2019 China Collegiate Programming Contest Qinhuangdao Onsite E Escape

题目描述

http://codeforces.com/gym/102361/problem/E

简要题意:

Solution

大概就是每个点只允许一个机器人拐弯一次

注意到要实现拐弯只需要将一个点拆成上下点和左右点即可

跑最大流判断是否所有机器人都能到达终点即可

建图:

$(i,j,0)$ 表示上下点,$(i,j,1)$ 表示左右点

$s$ 连 $(i,j,0)$,容量为 $1$

$(i,j,0)$ 连 四周的上下点,容量为 $1$

$(i,j,0)$ 连 $(i,j,1)$,容量为 $1$

$(i,j,1)$ 连 $(i,j,0)$,容量为 $1$

$(i,j,1)$ 连 四周的左右点,容量为 $1$

$(i,j,0)$ 连 $t$,容量为 $1$

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#include <iostream>
#include <cstring>
#include <queue>
#define maxn 20010
#define maxm 110
#define INF 1000000000
using namespace std;

int dx[] = { 0, 0, -1, 1 };
int dy[] = { 1, -1, 0, 0 };

int n, m, a, b;

int g[maxm][maxm];

inline int id(int x, int y) { return (x - 1) * m + y; }

char c[maxm];

struct Edge {
int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn], s, t;
bool bfs() {
fill(dep, dep + maxn, 0); dep[s] = 1;
queue<int> Q; Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (w > 0 && !dep[v]) {
dep[v] = dep[u] + 1;
if (v == t) return 1; Q.push(v);
}
}
}
return 0;
}

int dfs(int u, int _w) {
if (u == t || !_w) return _w;
int s = 0;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (dep[v] == dep[u] + 1 && w > 0) {
int di = dfs(v, min(_w - s, w));
e[i].w -= di; e[i ^ 1].w += di;
s += di; if (s == _w) break;
}
}
if (s < _w) dep[u] = 0;
return s;
}

int dinic() {
int mf = 0;
while (bfs()) mf += dfs(s, INF);
return mf;
}

void work() { fill(head, head + maxn, -1); c1 = 0;
cin >> n >> m >> a >> b; s = 0; t = 2 * n * m + 1;
for (int i = 1; i <= n; ++i) {
cin >> c + 1;
for (int j = 1; j <= m; ++j) g[i][j] = c[j] - '0';
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) {
if (g[i][j]) continue;
Add_edge(id(i, j), id(i, j) + n * m, 1);
Add_edge(id(i, j) + n * m, id(i, j), 1);
for (int k = 0; k < 4; ++k) {
int x = i + dx[k], y = j + dy[k];
if (x < 1 || x > n || y < 1 || y > m || g[x][y]) continue;
if (k >= 2) Add_edge(id(i, j), id(x, y), 1);
else Add_edge(id(i, j) + n * m, id(x, y) + n * m, 1);
}
}
for (int i = 1; i <= a; ++i) {
int x; cin >> x;
Add_edge(s, id(1, x), 1);
}
for (int i = 1; i <= b; ++i) {
int x; cin >> x;
Add_edge(id(n, x), t, 1);
}
cout << (dinic() == a ? "Yes" : "No") << "\n";
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

int T; cin >> T;
while (T--) work();
return 0;
}