HDU 2732 Leapin' Lizards
题意:有一些蜥蜴在一个迷宫里面,有一个跳跃力表示能跳到多远的柱子,然后每根柱子最多被跳一定次数,求这些蜥蜴还有多少是不管怎样都逃不出来的。
思路:把柱子拆点建图跑最大流就可以,还是挺明显的
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXNODE = 805;
const int MAXEDGE = 500005;
typedef int Type;
const Type INF = 0x3f3f3f3f;
struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};
struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;
void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}
bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}
Type dfs(int u, Type a) {
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
}
void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;
const int N = 25;
int T, n, m;
double d;
char str[N];
int main() {
int cas = 0;
scanf("%d", &T);
while (T--) {
scanf("%d%lf", &n, &d);
gao.init(n * 20 * 2 + 2);
int s = n * 20 * 2, t = n * 20 * 2 + 1;
for (int i = 0; i < n; i++) {
scanf("%s", str);
m = strlen(str);
for (int j = 0; j < m; j++) {
if (str[j] != '0')
gao.add_Edge(i * m + j, i * m + j + n * m, str[j] - '0');
if (i - d < 0 || i + d >= n || j - d < 0 || j + d >= m)
gao.add_Edge(i * m + j + n * m, t, INF);
}
}
int tot = 0;
for (int i = 0; i < n; i++) {
scanf("%s", str);
m = strlen(str);
for (int j = 0; j < m; j++) {
if (str[j] == 'L') {
tot++;
gao.add_Edge(s, i * m + j, 1);
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
for (int x = 0; x < n; x++) {
for (int y = 0; y < m; y++) {
int dx = i - x;
int dy = j - y;
double dis = sqrt(dx * dx * 1.0 + dy * dy);
if (dis > d) continue;
gao.add_Edge(i * m + j + n * m, x * m + y, INF);
}
}
}
}
printf("Case #%d: ", ++cas);
int ans = tot - gao.Maxflow(s, t);
if (ans == 0) printf("no ");
else printf("%d ", ans);
if (ans <= 1) printf("lizard was ");
else printf("lizards were ");
printf("left behind.\n");
}
return 0;
}