仙人掌
作者:互联网
任意一条边至多只出现在一条简单回路的无向连通图称为仙人掌
求仙人掌的最短路
大致做法:利用类似tarjan的方式将一个环缩到一起,建立圆方树,对于一个环找到头节点,建立一个方点,头节点向方点连一条权值为0的边,方点向环上的点再连一条权值为该点到头节点最短距离的边
然后再树上找最短路就是找到lca然后类似前缀和相减的方式,对于lca找到的如果方点需要特殊处理---把这个圆取一下两个点之间的最短距离
#include <iostream> #include <cstring> #include <algorithm> using namespace std; const int N = 12010, M = N * 3; int n, m, Q, new_n; int h1[N], h2[N], e[M], w[M], ne[M], idx; int dfn[N], low[N], cnt; int s[N], stot[N], fu[N], fw[N]; int fa[N][14], depth[N], d[N]; int A, B; void add(int h[], int a, int b, int c) { e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ; } void build_circle(int x, int y, int z) { int sum = z; for (int k = y; k != x; k = fu[k]) { s[k] = sum; sum += fw[k]; } s[x] = stot[x] = sum; add(h2, x, ++ new_n, 0); for (int k = y; k != x; k = fu[k]) { stot[k] = sum; add(h2, new_n, k, min(s[k], sum - s[k])); } } void tarjan(int u, int from) { dfn[u] = low[u] = ++ cnt; for (int i = h1[u]; ~i; i = ne[i]) { int j = e[i]; if (!dfn[j]) { fu[j] = u, fw[j] = w[i]; tarjan(j, i); low[u] = min(low[u], low[j]); if (dfn[u] < low[j]) add(h2, u, j, w[i]); } else if (i != (from ^ 1)) low[u] = min(low[u], dfn[j]); } for (int i = h1[u]; ~i; i = ne[i]) { int j = e[i]; if (dfn[u] < dfn[j] && fu[j] != u) build_circle(u, j, w[i]); } } void dfs_lca(int u, int father) { depth[u] = depth[father] + 1; fa[u][0] = father; for (int k = 1; k <= 13; k ++ ) fa[u][k] = fa[fa[u][k - 1]][k - 1]; for (int i = h2[u]; ~i; i = ne[i]) { int j = e[i]; d[j] = d[u] + w[i]; dfs_lca(j, u); } } int lca(int a, int b) { if (depth[a] < depth[b]) swap(a, b); for (int k = 13; k >= 0; k -- ) if (depth[fa[a][k]] >= depth[b]) a = fa[a][k]; if (a == b) return a; for (int k = 13; k >= 0; k -- ) if (fa[a][k] != fa[b][k]) { a = fa[a][k]; b = fa[b][k]; } A = a, B = b; return fa[a][0]; } int main() { scanf("%d%d%d", &n, &m, &Q); new_n = n; memset(h1, -1, sizeof h1); memset(h2, -1, sizeof h2); while (m -- ) { int a, b, c; scanf("%d%d%d", &a, &b, &c); add(h1, a, b, c), add(h1, b, a, c); } tarjan(1, -1); dfs_lca(1, 0); while (Q -- ) { int a, b; scanf("%d%d", &a, &b); int p = lca(a, b); if (p <= n) printf("%d\n", d[a] + d[b] - d[p] * 2); else { int da = d[a] - d[A], db = d[b] - d[B]; int l = abs(s[A] - s[B]); int dm = min(l, stot[A] - l); printf("%d\n", da + dm + db); } } return 0; }
标签:int,h1,fa,add,dfn,low,仙人掌 来源: https://www.cnblogs.com/cyq123/p/14230834.html