题目大意：同Petya and Graph，数据范围改成$n\leqslant5\times10^3,m\leqslant5\times10^4$
题解：同上
卡点：无
C++ Code：
#include <algorithm>
#include <cstdio>
#define maxn 5010
#define maxm 50010
const int N = maxn + maxm, M = N + 2 * maxm;
const int inf = 0x3f3f3f3f;namespace Network_Flow {int lst[N], head[N], cnt = 1;struct Edge {int to, nxt, w;} e[M << 1];inline void addedge(int a, int b, int c) {e[++cnt] = (Edge) { b, head[a], c }; head[a] = cnt;e[++cnt] = (Edge) { a, head[b], 0 }; head[b] = cnt;}int n, st, ed, MF;int GAP[N], d[N];int q[N], h, t;void init() {GAP[d[ed] = 1] = 1;for (int i = 0; i < n; ++i) lst[i] = head[i];q[h = t = 0] = ed;while (h <= t) {int u = q[h++];for (int i = head[u]; i; i = e[i].nxt) {int v = e[i].to;if (!d[v]) {d[v] = d[u] + 1;++GAP[d[v]];q[++t] = v;}}}}int dfs(int u, int low) {if (!low || u == ed) return low;int w, res = 0;for (int &i = lst[u]; i; i = e[i].nxt) if (e[i].w) {int v = e[i].to;if (d[u] == d[v] + 1) {w = dfs(v, std::min(low, e[i].w));res += w, low -= w;e[i].w -= w, e[i ^ 1].w += w;if (!low) return res;}}if (!(--GAP[d[u]])) d[st] = n + 1;++GAP[++d[u]], lst[u] = head[u];return res;}void ISAP(int S, int T) {st = S, ed = T;init();while (d[st] <= n) MF += dfs(st, inf);}
}
using Network_Flow::addedge;int n, m, sum;
int main() {scanf("%d%d", &n, &m); Network_Flow::n = n + m + 2;int st = 0, ed = n + m + 1;for (int i = 1, x; i <= n; ++i) {scanf("%d", &x);addedge(st, i, x);}for (int i = 1, a, b, c; i <= m; ++i) {scanf("%d%d%d", &a, &b, &c);addedge(a, n + i, inf);addedge(b, n + i, inf);addedge(n + i, ed, c);sum += c;}Network_Flow::ISAP(st, ed);printf("%d\n", sum - Network_Flow::MF);return 0;
}
--异常)
