CodeForces 472D Design Tutorial: Inverse the Problem (最小生成树)

题意：给出一个树中每两个节点的距离，问是否能构成一颗加权树。

思路：

以下几种情况可以直接输出NO：arr[i][i]!=0, arr[i][j] == 0(i != j), arr[i][j]!=arr[j][i].

否则，每次选最小的边，加入生成树，最后在算一遍节点两两之间的距离，如果与输入矩阵不同则输出NO。

节点两两距离dist[i][j] = dist[root][i] + dist[root][j] - 2 * dist[root][LCA[i][j]]

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>using namespace std;struct Edge{int from, to, v;Edge(int a, int b, int c) : from(a), to(b), v(c) {}bool operator < (const Edge &e) const {return v < e.v;}};int n;vector<Edge> edges;vector<Edge> tree[2005];int arr[2005][2005];int parent[2005];int father[2005];int vis[2005];int lca[2005][2005];int dp[2005];int find(int p) {if (p == parent[p]) return p;return parent[p] = find(parent[p]);}void un(int p, int q) {parent[find(p)] = find(q);}int getFather(int r) {if (r == father[r]) return r;else return getFather(father[r]);}void LCA(int x) {vis[x] = 1;for (int i = 0; i < tree[x].size(); i++) {Edge e = tree[x][i];if (!vis[e.to]) {LCA(e.to);father[e.to] = x;}}for (int i = 1; i <= n; i++) {if (vis[i] && !lca[i][x]) {lca[i][x] = lca[x][i] = getFather(i);}}}void dfs(int x, int pre) {for (int i = 0; i < tree[x].size(); i++) {Edge e = tree[x][i];if (e.to == pre) continue;dp[e.to] = dp[x] + e.v;dfs(e.to, x);}}int main() {scanf("%d", &n);for (int i = 1; i <= n; i++) {parent[i] = father[i] = i;for (int j = 1; j <= n; j++) {scanf("%d", &arr[i][j]);if ((i == j && arr[i][j]) || (i != j && !arr[i][j]) || (i > j && arr[i][j] != arr[j][i])) {printf("NO\n");return 0;}if (i < j)edges.push_back(Edge(i, j, arr[i][j]));}}sort(edges.begin(), edges.end());for (int i = 0; i < edges.size(); i++) {Edge e = edges[i];if (find(e.from) == find(e.to)) continue;else {un(e.from, e.to);tree[e.from].push_back(Edge(e.from, e.to, e.v));tree[e.to].push_back(Edge(e.to, e.from, e.v));}}LCA(1);dfs(1, 0);for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {int dist = dp[i] + dp[j] - dp[lca[i][j]] * 2;if (dist != arr[i][j]) {printf("NO\n");return 0;}}}printf("YES\n");return 0;}