继续畅通工程
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10206 Accepted Submission(s): 4451
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）。现得到城镇道路统计表，表中列出了任意两城镇间修建道路的费用，以及该道路是否已经修通的状态。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行给出村庄数目N ( 1< N < 100 )；随后的 N(N-1)/2 行对应村庄间道路的成本及修建状态，每行给4个正整数，分别是两个村庄的编号（从1编号到N），此两村庄间道路的成本，以及修建状态：1表示已建，0表示未建。
当N为0时输入结束。
当N为0时输入结束。
Output
每个测试用例的输出占一行，输出全省畅通需要的最低成本。
Sample Input
3
1 2 1 0
1 3 2 0
2 3 4 0
3
1 2 1 0
1 3 2 0
2 3 4 1
3
1 2 1 0
1 3 2 1
2 3 4 1
0
Sample Output
3
1
0
Author
ZJU
Source
1 #include <cstdio> 2 #include <queue> 3 #include <cstring> 4 5 using namespace std; 6 7 const int NV = 102; 8 const int NE = 10000; 9 const int INF = 1<<30; 10 int ne, nv, term, tot; 11 bool vis[NV]; 12 int dist[NV]; 13 struct Edge{ 14 int v, cost, next; 15 Edge(){} 16 Edge(int a, int c){v = a, cost = c;} 17 Edge(int a, int c, int d){v = a, cost = c, next = d;} 18 bool operator < (const Edge& x) const 19 { 20 return cost > x.cost; 21 } 22 }edge[NE]; 23 int eh[NV]; 24 25 int prim(int s = 1) 26 { 27 for(int i = 1; i <= nv; ++i) dist[i] = i == s ? 0 : INF; 28 priority_queue<Edge> que; 29 que.push(Edge(s, 0)); 30 while(!que.empty()) 31 { 32 Edge tmp = que.top(); 33 int u = tmp.v; 34 int cost = tmp.cost; 35 que.pop(); 36 if(vis[u]) continue; 37 vis[u] = true; 38 term += dist[u]; 39 40 for(int i = eh[u]; i != -1; i = edge[i].next) 41 { 42 int v = edge[i].v; 43 if(!vis[v] && dist[v] > edge[i].cost) 44 { 45 dist[v] = edge[i].cost; 46 que.push(Edge(v, dist[v])); 47 } 48 } 49 } 50 return term; 51 } 52 53 void addedge(int u, int v, int cost) 54 { 55 Edge e = Edge(v, cost, eh[u]); 56 edge[tot] = e; 57 eh[u] = tot++; 58 return; 59 } 60 61 int main() 62 { 63 #ifndef ONLINE_JUDGE 64 freopen("input.txt", "r", stdin); 65 #endif 66 while(scanf("%d", &nv) != EOF && nv) 67 { 68 memset(eh, -1, sizeof(eh)); 69 memset(vis, false, sizeof(vis)); 70 term = tot = 0; 71 int u, v, cost, state; 72 for(int i = 0; i < nv * (nv - 1) / 2; ++i) 73 { 74 scanf("%d%d%d%d", &u, &v, &cost, &state); 75 if(state) cost = 0; 76 addedge(u, v, cost); 77 addedge(v, u, cost); 78 } 79 prim(); 80 printf("%d ", term); 81 } 82 return 0; 83 }
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define SIZE 102 5 #define MAXN 10000 6 7 using namespace std; 8 9 const int INF = 1<<30; 10 int nv, ne; 11 struct Edge{ 12 int u, v, cost; 13 }edge[MAXN]; 14 15 int father[SIZE]; 16 17 bool cmp(const struct Edge &a, const struct Edge &b) 18 { 19 return a.cost < b.cost; 20 } 21 22 int find_father(int f) 23 { 24 return father[f] = f == father[f] ? f : find_father(father[f]); 25 } 26 27 int main() 28 { 29 #ifndef ONLINE_JUDGE 30 freopen("input.txt", "r", stdin); 31 #endif 32 while(scanf("%d", &nv) != EOF && nv) 33 { 34 int state; 35 for(int i=1; i<=nv; ++i) father[i] = i; 36 ne = nv*(nv-1)/2; 37 for(int i=0; i<ne; ++i) 38 { 39 scanf("%d%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost, &state); 40 if(state) 41 { 42 int u = find_father(edge[i].u); 43 int v = find_father(edge[i].v); 44 father[u] = v; 45 } 46 } 47 int term = 0; 48 sort(edge, edge+ne, cmp); 49 for(int i=0; i<ne; ++i) 50 { 51 52 int u = find_father(edge[i].u); 53 int v = find_father(edge[i].v); 54 if(u != v) 55 { 56 term += edge[i].cost; 57 father[u] = v; 58 } 59 } 60 printf("%d ", term); 61 } 62 return 0; 63 }