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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#155478 | #6870. Coin | neko_nyaa | AC ✓ | 217ms | 6640kb | C++23 | 2.8kb | 2023-09-01 17:51:59 | 2023-09-01 17:51:59 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
struct Dinic {
struct Edge {
int to, rev;
ll c, oc;
ll flow() { return max(oc - c, 0LL); } // if you need flows
};
vi lvl, ptr, q;
vector<vector<Edge>> adj;
Dinic(int n) : lvl(n), ptr(n), q(n), adj(n) {}
void addEdge(int a, int b, ll c, int rcap = 0) {
adj[a].push_back({b, sz(adj[b]), c, c});
adj[b].push_back({a, sz(adj[a]) - 1, rcap, rcap});
}
ll dfs(int v, int t, ll f) {
if (v == t || !f) return f;
for (int& i = ptr[v]; i < sz(adj[v]); i++) {
Edge& e = adj[v][i];
if (lvl[e.to] == lvl[v] + 1)
if (ll p = dfs(e.to, t, min(f, e.c))) {
e.c -= p, adj[e.to][e.rev].c += p;
return p;
}
}
return 0;
}
ll calc(int s, int t) {
ll flow = 0; q[0] = s;
rep(L,0,31) do { // 'int L=30' maybe faster for random data
lvl = ptr = vi(sz(q));
int qi = 0, qe = lvl[s] = 1;
while (qi < qe && !lvl[t]) {
int v = q[qi++];
for (Edge e : adj[v])
if (!lvl[e.to] && e.c >> (30 - L))
q[qe++] = e.to, lvl[e.to] = lvl[v] + 1;
}
while (ll p = dfs(s, t, LLONG_MAX)) flow += p;
} while (lvl[t]);
return flow;
}
bool leftOfMinCut(int a) { return lvl[a] != 0; }
};
void solve() {
int n, m, k; cin >> n >> m >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
vector<pair<int, int>> edges;
vector<vector<int>> needs(n);
for (int i = 0; i < n; i++) {
needs[i].push_back(0);
}
for (int i = 0; i < m; i++) {
int x, y; cin >> x >> y; x--, y--;
edges.emplace_back(x, y);
needs[x].push_back(i+1);
needs[y].push_back(i+1);
}
for (int i = 0; i < n; i++) {
needs[i].push_back(m+1);
}
vector<int> fr(n);
for (int i = 0; i < k; i++) {
int x; cin >> x; x--; fr[x] = 1;
}
int sz = 2 + 2*n + 4*m;
Dinic dc(sz);
map<pair<int, int>, int> id;
int source = 0, sink = sz-1;
int petr = 1; // source = 1, sink = sz-1
for (int i = 0; i < n; i++) {
for (int j = 0; j < needs[i].size(); j++) {
id[{i, needs[i][j]}] = petr;
petr++;
}
dc.addEdge(source, id[{i, needs[i][0]}], 1);
if (fr[i]) dc.addEdge(id[{i, needs[i].back()}], sink, a[i]);
for (int j = 1; j < needs[i].size(); j++) {
dc.addEdge(id[{i, needs[i][j-1]}], id[{i, needs[i][j]}], a[i]);
}
}
for (int i = 1; i <= m; i++) {
auto [x, y] = edges[i-1];
dc.addEdge(id[{x, i}], id[{y, i}], 1);
dc.addEdge(id[{y, i}], id[{x, i}], 1);
}
cout << dc.calc(source, sink) << '\n';
}
signed main() {
ios::sync_with_stdio(0); cin.tie(0);
int t; cin >> t;
while (t--) {
solve();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 217ms
memory: 6640kb
input:
10 2621 3000 873 542 5 3 3 1 2 8 2 2 4 1 1 2 5 3 1 5 1 2 4 3 1 1 6 3 3 3 1 1 1 1 5 6 2 4 2 2 2 1 5 2 1 3 3 2 2 1 3 1 1 4 3 3 2 4 2 2 1 1 3 1 1 1 2 2 1 2 3 3 2 1 1 5 1 3 3 1 1 3 2 2 2 3 4 4 4 3 1 1 1 1 2 1 1 1 1 1 1 4 2 2 2 2 1 6 2 3 1 3 2 3 1 3 2 1 2 1 3 1 2 1 6 1 3 1 3 2 1 2 2 2 2 4 3 1 3 1 4 1 2 1...
output:
1584 1441 1691 1673 1444 1481 1573 1399 1839 1788
result:
ok 10 lines