[网络流24题]P2754 [CTSC1999]家园 / 星际转移
最大流,边增广边算,分层解决
https://www.luogu.com.cn/problem/P2754
一开始我以为是分层费用流,但是很容易发现这样不知道要分多少层,这个层数其实是我们要求的答案。
因此不难想到一天一天建,当前最大流=需要运输的人数时即可停止
点击查看代码
#include
#define endl '\n'
#define IOS \
ios::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0)
#define P pair
typedef long long ll;
using namespace std;
const int N = 2 * 20 * 50 * 2 + 5;
const int M = N * N;
const int INF = 0x3f3f3f3f;
struct edge {
int v, w, to;
} e[M * 2];
int pre[N], cnt_edge, dep[N];
int S, T, z, head[N], sum;
int n, m, q[N], cur[N];
void add(int u, int v, int w) {
// cout << "add: " << u << " " << v << endl;
e[cnt_edge] = {v, w, head[u]};
head[u] = cnt_edge++;
e[cnt_edge] = {u, 0, head[v]};
head[v] = cnt_edge++;
}
bool bfs() {
for (int i = 0; i <= T; i++) dep[i] = 0;
dep[S] = 1;
int l = 0, r = 1;
q[r] = S;
while (l < r) {
int u = q[++l];
for (int i = head[u]; i != -1; i = e[i].to) {
int v = e[i].v;
if (!dep[v] && e[i].w) dep[v] = dep[u] + 1, q[++r] = v;
}
}
return dep[T];
}
int dfs(int u, int mi) {
int res = 0;
if (mi == 0 || u == T) return mi;
for (int &i = cur[u]; i != -1; i = e[i].to) {
int v = e[i].v;
if (dep[u] + 1 == dep[v] && e[i].w) {
int minn = dfs(v, min(mi - res, e[i].w));
e[i].w -= minn;
e[i ^ 1].w += minn;
res += minn;
if (res == mi) return res;
}
}
if (res == 0) dep[u] = 0;
return res;
}
int dinic() {
ll res = 0;
while (bfs()) {
memcpy(cur, head, sizeof(head));
// cout< s[25];
int id(int day, int x) {
if ((day - 1) * n + x >= S) flag = 1;
return min((day - 1) * n + x, T);
}
int main() {
memset(head, -1, sizeof head);
S = 4000 + 1, T = S + 1;
cin >> n >> m >> num;
for (int i = 1, tmp, o; i <= m; i++) {
cin >> cap[i] >> tmp;
for (int j = 0; j < tmp; j++) {
cin >> o;
s[i].push_back(o);
}
}
int ans = 0, now = 0;
while (now < num) {
if (flag) {
cout << 0 << endl;
return 0;
}
ans++;
// cout << ans << " " << now << " \n";
for (int i = 1; i <= m; i++) {
int sz = s[i].size();
int v = s[i][ans % sz], u = s[i][(ans - 1 + sz) % sz];
// cout << "a: " << u << " " << v << endl;
if(u == 0 && v == -1){
now += cap[i];
continue;
}
if (u == 0) {
add(S, id(ans + 1, v), cap[i]);
continue;
}
if (u == -1) continue;
if (v == 0) continue;
if (v == -1) {
add(id(ans, u), T, cap[i]);
continue;
}
if (ans == 1 && u != S) continue;
add(id(ans, u), id(ans + 1, v), cap[i]);
}
for (int i = 1; i <= n; i++) {
add(id(ans, i), id(ans + 1, i), INF);
}
// puts("----------");
now += dinic();
}
if (flag) {
cout << 0 << endl;
return 0;
}
cout << ans << endl;
return 0;
}