链接：
http://poj.org/problem?id=1511
题意：
给你一个有向图，输出从0到所有点的距离总和+从所有点到0的距离总和的最小值
题解：
先正向建图，求一个最短路，再反向建图，求一个最短路，一定要用spfa，dijkstra会超时
代码：
31 #define inf 1000000000000LL 32 struct Node { int u, v; ll c; }; 33 Node p[MAXN]; 34 int head[MAXN], next[MAXN]; 35 ll d[MAXN]; 36 bool used[MAXN]; 37 int now[MAXN]; 38 int e, top, n, m; 39 40 void add_node(int u, int v, ll c) { 41 p[e] = Node{ u,v,c }; 42 next[e] = head[u]; head[u] = e++; 43 } 44 45 bool relax(int u, int v, ll c) { 46 if (d[v]>d[u] + c) { 47 d[v] = d[u] + c; 48 return true; 49 } 50 return false; 51 } 52 53 void spfa(int t) { 54 memset(used, 0, sizeof(used)); 55 rep(i, 1, n + 1) d[i] = inf; 56 d[t] = 0, used[t] = 1, top = 0; 57 now[top++] = t; 58 while (top) { 59 int pre = now[--top]; 60 used[pre] = 0; 61 for (int j = head[pre]; j + 1; j = next[j]) { 62 if (relax(p[j].u, p[j].v, p[j].c) && used[p[j].v] == false) { 63 now[top++] = p[j].v; 64 used[p[j].v] = true; 65 } 66 } 67 } 68 } 69 70 int main() { 71 int T; 72 cin >> T; 73 while (T--) { 74 cin >> n >> m; 75 memset(head, -1, sizeof(head)); 76 memset(next, -1, sizeof(next)); 77 e = 0; 78 rep(i, 0, m) { 79 int u, v; 80 ll c; 81 scanf("%d%d%lld", &u, &v, &c); 82 add_node(u, v, c); 83 } 84 ll ans = 0; 85 spfa(1); 86 rep(i, 1, n + 1) ans += d[i]; 87 memset(head, -1, sizeof(head)); 88 memset(next, -1, sizeof(next)); 89 e = 0; 90 rep(i, 0, m) add_node(p[i].v, p[i].u, p[i].c); 91 spfa(1); 92 rep(i, 1, n + 1) ans += d[i]; 93 cout << ans << endl; 94 } 95 return 0; 96 }