21-5-22校賽G 自行車調度

n個點，m條邊的無向帶正權圖（編號1到n)。每個點初始有a[i]（0<=a[i]<=1e5,1<=i<=n)輛自行車，自行車管理員可以花費一個邊權的代價移動一輛自行車從邊的一端到另一端。求自行車管理員所需的最小的代價使每個點自行車數量相等（無解輸出-1）。

調度問題，百分之九十都是網絡流，這個題是圖上的費用流
直接在原圖跑費用流就好了啊，我在干什么，還建了個完全二分圖。。

const int N = 510, inf = 1e9; int a[N], head[N], num, now[N], s, t, dis[N]; long long ans = 0; bool vis[N];struct edge {int nxt, to, val, pri; }; edge e[1000010];void addEdge(int x, int y, int z, int w) {e[++ num].nxt = head[x];head[x] = num;e[num].to = y;e[num].val = z;e[num].pri = w; }bool spfa() {for (int i = 0; i <= t; i ++)dis[i] = inf;dis[s] = 0;queue <int> q;q.push(s);while (!q.empty()){int x = q.front(); q.pop();for (int i = head[x]; i != -1; i = e[i].nxt){int to = e[i].to;if (dis[to] > dis[x] + e[i].pri)if (e[i].val > 0){dis[to] = dis[x] + e[i].pri;q.push(to);}}}if (dis[t] != inf) return true;return false; }int dfs(int x, int flow) {if (vis[x]) return 0;vis[x] = 1;if (x == t)return flow;for (int i = now[x]; i != -1; i = e[i].nxt){now[x] = i;if (dis[e[i].to] == dis[x] + e[i].pri && e[i].val > 0){int val = dfs(e[i].to, min(e[i].val, flow));if (val > 0){e[i].val -= val;e[i ^ 1].val += val;ans += (long long) val * e[i].pri;return val;} }}return 0; }int main() {int T = 1;T = read();while (T --){int n, m, sum = 0;num = -1;n = read(); m = read();memset (head, -1, sizeof (head));for (int i = 1; i <= n; i ++){a[i] = read();sum += a[i];}for (int i = 1; i <= m; i ++){int u, v, w;u = read(); v = read(); w = read();addEdge(u, v, inf, w);addEdge(v, u, 0, -w);addEdge(v, u, inf, w);addEdge(u, v, 0, -w);}if (sum % n != 0){printf ("-1\n");continue;}sum /= n;int tot = 0;for (int i = 1; i <= n; i ++){a[i] -= sum;if (a[i] > 0){addEdge(0, i, a[i], 0);addEdge(i, 0, 0, 0);tot += a[i];}if (a[i] < 0){addEdge(i, n + 1, -a[i], 0);addEdge(n + 1, i, 0, 0);}}s = 0; t = n + 1;ans = 0; int val;while (spfa()){for (int i = 0; i <= t; i ++){vis[i] = 0;now[i] = head[i];}while (val = dfs(s, inf)){tot -= val;}}if (tot > 0){printf ("-1\n");continue;}printf ("%lld\n", ans);}return 0; } 與50位技術專家面對面20年技術見證，附贈技術全景圖

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