The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 19847 | Accepted: 6959 |
Description
Given a connected undirected graph, tell if its minimum spanning tree is unique.
Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.
Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.
Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.
Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.
Input
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <queue> 5 #define INF 0x3f3f3f3f 6 #define pii pair<int,int> 7 using namespace std; 8 const int maxn = 10000; 9 struct arc { 10 int to,cost,next; 11 arc(int x = 0,int y = 0,int z = -1) { 12 to = x; 13 cost = y; 14 next = z; 15 } 16 }; 17 arc e[maxn]; 18 int head[maxn],d[maxn],p[maxn],n,m,tot; 19 bool used[maxn]; 20 void add(int u,int v,int w) { 21 e[tot] = arc(v,w,head[u]); 22 head[u] = tot++; 23 } 24 int prim() { 25 priority_queue<pii,vector< pii >,greater< pii > >q; 26 for(int i = 1; i < maxn; ++i) { 27 d[i] = INF; 28 used[i] = false; 29 p[i] = -1; 30 } 31 d[1] = 0; 32 q.push(make_pair(d[1],1)); 33 int ans = 0; 34 bool isUnique = true; 35 while(!q.empty()) { 36 pair<int,int>now = q.top(); 37 q.pop(); 38 if(used[now.second]) continue; 39 used[now.second] = true; 40 ans += now.first; 41 for(int i = head[now.second]; ~i; i = e[i].next) { 42 if(!used[e[i].to] && d[e[i].to] > e[i].cost) { 43 d[e[i].to] = e[i].cost; 44 p[e[i].to] = now.second; 45 q.push(make_pair(d[e[i].to],e[i].to)); 46 } 47 if(used[e[i].to] && p[now.second] != -1 && p[now.second] != e[i].to && e[i].cost == now.first) isUnique = false; 48 } 49 } 50 return isUnique?ans:-1; 51 } 52 int main() { 53 int kase; 54 scanf("%d",&kase); 55 while(kase--) { 56 scanf("%d %d",&n,&m); 57 memset(head,-1,sizeof(head)); 58 for(int i = tot = 0; i < m; ++i) { 59 int u,v,w; 60 scanf("%d %d %d",&u,&v,&w); 61 add(u,v,w); 62 add(v,u,w); 63 } 64 int ans = prim(); 65 ans == -1?puts("Not Unique!"):printf("%d ",ans); 66 } 67 return 0; 68 } 69 /* 70 2 71 3 3 72 1 2 1 73 2 3 2 74 3 1 3 75 4 4 76 1 2 2 77 2 3 2 78 3 4 2 79 4 1 2 80 81 */
The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.
Output
For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.
Sample Input
2 3 3 1 2 1 2 3 2 3 1 3 4 4 1 2 2 2 3 2 3 4 2 4 1 2
Sample Output
3 Not Unique!
Source
解题：测试数据好像有问题啊，大大的问题。。。下面写法貌似只能过这题
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long 10 #define INF 0x3f3f3f 11 using namespace std; 12 int mp[101][101],d[101]; 13 int n,m; 14 bool vis[101]; 15 int prim(){ 16 int i,j,k,temp,index,ans = 0; 17 for(i = 1; i <= n; i++) d[i] = INF; 18 d[1] = 0; 19 memset(vis,false,sizeof(vis)); 20 for(i = 0; i < n; i++){ 21 temp = INF; 22 for(j = 1; j <= n; j++) 23 if(!vis[j] && d[j] < temp) temp = d[index = j]; 24 k = 0; 25 for(j = 1; j <= n; j++){ 26 if(vis[j] && mp[index][j] == temp) k++; 27 } 28 if(k > 1) return -1; 29 vis[index] = true; 30 ans += temp; 31 for(j = 1; j <= n; j++){ 32 if(!vis[j] && d[j] > mp[index][j]) d[j] = mp[index][j]; 33 } 34 } 35 return ans; 36 } 37 int main(){ 38 int ks,i,j,u,v,w; 39 scanf("%d",&ks); 40 while(ks--){ 41 scanf("%d%d",&n,&m); 42 for(i = 0; i <= n; i++) 43 for(j = 0; j <= n; j++) 44 mp[i][j] = INF; 45 for(i = 0; i < m; i++){ 46 scanf("%d%d%d",&u,&v,&w); 47 if(mp[u][v] > w){ 48 mp[u][v] = mp[v][u] = w; 49 } 50 } 51 w = prim(); 52 if(w < 0) puts("Not Unique!"); 53 else printf("%d ",w); 54 } 55 return 0; 56 }
上面的代码有些许问题
完整的正确代码
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 110; 18 int g[maxn][maxn],d[maxn],p[maxn],maxv[maxn][maxn],used[maxn][maxn],n,m; 19 bool vis[maxn]; 20 int prim() { 21 for(int i = 1; i <= n; ++i) { 22 d[i] = INF; 23 p[i] = -1; 24 vis[i] = false; 25 } 26 int sum = d[1] = 0; 27 for(int i = 1; i <= n; ++i) { 28 int minv = INF,index; 29 for(int j = 1; j <= n; ++j) 30 if(!vis[j] && d[j] < minv) 31 minv = d[index = j]; 32 if(p[index] > -1) { 33 used[index][p[index]] = used[p[index]][index] = 2; 34 for(int j = 1; j <= n; ++j) 35 if(vis[j]) maxv[j][index] = maxv[index][j] = max(maxv[j][p[index]],minv); 36 } 37 sum += minv; 38 vis[index] = true; 39 for(int j = 1; j <= n; ++j) 40 if(!vis[j] && g[index][j] < INF && g[index][j] < d[j]) { 41 d[j] = g[index][j]; 42 p[j] = index; 43 } 44 } 45 return sum; 46 } 47 int main() { 48 int t,u,v,w; 49 scanf("%d",&t); 50 while(t--) { 51 scanf("%d %d",&n,&m); 52 for(int i = 0; i <= n; ++i) 53 for(int j = 0; j <= n; ++j) { 54 maxv[i][j] = 0; 55 g[i][j] = INF; 56 used[i][j] = 0; 57 } 58 for(int i = 0; i < m; ++i){ 59 scanf("%d %d %d",&u,&v,&w); 60 if(g[u][v] > w){ 61 g[u][v] = g[v][u] = w; 62 used[u][v] = used[v][u] = 1; 63 } 64 } 65 int ans = prim(),ans2 = INF; 66 for(int i = 1; i <= n; ++i) 67 for(int j = i + 1; j <= n; ++j){ 68 if(used[i][j] == 1) ans2 = min(ans2,ans - maxv[i][j] + g[i][j]); 69 } 70 ans == ans2?puts("Not Unique!"):printf("%d ",ans); 71 } 72 return 0; 73 }
不用求次小生成树，也能判断是否唯一，prim版