描述：
就是求一个次小生成树的边权和
题解
我们先构造一个最小生成树， 把树上的边记录下来。
然后再枚举每条非树边(u, v, val)，在树上找出u 到v 路径上的最小边$g_0$ 和 严格次小边 $g_1$
如果$val > g_0$就可以考虑把$g_0$ 替换成$val$ 并记录答案。
如果$val = g_0$ 就把$g_1$替换成$val$ 记录答案。
然后我们就需要快速求出树链上的最小和次小边， 需要用树上倍增求LCA类似的方法求。
定义$g[0][ i ][ j ]$ 表示从$j$ 到第 $2^i$ 辈祖先中的最小边， $g[1][ i ][ j ] $表示从$j$ 到 第$2^i$ 辈祖先中的次小边， 满足以下关系：
1: $g[0][ i ][ j ] = max( g[0][ i - 1][ j ], g[0][ i - 1][ f[i - 1][ j ]]) $
2 :$ g[1][ i ][ j ] = max( g[1][ i - 1][ j ], g[1][ i - 1 ][ f[i - 1][ j ]]) $ 当$g[0][i - 1][ j ] = g[0][ i - 1][ f[i - 1][ j ]] $
3: $ g[1][ i ][ j ] = max(g[0][ i - 1][ j ], g[1][ i - 1][ f[i - 1][ j ]])$ 当$g[0][i - 1][ j ] < g[0][ i - 1][ f[i - 1][ j ]] $
4: $ g[1][ i ][ j ] = max(g[1][ i - 1][ j ], g[0][ i - 1][ f[i - 1][ j ]])$ 当$g[0][i - 1][ j ] > g[0][ i - 1][ f[i - 1][ j ]] $
倍增求树链上的最小和次小值时同理合并
题解
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define rep(i,a,b) for(register int i = (a); i <= (b); ++i) 6 #define per(i,a,b) for(register int i = (a); i >= (b); --i) 7 #define ll long long 8 using namespace std; 9 10 const int N = 5e5; 11 const int inf = ~0U >> 1; 12 13 int n, m; 14 ll sum, ans = 1e18; 15 int fa[N], f[30][N], g[2][30][N], dep[N]; 16 int head[N], tot; 17 18 struct edge { 19 int nxt, to, val; 20 }e[N << 2]; 21 22 struct node { 23 int u, v, val, mk; 24 }E[N << 2]; 25 26 inline int read() { 27 int X = 0, p = 1; char c = getchar(); 28 for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1; 29 for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0'; 30 return X * p; 31 } 32 33 inline void added(int u, int v, int val) { 34 e[++tot].to = v; 35 e[tot].nxt = head[u]; 36 e[tot].val = val; 37 head[u] = tot; 38 } 39 40 inline void add(int u, int v, int val) { 41 added(u, v, val); added(v, u, val); 42 } 43 44 inline int fd(int x) { 45 return fa[x] == x ? x : fa[x] = fd(fa[x]); 46 } 47 48 inline int cmp(const node &A, const node &B) { 49 return A.val < B.val; 50 } 51 52 inline void dfs(int u) { 53 for(int i = head[u]; i; i = e[i].nxt) { 54 int nt = e[i].to; 55 if(nt == f[0][u]) continue; 56 f[0][nt] = u; 57 g[0][0][nt] = e[i].val; 58 g[1][0][nt] = -inf; 59 dep[nt] = dep[u] + 1; 60 dfs(nt); 61 } 62 } 63 64 inline void LCA(int x, int y, int &a, int &b) { 65 int g0, g1; 66 if(dep[x] < dep[y]) swap(x, y); 67 for(int i = 20; ~i; --i) if(dep[f[i][x]] >= dep[y]) { 68 g0 = g[0][i][x]; g1 = g[1][i][x]; 69 if(g0 == a) b = max(b, g1); 70 if(g0 > a) b = max(a, g1); 71 if(g0 < a) b = max(g0, b); 72 a = max(a, g0); 73 x = f[i][x]; 74 } 75 for(int i = 20; ~i; --i) if(f[i][x] != f[i][y]) { 76 g0 = g[0][i][x]; g1 = g[1][i][x]; 77 if(g0 == a) b = max(b, g1); 78 if(g0 > a) b = max(a, g1); 79 if(g0 < a) b = max(g0, b); 80 a = max(a, g0); 81 82 g0 = g[0][i][y]; g1 = g[1][i][y]; 83 if(g0 == a) b = max(b, g1); 84 if(g0 > a) b = max(a, g1); 85 if(g0 < a) b = max(g0, b); 86 a = max(a, g0); 87 x = f[i][x]; y = f[i][y]; 88 } 89 90 g0 = g[0][0][x]; g1 = g[1][0][x]; 91 if(g0 == a) b = max(b, g1); 92 if(g0 > a) b = max(a, g1); 93 if(g0 < a) b = max(g0, b); 94 a = max(a, g0); 95 96 g0 = g[0][0][y]; g1 = g[1][0][y]; 97 if(g0 == a) b = max(b, g1); 98 if(g0 > a) b = max(a, g1); 99 if(g0 < a) b = max(g0, b); 100 a = max(a, g0); 101 } 102 103 int main() 104 { 105 n = rd; m = rd; 106 rep(i, 1, n) fa[i] = i; 107 rep(i, 1, m) { 108 int u = rd, v = rd, val = rd; 109 E[i].u = u; E[i].v = v; E[i].val = val; E[i].mk = 0; 110 } 111 sort(E+1, E+1+m, cmp); 112 rep(i, 1, m) { 113 int x = fd(E[i].u), y = fd(E[i].v); 114 if(x == y) continue; 115 sum += E[i].val; 116 fa[y] = x; 117 E[i].mk = 1; 118 add(E[i].u, E[i].v, E[i].val); 119 } 120 dep[1] = 1; 121 dfs(1); 122 rep(i, 1, 20) rep(j, 1, n) { 123 f[i][j] = f[i - 1][f[i - 1][j]]; 124 g[0][i][j] = max(g[0][i - 1][j], g[0][i - 1][f[i - 1][j]]); 125 int tmp = -inf; 126 if(g[0][i - 1][j] == g[0][i - 1][f[i - 1][j]]) tmp = max(g[1][i - 1][j], g[1][i - 1][f[i - 1][j]]); 127 if(g[0][i - 1][j] < g[0][i - 1][f[i - 1][j]]) tmp = max(g[0][i - 1][j], g[1][i - 1][f[i - 1][j]]); 128 if(g[0][i - 1][j] > g[0][i - 1][f[i - 1][j]]) tmp = max(g[1][i - 1][j], g[0][i - 1][f[i - 1][j]]); 129 g[1][i][j] = tmp; 130 } 131 rep(i, 1, m) if(!E[i].mk) { 132 int x = E[i].u, y = E[i].v, g0 = -inf, g1 = -inf; 133 LCA(x, y, g0, g1); 134 if(E[i].val == g0 && g1 != -inf) ans = min(ans, sum + E[i].val - g1); 135 if(E[i].val > g0) ans = min(ans, sum + E[i].val - g0); 136 } 137 printf("%lld ",ans); 138 }