March of the Penguins

Description

Somewhere near the south pole, a number of penguins are standing on a number of ice floes. Being social animals, the penguins would like to get together, all on the same floe. The penguins do not want to get wet, so they have use their limited jump distance to get together by jumping from piece to piece. However, temperatures have been high lately, and the floes are showing cracks, and they get damaged further by the force needed to jump to another floe. Fortunately the penguins are real experts on cracking ice floes, and know exactly how many times a penguin can jump off each floe before it disintegrates and disappears. Landing on an ice floe does not damage it. You have to help the penguins find all floes where they can meet.

A sample layout of ice floes with 3 penguins on them.

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

  • One line with the integer N (1 ≤ N ≤ 100) and a floating-point number D (0 ≤ D ≤ 100 000), denoting the number of ice pieces and the maximum distance a penguin can jump.

  • N lines, each line containing xi, yi, ni and mi, denoting for each ice piece its X and Y coordinate, the number of penguins on it and the maximum number of times a penguin can jump off this piece before it disappears (−10 000 ≤ xi, yi ≤ 10 000, 0 ≤ ni ≤ 10, 1 ≤ mi ≤ 200).

Output

Per testcase:

  • One line containing a space-separated list of 0-based indices of the pieces on which all penguins can meet. If no such piece exists, output a line with the single number −1.

Sample Input

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

Sample Output

1 2 4
-1
题意：海上有些冰块，冰块上有些企鹅，现在所有企鹅要聚在一起，企鹅都有一样的跳远距离。一个企鹅登陆冰块时，冰块不会受损，但一个企鹅跳离冰块，冰块会受损，受损度为1。给出每个冰块的坐标、能承受的受损度，还有企鹅的跳远距离。问所有企鹅最后所在的位置。
分析：每个冰块拆成两点u,u'，容量为它的受损度x，表示可以让x只企鹅跳出去。源到每个点的容量为每个冰块开始的企鹅数量，接下来如果u,v两点的距离小于等于企鹅的跳远距离，则u'到v容量为无穷，表示v可以让无数只企鹅登陆（因为登陆不受损）。然后枚举汇点。最大流......
代码：
#include <stdio.h>
#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;
const int N=20010;
const int MAXE=200010;
const int inf = 1<<29;
struct Edge
{
    int v, next, w;
}edge[MAXE*8];
int head[N], pre[N], cur[N], level[N], gap[N];
int cnt;
void addedge(int u, int v, int w)
{
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
    edge[cnt].v = u;
    edge[cnt].w = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}
int sap(int s, int t, int n)
{
    int flow = 0, aug = inf;
    int u;
    bool flag;
    for(int i = 0; i <= n; i++)
    {
        cur[i] = head[i];
        gap[i] = level[i] = 0;
    }
    gap[s] = n;
    u = pre[s] = s;
    while (level[s] < n)
    {
        flag = 0;
        for (int &j = cur[u]; j != -1; j = edge[j].next)  
        {
            int v = edge[j].v;
            if (edge[j].w > 0 && level[u] == level[v] + 1)
            {
                flag = 1;
                if (edge[j].w < aug) aug = edge[j].w;
                pre[v] = u;
                u = v;
                if (u == t)
                {
                    flow += aug;
                    while (u != s)
                    {
                        u = pre[u];
                        edge[cur[u]].w -= aug;
                        edge[cur[u]^1].w += aug;
                    }
                    aug = inf;
                }
                break;
            }
        }
        if(flag)
            continue;
        int mindis = n;
        for(int j = head[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].v;
            if(edge[j].w > 0 && level[v] < mindis)
            {
                mindis = level[v];
                cur[u] = j;
            }
        }
        if((--gap[level[u]]) == 0)
            break;
        gap[level[u]=mindis+1]++;
        u = pre[u];
    }
    return flow;
}
struct P
{
    int x, y, cap, time; 
}floes[110];
int redis(P& a, P& b)
{
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
void build (int s, int t, int ver, int n, double maxdis)
{
    int i, j;
    cnt = 0;
    for (i = 0; i <= ver; i++)
    {
        head[i] = -1;
    }
    for (i = 1; i <= n; i++)
    {
        addedge(s, i, floes[i].cap);
        addedge(i, i + n, floes[i].time);
    }
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (maxdis * maxdis > (double)redis(floes[i], floes[j]))
            {
                addedge(i + n, j, inf);
            }
        }
    }
}
int main()
{
    int ca, n, sum, i, s, t, ver;
    double maxdis;
    scanf("%d", &ca);
    while (ca--)
    {
        scanf("%d%lf", &n, &maxdis);
        int s = 0, t, ver = n + n + 1;
        for (i = 1, sum = 0; i <= n; i++)
        {
            scanf("%d%d%d%d", &floes[i].x, &floes[i].y, &floes[i].cap, &floes[i].time);
            sum += floes[i].cap;
        }
        int sign;
        for (i = 1, sign = 0; i <= n; i++)
        {
            t = i;
            build (s, t, ver, n, maxdis);
            int ans = sap(s, t, ver);        
            if (ans == sum)
            {
                sign = 1;
                printf("%d ", t - 1);
            }
        }
        if (!sign) printf("-1");
        printf("\n");
        
    }
    return 0;
}