s + 1);
int N = strlen(s + 1);
for(int i = 1 ; i <= N ; ++i) {
cnt[s[i] - 'a']++;
}
int minn = min(min(cnt[0],cnt[1]),cnt[2]);
int mmax = max(max(cnt[0],cnt[1]),cnt[2]);
if(mmax - minn <= 1) puts("YES");
else puts("NO");
return 0;
}
C - Time Gap
题解
显然每个时间有超过三个人答案一定是0
之后对于每个时间里的人进行枚举是d还是24 - d即可
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int cnt[15];
int N,D[55];
bool vis[25],mark[15],c[25];
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) {read(D[i]);cnt[D[i]]++;}
}
void Solve() {
if(cnt[0] || cnt[12] >= 2) {puts("0");return;}
int S = 0;
int ans = 0,d = 12;
for(int i = 1 ; i <= 12 ; ++i) {
if(cnt[i] > 2) {puts("0");return;}
if(cnt[i] == 2) {
vis[i] = 1;vis[24 - i] = 1;
}
else if(cnt[i]) {d = min(d,i);S |= 1 << i - 1;mark[i] = 1;}
}
for(int T = S ; ; T = S & (T - 1)) {
memcpy(c,vis,sizeof(vis));
for(int i = 1 ; i <= 12 ; ++i) {
if(mark[i]) {
if(T >> (i - 1) & 1) c[24 - i] = 1;
else c[i] = 1;
}
}
int pre = 0,t = d;
for(int i = 1 ; i <= 24 ; ++i) {
if(c[i]) {t = min(t,i - pre);pre = i;}
}
ans = max(ans,t);
if(T == 0) break;
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}
D - Zabuton
题解
如果两个相邻的点\(a\)和\(b\)，前面的前缀和是\(x\)
那么我们有
\(min(H[a],H[b] - P[a]) >= x\)
\(min(H[b],H[a] - P[b]) >= x\)
我们希望允许尽量宽松的x
如果\(min(H[a],H[b] - P[a]) < min(H[b],H[a] - P[b])\)的话，a在前，否则b在前
排序后进行\(dp[i][j]\)表示到第\(i\)个点选了\(j\)个点最小前缀和即可，用前缀min优化更新
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int H[MAXN],P[MAXN],id[MAXN],N;
int64 sum[MAXN];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(H[i]);read(P[i]);id[i] = i;
}
sort(id + 1,id + N + 1,[](int a,int b) {
if(H[a] < H[b] - P[a]) return true;
if(H[b] < H[a] - P[b]) return false;
return H[a] + P[a] < H[b] + P[b];
});
for(int i = 1 ; i <= N ; ++i) sum[i] = 1e18;
for(int i = 1 ; i <= N ; ++i) {
int u = id[i];
for(int j = N ; j >= 1 ; --j) {
if(H[u] >= sum[j - 1]) {
sum[j] = min(sum[j],sum[j - 1] + P[u]);
}
}
}
for(int i = N ; i >= 1 ; --i) {
if(sum[i] != 1e18) {out(i);enter;return;}
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
E - Combination Lock
题解
处理成差分，对称位置的差分和为0
例如abcba
可以得到的差分是
相当于数列
0123210
111-1-1-1
而有区间加呢，相当于在前面打了一个+1，后面打了一个-1
我们把这两个位置连边
并且给所有对称位置连边
合法的情况仅当一个联通块和为0
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
char s[MAXN];
int a[MAXN],N,M;
struct node {
int to,next;
}E[MAXN * 10];
int head[MAXN],sumE,sum;
bool vis[MAXN];
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void Init() {
scanf("%s",s + 1);
N = strlen(s + 1);
s[0] = 'a';s[N + 1] = 'a';
for(int i = 1 ; i <= N + 1; ++i) a[i] = (s[i] - s[i - 1] + 26) % 26;
read(M);
int L,R;
for(int i = 1 ; i <= M ; ++i) {
read(L);read(R);
add(L,R + 1);add(R + 1,L);
}
for(int i = 1 ; i <= N + 1 ; ++i) {
add(i,N + 2 - i);
}
}
void dfs(int u) {
vis[u] = 1;
sum = (sum + a[u]) % 26;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) {
dfs(v);
}
}
}
void Solve() {
bool flag = 1;
for(int i = 1 ; i <= N + 1 ; ++i) {
if(!vis[i]) {
sum = 0;
dfs(i);
if(sum != 0) {flag = 0;break;}
}
}
if(flag) puts("YES");
else puts("NO");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}
F - Distribute Numbers
题解
给第一行填上1 - K，很容易发现N的下界是\(K(K - 1) + 1\)
然后给第2到\(K\)行全填上1，\(K + 1\)到\(2K\)全填上2，以此类推
然后我们再把剩余未分配的数全填到\(2\)行到\(K\)行大小为\(K - 1\)的正方形矩阵里
然后我们要把这个正方形矩阵划分成\(K - 1\)种，每种\(K - 1\)条不相交的链
把\(K - 1\)选一个质数可以完成这个操作
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
vector<int> v[1500];
int N = 1407;
int K = 38;
int a[45][45];
void Solve() {
for(int i = 1 ; i <= K ; ++i) v[1].pb(i);
int t = 0;
for(int i = 2 ; i <= N ; i += (K - 1)) {
++t;
for(int j = i ; j <= i + (K - 1) - 1 ; ++j) v[j].pb(t);
}
t = K;
for(int i = 0 ; i < K - 1 ; ++i) {
for(int j = 0 ; j < K - 1 ; ++j) {
a[i][j] = ++t;
v[i + 2].pb(t);
}
}
t = 0;
for(int i = K + 1 ; i <= N ; i += (K - 1)) {
for(int j = 0 ; j < K - 1 ; ++j) {
int p = j;
for(int h = 0 ; h < K - 1 ; ++h) {
v[i + j].pb(a[h][p]);
p = (p + t) % (K - 1);
}
}
++t;
}
out(N);space;out(K);enter;
for(int i = 1 ; i <= N ; ++i) {
for(auto b : v[i]) {
out(b);space;
}
enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
G - Mancala
题解
很容易想到\(sum[i][j]\)表示第\(i\)个点还需要后面给加\(j\)次的价值总和，\(cnt[i][j]\)表示第\(i\)个点还需要后面给加\(j\)次的方案数
由于能用到的状态不多，可以记忆化搜索，答案是\(dp[N][0]\)
转移就枚举这第\(i\)位放了\(p\)个，需要后面加\(d\)个
简单列个方程可以知道后面需要加\(d + \lfloor \frac{d + p}{i} \rfloor\)
然后加上\(cnt[i - 1][d + \lfloor \frac{d + p}{i} \rfloor] * (p - \frac{d + p}{i})\)
边界是对于1，\(sum[1][d] = \frac{(K - 1) * (K + 2)}{2} - 2 * d,cnt[1][d] = K + 1\)
因为1如果不满的话加多少都会扔掉
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
bool vis[105][100005];
int sum[105][100005],cnt[105][100005];
int N,K;
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &x,int y) {
x = inc(x,y);
}
int fpow(int x,int c) {
int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
void dfs(int p,int d) {
if(vis[p][d]) return;
vis[p][d] = 1;
if(p == 1) {
cnt[p][d] = K + 1;sum[p][d] = (mul(2,MOD - d) + (K + 2) * (K - 1) / 2) % MOD;
return ;
}
for(int i = 0 ; i < p ; ++i) {
if(i > K) break;
dfs(p - 1,d + (d + i) / p);
update(cnt[p][d],cnt[p - 1][d + (d + i) / p]);
update(sum[p][d],inc(sum[p - 1][d + (d + i) / p] ,mul(cnt[p - 1][d + (d + i) / p] ,inc(i, MOD - (d + i) / p))));
}
if(p <= K) {
dfs(p - 1,d + 1 + d / p);
update(cnt[p][d],cnt[p - 1][d + 1 + d / p]);
update(sum[p][d],inc(sum[p - 1][d + 1 + d / p], mul(cnt[p - 1][d + 1 + d / p],inc(p, MOD - d / p - 1))));
}
for(int i = p + 1 ; i <= K ; ++i) {
dfs(p - 1,d);
update(cnt[p][d],cnt[p - 1][d]);
update(sum[p][d],inc(sum[p - 1][d],mul(cnt[p - 1][d], i)));
}
}
void Solve() {
read(N);read(K);
dfs(N,0);
out(sum[N][0]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
H - Poor Penguin
题解
题解里的图画的挺好的
就是我们考虑把大矩形分成小矩形的代价是什么
例如一个矩形\([lx,ly,rx,ry]\)
我把它从\([lx,ly,i,j]\)里分出来
需要就是把
\([lx,ry + 1,rx,j]\)和\([rx + 1,ly,i,ry]\)里所有的障碍都扣去
然后我们对于每个包含p点的矩形，计算使得从P开始左上右上左下右下的一个角都扣去的最小代价
其实这是\(n^6\)的，怂的一批，不过记忆化搜索加上一点剪枝和AtCoder强大的评测机，好像还是不到1s的样子
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
bool vis[45][45][45][45];
int dp[45][45][45][45];
char s[45][45];
int H,W,sum[45][45];
void Init() {
read(H);read(W);
for(int i = 1 ; i <= H ; ++i) scanf("%s",s[i] + 1);
}
int query(int lx,int ly,int rx,int ry) {
return sum[rx][ry] + sum[lx - 1][ly - 1] - sum[rx][ly - 1] - sum[lx - 1][ry];
}
int dfs(int lx,int ly,int rx,int ry) {
if(lx == 1 && ly == 1 && rx == H && ry == W) return 0;
if(vis[lx][ly][rx][ry]) return dp[lx][ly][rx][ry];
int res = 100000;
for(int i = rx + 1 ; i <= H ; ++i) {
for(int j = ry + 1 ; j <= W ; ++j) {
int t = query(lx,ry + 1,rx,j) + query(rx + 1,ly,i,ry);
if(t >= res) break;
res = min(t + dfs(lx,ly,i,j),res);
}
}
for(int i = rx + 1 ; i <= H ; ++i) {
for(int j = ly - 1 ; j >= 1 ; --j) {
int t = query(lx,j,rx,ly - 1) + query(rx + 1,ly,i,ry);
if(t >= res) break;
res = min(t + dfs(lx,j,i,ry),res);
}
}
for(int i = lx - 1 ; i >= 1 ; --i) {
for(int j = ly - 1 ; j >= 1 ; --j) {
int t = query(i,ly,lx - 1,ry) + query(lx,j,rx,ly - 1);
if(t >= res) break;
res = min(t + dfs(i,j,rx,ry),res);
}
}
for(int i = lx - 1 ; i >= 1 ; --i) {
for(int j = ry + 1 ; j <= W ; ++j) {
int t = query(i,ly,lx - 1,ry) + query(lx,ry + 1,rx,j);
if(t >= res) break;
res = min(t + dfs(i,ly,rx,j),res);
}
}
vis[lx][ly][rx][ry] = 1;
return dp[lx][ly][rx][ry] = res;
}
void Solve() {
pii p;
for(int i = 1 ; i <= H ; ++i) {
for(int j = 1 ; j <= W ; ++j) {
if(s[i][j] == 'P') p = mp(i,j);
if(s[i][j] == '#') sum[i][j]++;
sum[i][j] = sum[i][j] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
}
}
int ans = 100000;
for(int i = 1 ; i <= H ; ++i) {
for(int j = 1 ; j <= W ; ++j) {
for(int k = i ; k <= H ; ++k) {
for(int h = j ; h <= W ; ++h) {
if(p.fi >= i && p.fi <= k && p.se >= j && p.se <= h) {
int t = query(i,j,p.fi,p.se);
t = min(t,query(p.fi,p.se,k,h));
t = min(t,query(i,p.se,p.fi,h));
t = min(t,query(p.fi,j,k,p.se));
if(t >= ans) continue;
ans = min(t + dfs(i,j,k,h),ans);
}
}
}
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}
I - Full Tournament
题解
观察一下可得
\(a_{i} < a_{i | 2^{k}}\)
我们把这个当做一条边，会连出一个dag，相当于给出拓扑序的一部分回复全部
可以对每个点求一个放的位置取值范围，如果这个点固定了就把左右端点都设成那个值
然后每次选一个右端点最小的放进去，不存在或右端点不合法就是无解了
恢复成原来的观察一下就是二进制反转
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int A[(1 << 18) + 5],N;
vector<int> e[2][(1 << 18) + 5];
int d[2][(1 << 18) + 5];
int L[(1 << 18) + 5],R[(1 << 18) + 5];
int que[2][(1 << 18) + 5],ql,qr,ans[(1 << 18) + 5];
vector<int> st[(1 << 18) + 5],ed[(1 << 18) + 5];
struct cmp {
bool operator () (const int &a,const int &b) const {
return R[a] < R[b] || (R[a] == R[b] && a < b);
}
};
set<int,cmp> S;
void Init() {
read(N);
for(int i = 0 ; i < (1 << N) ; ++i) {
read(A[i]);--A[i];
for(int j = 0 ; j < N ; ++j) {
if(!(i >> j & 1)) {
e[0][i].pb(i + (1 << j));
d[0][i + (1 << j)]++;
e[1][i + (1 << j)].pb(i);
d[1][i]++;
}
}
L[i] = 0;R[i] = (1 << N) - 1;
}
}
void Solve() {
L[0] = R[0] = 0;
for(int i = 0 ; i < (1 << N) ; ++i) {
if(A[i] != -1) L[i] = A[i];
int u = i;
for(auto v : e[0][u]) {
L[v] = max(L[v],L[u] + 1);
}
}
R[(1 << N) - 1] = L[(1 << N) - 1] = (1 << N) - 1;
for(int i = (1 << N) - 1 ; i >= 0 ; --i) {
if(A[i] != -1) R[i] = A[i];
int u = i;
for(auto v : e[1][u]) {
R[v] = min(R[u] - 1,R[v]);
}
}
for(int i = 0 ; i < (1 << N) ; ++i) {
st[L[i]].pb(i);
}
for(int i = 0 ; i < (1 << N) ; ++i) {
for(auto t : st[i]) S.insert(t);
if(S.empty()) {puts("NO");return;}
int t = *S.begin();
S.erase(S.begin());
if(R[t] < i) {puts("NO");return;}
ans[t] = i;
}
puts("YES");
for(int i = 1 , j = (1 << N - 1) ; i < (1 << N) - 1 ; ++i) {
if(i < j) swap(ans[i],ans[j]);
int k = (1 << N - 1);
while(j >= k) {
j -= k;
k >>= 1;
}
j += k;
}
for(int i = 0 ; i < (1 << N) ; ++i) {
out(ans[i] + 1);space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}
J - Tree MST
题解
好像直接点分可以爆艹
就是考虑点分每个点向别的子树里距离最短的点连边即可
然后直接kruskal
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int64 X[MAXN];
struct node {
int to,next;int64 val;
}E[MAXN * 2];
int head[MAXN],sumE;
int que[MAXN],ql,qr,fa[MAXN],siz[MAXN],son[MAXN];
int64 dis[MAXN];
pair<int64,int> pre[MAXN],suf[MAXN];
vector<int> ver[MAXN];
bool vis[MAXN];
struct Enode {
int u,v;int64 c;
friend bool operator < (const Enode &a,const Enode &b) {
return a.c < b.c;
}
}edge[MAXN * 30];
int tot;
void add(int u,int v,int64 c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].val = c;
head[u] = sumE;
}
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(X[i]);
int a,b;int64 c;
for(int i = 1 ; i < N ; ++i) {
read(a);read(b);read(c);
add(a,b,c);add(b,a,c);
}
}
int getfa(int u) {
return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
int Calc(int st) {
que[ql = qr = 1] = st;
fa[st] = 0;dis[st] = 0;
while(ql <= qr) {
int u = que[ql++];
siz[u] = 1;son[u] = 0;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa[u] && !vis[v]) {
que[++qr] = v;
fa[v] = u;
}
}
}
int res = que[qr];
for(int i = qr ; i >= 1 ; --i) {
int u = que[i];
if(fa[u]) {son[fa[u]] = max(son[fa[u]],siz[u]);siz[fa[u]] += siz[u];}
son[u] = max(son[u],qr - siz[u]);
if(son[u] < son[res]) res = u;
}
return res;
}
void dfs(int u) {
int G = Calc(u);
vis[G] = 1;
int cnt = 0;
dis[G] = 0;
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) {
++cnt;
ver[cnt].clear();
ver[cnt].pb(v);
fa[v] = G;dis[v] = dis[G] + E[i].val;
int s = 0;pair<int64,int> val = mp(dis[v] + X[v],v);
while(s < ver[cnt].size()) {
int n = ver[cnt][s];++s;
for(int k = head[n] ; k ; k = E[k].next) {
int h = E[k].to;
if(!vis[h] && h != fa[n]) {
fa[h] = n;
ver[cnt].pb(h);
dis[h] = dis[n] + E[k].val;
val = min(mp(dis[h] + X[h],h),val);
}
}
}
pre[cnt] = val;
suf[cnt] = val;
}
}
pre[0] = mp(1e18,0);
for(int i = 1 ; i <= cnt ; ++i) pre[i] = min(pre[i - 1],pre[i]);
suf[cnt + 1] = mp(1e18,0);
for(int i = cnt ; i >= 1 ; --i) suf[i] = min(suf[i + 1],suf[i]);
for(int i = 1 ; i <= cnt ; ++i) {
pair<int64,int> t = min(pre[i - 1],suf[i + 1]);
t = min(t,mp(X[G],G));
for(auto v : ver[i]) {
edge[++tot] = (Enode){v,t.se,t.fi + dis[v] + X[v]};
}
}
edge[++tot] = (Enode){G,pre[cnt].se,pre[cnt].fi + dis[G] + X[G]};
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) dfs(v);
}
}
void Solve() {
dfs(1);
sort(edge + 1,edge + tot + 1);
for(int i = 1 ; i <= N ; ++i) fa[i] = i;
int64 ans = 0;
int cnt = 0;
for(int i = 1 ; i <= tot ; ++i) {
if(getfa(edge[i].u) != getfa(edge[i].v)) {
ans += edge[i].c;
fa[getfa(edge[i].u)] = getfa(edge[i].v);
++cnt;
if(cnt == N - 1) break;
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}
