【题目链接】
【思路要点】
- 建出GomoryHu-Tree ,剩余部分用DFS即可解决。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h>
using namespace std;
#define MAXN 205
#define INF 1e9
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
struct edge {int dest, flow; unsigned home; };
struct treeedge {int dest, index; };
int n, m, s, t, Q, cnt, value;
int dist[MAXN], p[MAXN], q[MAXN];
unsigned curr[MAXN];
bool visited[MAXN];
vector <edge> a[MAXN];
vector <treeedge> b[MAXN];
bool bfs() {
memset(dist, 0, sizeof(dist));
static int q[MAXN];
int l = 0, r = 0;
dist[s] = 1, q[0] = s;
while (l <= r) {
int tmp = q[l++];
for (unsigned i = 0; i < a[tmp].size(); i++)
if (a[tmp][i].flow != 0 && dist[a[tmp][i].dest] == 0) {
dist[a[tmp][i].dest] = dist[tmp] + 1;
q[++r] = a[tmp][i].dest;
}
}
return dist[t] != 0;
}
int dinic(int pos, int limit) {
if (pos == t) return limit;
int used = 0, tmp;
for (unsigned &i = curr[pos]; i < a[pos].size(); i++)
if (a[pos][i].flow != 0 && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)))) {
used += tmp;
a[pos][i].flow -= tmp;
a[a[pos][i].dest][a[pos][i].home].flow += tmp;
if (used == limit) return limit;
}
return used;
}
void addedge(int s, int t, int flow) {
a[s].push_back((edge) {t, flow, a[t].size()});
a[t].push_back((edge) {s, flow, a[s].size() - 1});
}
void ClearGraph() {
for (int i = 1; i <= n; i++)
for (unsigned j = 0; j < a[i].size(); j++)
a[i][j].flow = a[a[i][j].dest][a[i][j].home].flow = (a[i][j].flow + a[a[i][j].dest][a[i][j].home].flow ) / 2;
}
void solve(int l, int r) {
if (l == r) return;
s = p[l], t = p[r];
int flow = 0;
ClearGraph();
while (bfs()) {
memset(curr, 0, sizeof(curr));
flow += dinic(s, INF);
}
b[s].push_back((treeedge) {t, flow});
b[t].push_back((treeedge) {s, flow});
int tl = l - 1, tr = r + 1;
for (int i = l; i <= r; i++)
if (dist[p[i]]) q[++tl] = p[i];
else q[--tr] = p[i];
for (int i = l; i <= r; i++)
p[i] = q[i];
solve(l, tl); solve(tr, r);
}
void work(int pos) {
visited[pos] = true; cnt++;
for (unsigned i = 0; i < b[pos].size(); i++)
if (!visited[b[pos][i].dest] && b[pos][i].index > value) work(b[pos][i].dest);
}
int main() {
int T; read(T);
while (T--) {
read(n), read(m);
for (int i = 1; i <= n; i++) {
a[i].clear();
b[i].clear();
}
for (int i = 1; i <= m; i++) {
int x, y, z;
read(x), read(y), read(z);
if (x != y) addedge(x, y, z);
}
for (int i = 1; i <= n; i++)
p[i] = i;
solve(1, n);
read(Q);
for (int i = 1; i <= Q; i++) {
memset(visited, false, sizeof(visited));
read(value);
int ans = n * (n - 1) / 2;
for (int j = 1; j <= n; j++) {
if (visited[j]) continue;
cnt = 0; work(j);
ans -= cnt * (cnt - 1) / 2;
}
printf("%d\n", ans);
}
printf("\n");
}
return 0;
}