题意
T组测试数据,每组先输入N,d,表示有N块浮冰。然后输入N行表式N块浮冰的信息编号为0~(N-1),(x[i],y[i],n[i],m[i])表示第 i 块浮冰在(x[i],y[i])位置,上面有n[i]只企鹅,m[i]表示这块浮冰最多允许跳跃m[i]只企鹅从这里离开。每块浮冰都可以承载无穷的企鹅,问哪几块浮冰可以使全部企鹅都可以到达,输出这几块浮冰的编号,否则输出-1.
思路
其实就是每个点有一个结点容量,这类问题,我们把每个点拆做两个点,左边管进,右边管出,左到右容量为结点容量。建图如下:
左边的点编号为[0,N),右边的点编号为[N,N2),源点为N2。
①左点跟右点建边,容量为该点容量
②超级源点与有企鹅的点的左点建边,容量为企鹅数量。
枚举每块浮冰作为汇点,记得汇点是左点,而不是右点。
每次要把所有边的流量清0,或者重新构建一次图
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, cap, flow; //起点,终点,容量,流量
Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct Dinic
{
int n, m, s, t; //结点数,边数(包括反向弧),源点s,汇点t
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表,G[i][j]表示结点i的第j条边在edges数组中的序号
int d[MAXN]; //从起点到i的距离(层数差)
int cur[MAXN]; //当前弧下标
bool vis[MAXN]; //BFS分层使用
void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS()//构造分层网络
{
memset(vis, 0, sizeof(vis));
queue<int> Q;
d[s] = 0;
vis[s] = true;
Q.push(s);
while (!Q.empty())
{
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int a)//沿阻塞流增广
{
if (x == t || a == 0) return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)//从上次考虑的弧
{
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增广
{
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
int MaxFlow(int s, int t)
{
this->s = s; this->t = t;
int flow = 0;
while (BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
}gao;
const int maxn = 105;
const double eps = 1e-7;
int x[MAXN], y[MAXN], n[MAXN], m[MAXN];
double Distance(int i, int j)
{
return hypot(fabs(x[i]-x[j]), fabs(y[i]-y[j]));
}
int main()
{
int T, CASE = 1; scanf("%d", &T);
while (T--)
{
int N; double d; scanf("%d%lf", &N, &d);
int s = N*2;
gao.init(s);
int sum = 0;
for (int i = 0; i < N; i++)
{
scanf("%d%d%d%d", &x[i], &y[i], &n[i], &m[i]);
gao.AddEdge(s, i, n[i]);
gao.AddEdge(i, i+N, m[i]);
sum += n[i];
}
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
if (Distance(i, j)-d < eps)
{
gao.AddEdge(i+N, j, INF);
}
}
}
vector<int> ans;
for (int i = 0; i < N; i++)
{
//每次要把所有边的流量清0,或者重新构建一次图
for (int i = 0; i < gao.m; i++) gao.edges[i].flow = 0;
if (gao.MaxFlow(s, i) == sum) ans.push_back(i);
}
if (!ans.size()) printf("-1\n");
else
{
for (int i = 0; i < ans.size(); i++)
printf("%d%c", ans[i], i==ans.size()-1?'\n':' ');
}
}
return 0;
}
/*
2
5 3.5
1 1 1 1
2 3 0 1
3 5 1 1
5 1 1 1
5 4 0 1
3 1.1
-1 0 5 10
0 0 3 9
2 0 1 1
*/