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@ -63,18 +63,19 @@ $1≤n≤300,0≤v_i,p_i,j≤10^5$
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 310;
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const int INF = 0x3f3f3f3f;
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int n; // n条顶点
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int res; // 最小生成树的权值和
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int el; // 边数
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// Kruskal用到的结构体
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const int M = 2 * N * N; // 无向图*2，稠密图N*N
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struct Edge {
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int a, b, w;
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int a, b, c;
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const bool operator<(const Edge &t) const {
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return w < t.w;
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return c < t.c;
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}
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} e[M];
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} edge[M];
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int el; // 边数
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// 并查集
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int p[N];
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int find(int x) {
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@ -84,39 +85,40 @@ int find(int x) {
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// Kruskal算法
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int kruskal() {
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// 按边的权重排序
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sort(e, e + el);
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sort(edge, edge + el);
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// 初始化并查集,注意并查集的初始是从0开始的，因为0号是超级源点
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for (int i = 0; i <= n; i++) p[i] = i;
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// 枚举每条边
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for (int i = 0; i < el; i++) {
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int a = e[i].a, b = e[i].b, w = e[i].w;
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int a = edge[i].a, b = edge[i].b, c = edge[i].c;
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a = find(a), b = find(b);
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if (a != b)
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p[a] = b, res += w;
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p[a] = b, res += c;
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}
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return res;
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}
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int main() {
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cin >> n;
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// 建立超级源点(0 <-> 1~n )
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int w;
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int c;
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for (int i = 1; i <= n; i++) {
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cin >> w; // 点权
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e[el++] = {0, i, w};
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e[el++] = {i, 0, w};
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cin >> c; // 点权转边权
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edge[el++] = {0, i, c};
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edge[el++] = {i, 0, c};
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}
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// 本题是按矩阵读入的，不是按a,b,c方式读入的
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++) {
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cin >> w;
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e[el++] = {i, j, w};
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e[el++] = {j, i, w};
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cin >> c;
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edge[el++] = {i, j, c};
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edge[el++] = {j, i, c};
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}
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// 利用Kruskal计算最小生成树
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printf("%d\n", kruskal());
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cout << kruskal() << endl;
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return 0;
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}
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@ -131,34 +133,36 @@ const int N = 310;
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int n;
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int g[N][N];
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int dist[N];
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int dis[N];
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bool st[N];
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int res; // 最小生成树里面边的长度之和
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int prim() {
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memset(dist, 0x3f, sizeof dist); // 初始化所有距离为INF
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dist[0] = 0; // 超级源点是在生成树中的
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void prim() {
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memset(dis, 0x3f, sizeof dis); // 初始化所有距离为INF
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dis[0] = 0; // 超级源点是在生成树中的
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for (int i = 0; i <= n; i++) { // 注意：这里因为引入了超级源点，所以点的个数是n+1
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int t = -1;
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for (int j = 0; j <= n; j++)
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if (!st[j] && (t == -1 || dist[t] > dist[j]))
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t = j;
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st[t] = true;
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res += dist[t];
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if (!st[j] && (t == -1 || dis[t] > dis[j])) t = j;
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if (i) res += dis[t];
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// 有超级源点的题，是必然存在最小生成树的
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// 注意这里也是需要从0~n共n+1个
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for (int j = 0; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
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for (int j = 0; j <= n; j++)
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if (!st[j] && dis[j] > g[t][j])
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dis[j] = g[t][j];
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st[t] = true;
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}
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return res;
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}
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int main() {
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cin >> n;
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// 建立超级源点(0 <-> 1~n )，点权转化为超级源点到此节点的边权
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for (int i = 1; i <= n; i++) {
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cin >> g[0][i];
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g[i][0] = g[0][i];
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int c;
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cin >> c;
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g[i][0] = g[0][i] = c;
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}
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// 本题是按矩阵读入的，不是按a,b,c方式读入的
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for (int i = 1; i <= n; i++)
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@ -166,7 +170,8 @@ int main() {
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cin >> g[i][j];
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// 利用prim计算最小生成树
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printf("%d\n", prim());
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prim();
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cout << res << endl;
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return 0;
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}
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