UVA 12125 - March of the Penguins
题目链接
题意:给定一些冰块,每个冰块上有一些企鹅,每个冰块有一个可以跳出的次数限制,每个冰块位于一个坐标,现在每个企鹅跳跃力为d,问所有企鹅能否跳到一点上,如果可以输出所有落脚冰块,如果没有方案就打印-1
思路:最大流,拆点表示冰块次数限制,然后枚举落脚冰块建图跑最大流即可
代码:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXNODE = 105 * 2;
const int MAXEDGE = 200005;
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;
}
bool Maxflow(int s, int t, int tot) {
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 == tot;
}
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 = 105;
int t, n;
double d;
struct Point {
int x, y, n, m;
void read() {
scanf("%d%d%d%d", &x, &y, &n, &m);
}
} p[N];
double dis(Point a, Point b) {
int dx = a.x - b.x;
int dy = a.y - b.y;
return sqrt(dx * dx * 1.0 + dy * dy);
}
int tot;
bool solve(int u) {
gao.init(2 * n + 2);
for (int i = 1; i <= n; i++)
gao.add_Edge(0, i, p[i].n);
gao.add_Edge(u + n, 2 * n + 1, tot);
for (int i = 1; i <= n; i++) {
if (i == u) gao.add_Edge(i, i + n, tot);
else gao.add_Edge(i, i + n, p[i].m);
for (int j = i + 1; j <= n; j++) {
if (dis(p[i], p[j]) > d) continue;
gao.add_Edge(i + n, j, tot);
gao.add_Edge(j + n, i, tot);
}
}
return gao.Maxflow(0, 2 * n + 1, tot);
}
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%lf", &n, &d);
tot = 0;
for (int i = 1; i <= n; i++) {
p[i].read();
tot += p[i].n;
}
int bo = 0;
for (int i = 1; i <= n; i++) {
if (!solve(i)) continue;
if (bo) printf(" ");
else bo = 1;
printf("%d", i - 1);
}
if (bo == 0) printf("-1");
printf("\n");
}
return 0;
}