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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 |
---|---|---|---|---|---|---|---|---|---|
#651059 | #4775. Pool construction | SGColin | AC ✓ | 21ms | 4304kb | C++20 | 3.1kb | 2024-10-18 17:20:02 | 2024-10-18 17:20:02 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
inline int rd() {
int x = 0;
bool f = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) f |= (c == '-');
for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return f ? -x : x;
}
#define eb emplace_back
#define all(s) (s).begin(), (s).end()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define per(i, a, b) for (int i = (a); i >= (b); --i)
inline bool getmin(int &a, int b) {return (a > b ? (a = b, true) : false);}
inline bool getmax(int &a, int b) {return (a < b ? (a = b, true) : false);}
// F is the type of flow
template<const int V, const int E, class F, const F flowInf>
struct MF {
int tot = 1, S, T, hd[V], cur[V], dis[V];
struct edge{int to, nxt; F cap;} e[E << 1];
void clear() {tot = 1; memset(hd, 0, sizeof(hd));}
void add(int u, int v, F w) {
e[++tot].nxt = hd[u], hd[u] = tot, e[tot].to = v, e[tot].cap = w;
e[++tot].nxt = hd[v], hd[v] = tot, e[tot].to = u, e[tot].cap = 0;
}
inline bool bfs() {
static int q[V], qhd, qtl;
memcpy(cur, hd, sizeof(hd));
memset(dis, -1, sizeof(dis));
q[qhd = qtl = 1] = S; dis[S] = 0;
while (qhd <= qtl) {
int u = q[qhd++];
for (int i = hd[u], v; i; i = e[i].nxt)
if (dis[v = e[i].to] == -1 && e[i].cap != 0) {
dis[v] = dis[u] + 1; q[++qtl] = v;
}
}
return dis[T] != -1;
}
F dfs(int u, F rem) {
if (u == T) return rem;
F flow = 0;
for (int i = cur[u], v; i && rem; i = e[i].nxt) {
cur[u] = i; v = e[i].to;
F nw = min(rem, e[i].cap);
if (nw != 0 && dis[v] == dis[u] + 1) {
int ret = dfs(v, nw);
flow += ret; rem -= ret;
e[i].cap -= ret; e[i ^ 1].cap += ret;
}
}
if (flow == 0) dis[u] = -1;
return flow;
}
F dinic(int source, int sink) {
S = source; T = sink; F flow = 0;
while (bfs()) flow += dfs(S, flowInf);
return flow;
}
};
const ll inf = 1000000000000000000ll;
MF<2510, 1000000, ll, inf> G;
const int N = 51;
bool ty[N][N];
int id[N][N];
inline void work() {
G.clear(); G.S = 2501; G.T = 2502;
int h = rd(), w = rd(), d = rd(), f = rd(), b = rd();
rep(i, 1, w) rep(j, 1, h) {
id[i][j] = (i - 1) * h + j;
char c = getchar();
while (c != '.' && c != '#') c = getchar();
ty[i][j] = (c == '#');
G.add(G.S, id[i][j], (ty[i][j] ? 0 : f));
if (i != 1 && i != w && j != 1 && j != h) G.add(id[i][j], G.T, (ty[i][j] ? d : 0));
else G.add(id[i][j], G.T, inf);
if (i > 1) {G.add(id[i][j], id[i - 1][j], b); G.add(id[i - 1][j], id[i][j], b);}
if (j > 1) {G.add(id[i][j], id[i][j - 1], b); G.add(id[i][j - 1], id[i][j], b);}
}
printf("%lld\n", G.dinic(G.S, G.T));
}
int main() {
per(t, rd(), 1) work();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3916kb
input:
3 3 3 5 5 1 #.# #.# ### 5 4 1 8 1 #..## ##.## #.#.# ##### 2 2 27 11 11 #. .#
output:
9 27 22
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 21ms
memory: 4304kb
input:
56 3 3 5 5 1 #.# #.# ### 5 4 1 8 1 #..## ##.## #.#.# ##### 2 2 1 1 1 ## ## 2 2 1 10000 1 .. .. 5 4 20 41 9 ##### ##.## #.#.# ##### 5 4 20 41 10 ##### ##.## #.#.# ##### 5 4 20 41 11 ##### ##.## #.#.# ##### 5 4 20 39 10 ##### ##.## #.#.# ##### 3 3 9760 9015 711 .#. #.# ### 5 5 7415 7931 2080 ..... #.....
output:
9 27 0 40000 108 120 123 117 20874 100110 112364 203900 271440 462119 490330 1746528 1067774 1055196 2609818 2094932 5199902 13978 73960 99018 262976 224632 78984 167795 392774 649054 1232290 135876 318982 413042 1479538 1680354 349557 540100 2101110 335884 2245998 170698 780013 1804351 2998519 3661...
result:
ok 56 lines