【HDU 1863】畅通工程

畅通工程

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33115    Accepted Submission(s): 14628

Problem Description

省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可）。经过调查评估，得到的统计表中列出了有可能建设公路的若干条道路的成本。现请你编写程序，计算出全省畅通需要的最低成本。

 

Input

测试输入包含若干测试用例。每个测试用例的第1行给出评估的道路条数 N、村庄数目M ( < 100 )；随后的 N 
行对应村庄间道路的成本，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间道路的成本（也是正整数）。为简单起见，村庄从1到M编号。当N为0时，全部输入结束，相应的结果不要输出。

 

Output

对每个测试用例，在1行里输出全省畅通需要的最低成本。若统计数据不足以保证畅通，则输出“?”。

 

Sample Input

3 31 2 11 3 22 3 41 32 3 20 100

 

Sample Output

3?

 

 

Prim O(n^2)

#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3f#define maxn 105typedef long long ll;using namespace std;int cost[maxn][maxn];int closedge[maxn];bool vis[maxn];int Prim(int n){memset(vis, false, sizeof(vis));vis[1] = true;int ans = 0;for (int i = 2; i <= n; i++) closedge[i] = cost[1][i];for (int i = 2; i <= n; i++) {int p = -1, minc = INF;for (int j = 1; j <= n; j++) {if (!vis[j] && closedge[j] < minc) {minc = closedge[j];p = j;}}if (p == -1) return -1;vis[p] = true;ans += minc;for (int j = 1; j <= n; j++) {if (!vis[j] && cost[p][j] < closedge[j])closedge[j] = cost[p][j];}}return ans;}int main(){int n, m, u, v, w;while (~scanf("%d%d", &n, &m) && n) {memset(cost, INF, sizeof(cost));while (n--) {scanf("%d%d%d", &u, &v, &w);cost[u][v] = w;cost[v][u] = w;}int ans = Prim(m);if (ans == -1)printf("?\n");elseprintf("%d\n", ans);}return 0;}

Kruskal O(eloge)

#include <cstdio>#include <cstring>#include <algorithm>#define maxn 105#define INF 0x3f3f3f3ftypedef long long ll;using namespace std;int cost[maxn][maxn];int fa[maxn];struct Edge{int u, v, w;}edge[maxn*maxn];int tol;void add(int u, int v, int w){edge[tol].u = u;edge[tol].v = v;edge[tol++].w = w;}void Init(){tol = 0;memset(fa, -1, sizeof(fa));memset(cost, INF, sizeof(cost));}bool cmp(Edge a, Edge b){return a.w < b.w;}int Find(int x){if (fa[x] == -1) return x;return fa[x] = Find(fa[x]);}int Kruskal(int n){sort(edge, edge + tol, cmp);int ans = 0;int cnt = 0;for (int i = 0; i < tol; i++) {int u = edge[i].u, v = edge[i].v, w = edge[i].w;int f1 = Find(u), f2 = Find(v);if (f1 != f2) {fa[f1] = f2;ans += w;cnt++;}if (cnt == n - 1) break;}if (cnt < n - 1) return -1;return ans;}int main(){int n, m, u, v, w;while (~scanf("%d%d", &n, &m) && n) {Init();while (n--) {scanf("%d%d%d", &u, &v, &w);if (w < cost[u][v]) {cost[u][v] = cost[v][u] = w;add(u, v, w);}}int ans = Kruskal(m);if (ans == -1) printf("?\n");else printf("%d\n", ans);}return 0;}