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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24998 Accepted Submission(s): 10867
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 5 int N, M; 6 7 struct Edge{ 8 int u, v, w; 9 bool operator() (Edge a, Edge b) 10 { 11 return a.w<b.w; 12 } 13 }; 14 Edge e[5050]; 15 int f[105]; 16 17 int getf(int x) 18 { 19 return x = f[x] ? f[x] : f[x] = getf(f[x]); 20 } 21 22 bool merge(int a, int b) 23 { 24 int t1 = getf(a); 25 int t2 = getf(b); 26 if(t1 != t2){ 27 f[t2] = t1; 28 return true; 29 } 30 return false; 31 } 32 33 void kruskal() 34 { 35 for(int i = 1; i <= M; ++i) 36 f[i] = i; 37 sort(e, e+N, Edge()); 38 int cnt = 0; 39 int ans = 0; 40 bool success = false; 41 for(int i = 0; i < N; ++i){ 42 if(merge(e[i].u, e[i].v)){ 43 ans += e[i].w; 44 ++cnt; 45 } 46 if(cnt == M-1){ 47 success = true; 48 break; 49 } 50 } 51 if(success) 52 printf("%d\n", ans); 53 else 54 printf("?\n"); 55 } 56 57 int main() 58 { 59 while(scanf("%d%d", &N, &M), N){ 60 for(int i = 0; i < N; ++i){ 61 scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); 62 } 63 if(N < M-1) //明显统计数据不足 64 printf("?\n"); 65 else 66 kruskal(); 67 } 68 return 0; 69 }
Prim
1 #include <cstdio> 2 #include <cstring> 3 4 const int INF = 0x3f3f3f3f; 5 int N, M; 6 int e[105][105]; 7 int dis[105]; 8 bool vis[105]; 9 10 void prim() 11 { 12 13 for(int i = 1; i <= M; ++i) 14 dis[i] = e[1][i]; 15 vis[1] = true; 16 int ans = 0; 17 int cnt = 1; 18 bool success = true; 19 while(cnt<M){ 20 int min_dis = INF, u; 21 for(int i = 1; i <= M; ++i){ 22 if(!vis[i] && dis[i]<min_dis){ 23 min_dis = dis[u = i]; 24 } 25 } 26 if(min_dis == INF){ 27 success = false; 28 break; 29 } 30 vis[u] = true; 31 ++cnt; 32 ans += dis[u]; 33 for(int i = 1; i <= M; ++i){ 34 if(!vis[i] && e[u][i]<dis[i]){ 35 dis[i] = e[u][i]; 36 } 37 } 38 } 39 if(success) 40 printf("%d\n", ans); 41 else 42 printf("?\n"); 43 } 44 45 int main() 46 { 47 while(scanf("%d%d", &N, &M), N){ 48 memset(vis, 0, sizeof(vis)); 49 for(int i = 1; i <= M; ++i) 50 for(int j = 1; j <= M; ++j) 51 e[i][j] = (i == j ? 0 : INF); 52 int u, v, w; 53 for(int i = 1; i <= N; ++i){ 54 scanf("%d%d%d", &u, &v, &w); 55 e[u][v] = e[v][u] = w; 56 } 57 if(N < M-1) 58 printf("?\n"); 59 else 60 prim(); 61 } 62 return 0; 63 }
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原文地址:http://www.cnblogs.com/inmoonlight/p/5671818.html