[USACO08NOV]Cheering up the Cow

嘟嘟嘟

这道题删完边后是一棵树，那么一定和最小生成树有关啦。

考虑最后的生成树，无论从哪一个点出发，每一条边都会访问两次，而且在看一看样例，会发现走第w条边(u, v)的代价是L[w] * 2 + c[u] + c[v],所以说把每一条边的边权改为这个，然后跑裸的最小生成树就行了。然后答案还要加上出发的点的权值，那自然要加权值最小的点。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<stack>
 #include<queue>
 #include<vector>
 #include<cctype>
 using namespace std;
 #define space putchar(' ')
 #define enter puts("")
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const  db eps = 1e-;
 const int maxn = 1e4 + ;
 const int max_size = 1e5 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = (ans << ) + (ans << ) + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) putchar('-'), x = -x;
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, m, a[maxn];
 struct Node
 {
     int x, y, cost;
     bool operator < (const Node& other)const
     {
         return cost < other.cost;
     }
 }t[max_size];

 int p[maxn];
 void init()
 {
     for(int i = ; i <= n; ++i) p[i] = i;
 }
 int Find(int x)
 {
     return x == p[x] ? x : p[x] = Find(p[x]);
 }

 int ans = , cnt = , Min = INF;

 int main()
 {
     n = read(); m = read();
     for(int i = ; i <= n; ++i) a[i] = read(), Min = min(Min, a[i]);
     for(int i = ; i <= m; ++i)
     {
         t[i].x = read(); t[i].y = read(); t[i].cost = read();
         t[i].cost = (t[i].cost << ) + a[t[i].x] + a[t[i].y];
     }
     sort(t + , t + m + );
     init();
     for(int i = ; i <= m; ++i)
     {
         int px = Find(t[i].x), py = Find(t[i].y);
         if(px != py)
         {
             p[px] = py;
             ans += t[i].cost; cnt++;
             if(cnt == n - ) break;
         }
     }
     write(ans + Min); enter;
     return ;
 }