LightOJ - 1040 Donation(最小生成树)

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题目大意：给你一个邻接矩阵，表示两点之间的花费，现在问，花费总和-最小生成树的值
如果最小生成树不存在，另外输出

解题思路：模版题

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int MAXNODE = 110;const int MAXEDGE = 50010;typedef int Type;struct Edge{    int u, v;    Type d;    Edge() {}    Edge(int u, int v, Type d): u(u), v(v), d(d) {}}E[MAXEDGE];int n, m, tot, sum, cas = 1;int f[MAXNODE];Type maxcost[MAXNODE][MAXNODE];vector<Edge> G[MAXNODE];//初始化并查集和最小生成树的边void init() {    scanf("%d", &n);    for (int i = 0; i < n; i++) {        f[i] = i;        G[i].clear();    }    sum = m = 0;    int val;    for (int i = 0; i < n; i++)        for (int j = 0; j < n; j++) {            scanf("%d", &val);            if (val != 0) {                E[m++] = Edge(i, j, val);                sum += val;            }        }}int find(int x) {    return x == f[x] ? x : f[x] = find(f[x]);}bool cmp(const Edge &a, const Edge &b) {    return a.d < b.d;}//dfs找路径最大值,maxcost[i][j]维护的是树上的i到j点的路径上，最长的那条边的权值 void dfs(int s, int u, Type Max, int fa) {    maxcost[s][u] = max(maxcost[s][u], Max);    for (int i = 0; i < G[u].size(); i++) {        int v = G[u][i].v;        if (v == fa) continue;        double tmp = max(Max, G[u][i].d);        dfs(s, v, tmp, u);    }}//Kruskal找到最小生成树，并将最小生成树记录下来，以便后面用来求两点之间的最长边void solve() {    sort(E, E + m, cmp);    Type Sum = 0;    int num = 0;    for (int i = 0; i < m; i++) {        int fx = find(E[i].u);        int fy = find(E[i].v);        if (fx != fy) {            f[fx] = fy;            Sum += E[i].d;            G[E[i].u].push_back(E[i]);            swap(E[i].u, E[i].v);            G[E[i].u].push_back(E[i]);            num++;        }    }    if (num != n - 1)         printf("Case %d: %d\n", cas++ , -1);    else         printf("Case %d: %d\n", cas++, sum - Sum);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}

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