题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2121
Ice_cream’s world II
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5832 Accepted Submission(s): 1493
Problem Description
After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it also a very difficult problem, because the world have N cities and M roads, every road was directed. Wiskey is a chief engineer in ice_cream world. The queen asked Wiskey must find a suitable location to establish the capital, beautify the roads which let capital can visit each city and the project’s cost as less as better. If Wiskey can’t fulfill the queen’s require, he will be punishing.
Input
Every case have two integers N and M (N<=1000, M<=10000), the cities numbered 0…N-1, following M lines, each line contain three integers S, T and C, meaning from S to T have a road will cost C.
Output
If no location satisfy the queen’s require, you must be output “impossible”, otherwise, print the minimum cost in this project and suitable city’s number. May be exist many suitable cities, choose the minimum number city. After every case print one blank.
Sample Input
3 1
0 1 1
4 4
0 1 10
0 2 10
1 3 20
2 3 30
Sample Output
impossible
40 0
Author
Wiskey
Source
题解：
1.题目要求：给定一幅有向图，求最小树形图（根节点不确定）。
2.一开始想枚举每个结点作为根节点，然后跑zhuliu算法，求出最小值。结果发现复杂度太大。
3.可行做法：设置一个超级点，作为虚拟的根节点，把超级点连向每一个题目中的点。然后跑zhuliu算法，如果所得的最小树形图中只有一条超级边（超级点连向题目中的点，这个点就是实际的根节点），那么就求出实际了最小树形图；如果有多条超级边（实际得到的为最小树形图森林），则无解。
4.那么超级边的权值应该设为多少呢？由于我们需要从zhuliu算法返回的数据中判断出有多少条超级边，所以超级边就应该设置的足够大，以方便检测，但又不能溢出。所以我们将其设置为题目中所有边的权值之和+1。这样，只要zhuliu()返回来的数据：ans<2*super_edge，就表明只含有一条超级边，所以最终答案为ans-super_edge（减去人工设置的超级边）。否则，如果ans>=2*super_edge，则表明至少有两条超级边，也就说明了：在实际的图中（没有超级点），至少有两个结点是没有入边的。然而没有入边的结点只能有1个或者没有（作为根节点），所以无解。
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 typedef long long LL; 8 const double EPS = 1e-6; 9 const int INF = INT_MAX; 10 const LL LNF = 9e18; 11 const int MOD = 1e9+7; 12 const int MAXN = 1e3+10; 13 14 struct Edge 15 { 16 int u, v, w; 17 }edge[10010]; 18 19 //super_edge为超级点连向每个普通点的边权， root_pos用于记录实际的根节点。 20 int super_edge, root_pos; 21 int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN]; 22 23 int zhuliu(int root, int n, int m) 24 { 25 int res = 0; 26 while(1) 27 { 28 for(int i = 0; i<n; i++) 29 in[i] = INF; 30 for(int i = 0; i<m; i++) 31 if(edge[i].u!=edge[i].v && edge[i].w<in[edge[i].v]) 32 { 33 pre[edge[i].v] = edge[i].u; 34 in[edge[i].v] = edge[i].w; 35 //为什么可以这样记录实际的根节点呢？因为在main()函数中，我们设置超级点连向普通点的时候， 36 //边的下标从m开始，对应着结点0， m+1对应着结点1，………所以我们可以根据边的下标得出边的终点。 37 if(edge[i].u==root) 38 root_pos = i; 39 40 } 41 42 for(int i = 0; i<n; i++) 43 if(i!=root && in[i]==INF) 44 return -1; 45 46 int tn = 0; 47 memset(id, -1, sizeof(id)); 48 memset(vis, -1, sizeof(vis)); 49 in[root] = 0; 50 for(int i = 0; i<n; i++) 51 { 52 res += in[i]; 53 int v = i; 54 while(vis[v]!=i && id[v]==-1 && v!=root) 55 { 56 vis[v] = i; 57 v = pre[v]; 58 } 59 if(v!=root && id[v]==-1) 60 { 61 for(int u = pre[v]; u!=v; u = pre[u]) 62 id[u] = tn; 63 id[v] = tn++; 64 } 65 } 66 if(tn==0) break; 67 for(int i = 0; i<n; i++) 68 if(id[i]==-1) 69 id[i] = tn++; 70 71 for(int i = 0; i<m; i++) 72 { 73 int v = edge[i].v; 74 edge[i].u = id[edge[i].u]; 75 edge[i].v = id[edge[i].v]; 76 if(edge[i].u!=edge[i].v) 77 edge[i].w -= in[v]; 78 } 79 n = tn; 80 root = id[root]; 81 } 82 return res; 83 } 84 85 int main() 86 { 87 int n, m; 88 while(scanf("%d%d", &n, &m)!=EOF) 89 { 90 super_edge = 0; 91 for(int i = 0; i<m; i++) 92 { 93 scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w); 94 super_edge += edge[i].w; 95 } 96 97 super_edge++; 98 for(int i = 0; i<n; i++) //n为超级点，将超级点连向每一个题目中的点 99 { 100 edge[m+i].u = n; 101 edge[m+i].v = i; 102 edge[m+i].w = super_edge; 103 } 104 105 int ans = zhuliu(n, n+1, m+n); 106 if(ans==-1 || ans>=2*super_edge) printf("impossible "); 107 else printf("%d %d ", ans-super_edge, root_pos-m); 108 } 109 }