POJ1258 Agri-Net MST最小產生樹題解

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搭建一個最小代價的網路，最原始的最小產生樹的應用。

這裡使用Union find和Kruskal演算法求解.

注意：

1 給出的資料是原始的矩陣圖，可是須要轉化為邊表示的圖，方便運用Kruskal，由於須要sort

2 降低邊。一個矩陣最多須要(N*N-N)>>1條邊，有人討論本題是否有向，那是無意義的。由於本題的最小產生樹和方向無關。

3 使用Union find是為了推斷是否有環。比原始推斷快非常多。

#include <stdio.h>#include <stdlib.h>const int MAX_VEC = 101;int N;//number of verticesstruct SubSet{int p, rank;};struct Edge{int src, des, wei;};struct Graph{int V, E;Edge *edge;Graph(int v, int e) : V(v), E(e){edge = new Edge[E];}~Graph(){if (edge) delete[]edge; edge = NULL;}};static int cmp(const void *a, const void *b){Edge *a1 = (Edge *) a;Edge *b1 = (Edge *) b;return a1->wei - b1->wei;}SubSet *subs;Edge *res;Graph *gra;void initResource(){subs = new SubSet[N];for (int i = 0; i < N; i++){subs[i].p = i;subs[i].rank = 0;}res = new Edge[N-1];}inline void releaseResource(){if (subs) delete [] subs;if (res) delete [] res;}int find(int node){if (subs[node].p != node)subs[node].p = find(subs[node].p);return subs[node].p;}inline void unionTwo(int x, int y){int xroot = find(x);int yroot = find(y);if (subs[xroot].rank < subs[yroot].rank) subs[xroot].p = yroot;else if (subs[xroot].rank > subs[yroot].rank) subs[yroot].p = xroot;else{subs[xroot].rank++;subs[yroot].p = xroot;}}int mst(){initResource();qsort(gra->edge, gra->E, sizeof(Edge), cmp);for (int i = 0, v = 0; i < gra->E && v < gra->V - 1; i++){int xroot = find(gra->edge[i].src);int yroot = find(gra->edge[i].des);if (xroot != yroot){unionTwo(xroot, yroot);res[v++] = gra->edge[i];}}int ans = 0;for (int i = 0; i < N-1; i++){ans += res[i].wei;}releaseResource();return ans;}int main(){int w;while (~scanf("%d", &N)){gra = new Graph(N, (N*N-N)>>1);int e = 0;for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){scanf("%d", &w);if (j <= i) continue;//下三角形的值不入邊gra->edge[e].src = i;gra->edge[e].des = j;gra->edge[e++].wei = w;}}printf("%d\n", mst());delete gra;}return 0;}

POJ1258 Agri-Net MST最小產生樹題解