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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
#402173#1076. Flooding Fieldsstasio6#AC ✓479ms158900kbC++143.6kb2024-04-30 04:41:402024-04-30 04:41:41

Judging History

你现在查看的是最新测评结果

  • [2024-04-30 04:41:41]
  • 评测
  • 测评结果:AC
  • 用时:479ms
  • 内存:158900kb
  • [2024-04-30 04:41:40]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
#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()
#define PB push_back
#define FS first
#define SD second
typedef pair<int, int> pii;
typedef vector<int> vi;
template<class X, class Y> X cmx(X &a, Y b) { a = max<X>(a, b); }
#define ary(k) array<int, k>

void out(int k) {
    cout << k%2 << " ";
    k /= 2;
    cout << k/9 << " " << k/3%3 << " " << k%3 << "  ";
}

struct Dinic {
    struct Edge {
        int to, rev;
        ll c, oc;
        ll flow() { return max(oc - c, 0LL);}
    };
    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, ll rcap = 0) {
//        out(a); out(b); cout << "\n";
        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 {
                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;}
};

int height[102][102];
int flood[25];

int dirx[5] = {0, 0, 1, 0, -1};
int diry[5] = {0, 1, 0, -1, 0};

signed main() {
	cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit);
    while (true) {
        int n, m, h;
        cin >> n >> m >> h;
        if (n == 0 && m == 0 && h == 0) {
            break;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> height[i][j];
            }
        }
        int source = n*n*(h+1)*2, sink = n*n*(h+1)*2+1;
        Dinic dinic(sink+1);
        for (int i = 0; i < m; i++) {
            int x, y;
            cin >> x >> y;
            dinic.addEdge(source, (x*n+y)*2, 1);
        }
        flood[0] = -1;
        for (int k = 1; k <= h; k++) {
            cin >> flood[k];
        }
        for (int k = 0; k <= h; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    int cur = (k*n*n+n*i+j)*2;
                    if (flood[k] < height[i][j])
                        dinic.addEdge(cur, cur+1, 1);
                    if (k == h) {
                        dinic.addEdge(cur+1, sink, 1);
                        continue;
                    }
                    for (int hh = 0; hh < 5; hh++) {
                        int ni = i + dirx[hh], nj = j + diry[hh];
                        if (ni < 0 || ni >= n || nj < 0 || nj >= n)
                            continue;
                        int nast = ((k+1)*n*n+n*ni+nj)*2;
                        dinic.addEdge(cur+1, nast, 1);
                    }
                }
            }
        }
        cout << dinic.calc(source, sink) << "\n";
    }
}

详细

Test #1:

score: 100
Accepted
time: 479ms
memory: 158900kb

input:

3 1 3
2 1 2
3 1 3
2 4 2
1 1
0
2
1
2 4 2
0 1
1 2
0 0
0 1
1 1
1 0
1
0
50 50 24
5 20 21 3 2 9 22 15 19 22 18 14 19 14 15 4 22 22 7 12 7 2 2 0 25 0 7 6 4 3 2 17 1 12 7 12 1 19 12 3 23 5 5 10 22 15 17 3 13 4
10 21 7 12 10 22 23 18 17 5 17 18 2 14 12 13 13 19 19 18 8 11 3 0 24 0 6 1 9 12 14 13 20 22 5 13 ...

output:

1
1
1
2
2
2
0
92
99
71

result:

ok 10 lines