给出一个带权有向图，要使整个图连通。SCC中的点之间花费为0，所以就先缩点，然后缩点后两点之间的权值为最小边的权值，把这些权值累加起来就是答案。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #include <stack> 7 using namespace std; 8 9 int n, m; 10 11 const int maxn = 50000 + 10; 12 13 vector<int> G[maxn], C[maxn]; 14 15 stack<int> S; 16 int pre[maxn], lowlink[maxn], sccno[maxn]; 17 int dfs_clock, scc_cnt; 18 19 void dfs(int u) 20 { 21 pre[u] = lowlink[u] = ++dfs_clock; 22 S.push(u); 23 for(int i = 0; i < G[u].size(); i++) 24 { 25 int v = G[u][i]; 26 if(!pre[v]) 27 { 28 dfs(v); 29 lowlink[u] = min(lowlink[u], lowlink[v]); 30 } 31 else if(!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]); 32 } 33 34 if(lowlink[u] == pre[u]) 35 { 36 scc_cnt++; 37 for(;;) 38 { 39 int x = S.top(); S.pop(); 40 sccno[x] = scc_cnt; 41 if(x == u) break; 42 } 43 } 44 } 45 46 void find_scc() 47 { 48 dfs_clock = scc_cnt = 0; 49 memset(pre, 0, sizeof(pre)); 50 memset(sccno, 0, sizeof(sccno)); 51 for(int i = 0; i < n; i++) if(!pre[i]) dfs(i); 52 } 53 54 int cost[maxn]; 55 56 int main() 57 { 58 while(scanf("%d%d", &n, &m) == 2) 59 { 60 for(int i = 0; i < n; i++) { G[i].clear(); C[i].clear(); } 61 while(m--) 62 { 63 int u, v, d; scanf("%d%d%d", &u, &v, &d); 64 G[u].push_back(v); C[u].push_back(d); 65 } 66 find_scc(); 67 68 memset(cost, -1, sizeof(cost)); 69 for(int i = 0; i < n; i++) 70 for(int j = 0; j < G[i].size(); j++) 71 { 72 int u = sccno[i], v = sccno[G[i][j]]; 73 if(u == v) continue; 74 if(cost[v] == -1) cost[v] = C[i][j]; 75 else cost[v] = min(cost[v], C[i][j]); 76 } 77 78 int ans = 0; 79 for(int i = 1; i <= scc_cnt; i++) if(cost[i] != -1) ans += cost[i]; 80 printf("%d\n", ans); 81 } 82 83 return 0; 84 }
