lightoj 1154 - Penguins 【拆点建图后枚举汇点 满流判可行解】

1154 - Penguins

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Time Limit: 4 second(s)Memory Limit: 32 MB

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

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case starts with the integer N (1 ≤ N ≤ 100) and a floating-point number D (0 ≤ D ≤ 10⁵), denoting the number of ice pieces and the maximum distance a penguin can jump. After that there will be N lines, each line containing x_i, y_i, n_i and m_i, 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 (-10000 ≤ x_i, y_i ≤ 10000, 0 ≤ n_i ≤ 10, 1 ≤ m_i ≤ 200).

Output

For each case of input, print the case number and a space-separated list of 0-based indices of the pieces on which all penguins can meet. If no such piece exists, output -1.

Sample Input

Output for 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

Case 1: 1 2 4

Case 2: -1

 

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SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

定义：浮冰承受限度——可以从该浮冰上跳出的企鹅的数目。

题意：有n块大浮冰，现在已经给出每个浮冰的位置坐标、上面企鹅的数目以及可以承受的限度。现在你已经知道企鹅能够跳跃的最大距离，问你所有企鹅能不能在某一块浮冰上相聚，若可以则输出所有可能相聚的浮冰编号（从0开始），反之输出-1。

设置超级源点S，超级汇点T

思路：S向每块浮冰建边，容量为上面的企鹅数目。下面对每块浮冰，拆点建图，给每块浮冰增加限制，然后对于可达的两块浮冰，相互建边。最后枚举汇点，判断是否满流。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#define MAXN 200+10#define MAXM 100000+10#define INF 0x3f3f3f3fusing namespace std;struct MAXFLOW{    struct Edge{        int from, to, cap, flow, next;    };    Edge edge[MAXM], Redge[MAXM];    int head[MAXN], edgenum;    int Rhead[MAXN];    int dist[MAXN], cur[MAXN];    bool vis[MAXN];    void init(){        edgenum = 0;        memset(head, -1, sizeof(head));    }    void addEdge(int u, int v, int w)    {        Edge E1 = {u, v, w, 0, head[u]};        edge[edgenum] = E1;        head[u] = edgenum++;        Edge E2 = {v, u, 0, 0, head[v]};        edge[edgenum] = E2;        head[v] = edgenum++;    }    bool BFS(int s, int t)    {        queue<int> Q;        memset(dist, -1, sizeof(dist));        memset(vis, false, sizeof(vis));        dist[s] = 0, vis[s] = true;        Q.push(s);        while(!Q.empty())        {            int u = Q.front();            Q.pop();            for(int i = head[u]; i != -1; i = edge[i].next)            {                Edge E = edge[i];                if(!vis[E.to] && E.cap > E.flow)                {                    dist[E.to] = dist[u] + 1;                    if(E.to == t) return true;                    vis[E.to] = true;                    Q.push(E.to);                }            }        }        return false;    }    int DFS(int x, int a, int t)    {        if(x == t || a == 0) return a;        int flow = 0, f;        for(int &i = cur[x]; i != -1; i = edge[i].next)        {            Edge &E = edge[i];            if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap-E.flow), t)) > 0)            {                edge[i].flow += f;                edge[i^1].flow -= f;                flow += f;                a -= f;                if(a == 0) break;            }        }        return flow;    }    int Maxflow(int s, int t)    {        int flow = 0;        while(BFS(s, t))        {            memcpy(cur, head, sizeof(head));            flow += DFS(s, INF, t);        }        return flow;    }};MAXFLOW dinic;int n;double limit;int S, T;double dis(double x1, double y1, double x2, double y2){    return sqrt((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2));}double x[110], y[110];int num[110], can[110];void getMap(){    dinic.init();    for(int i = 1; i <= n; i++)    {        if(num[i])            dinic.addEdge(S, i, num[i]);        dinic.addEdge(i, i+n, can[i]);        for(int j = i+1; j <= n; j++)            if(dis(x[i], y[i], x[j], y[j]) <= limit)                dinic.addEdge(i+n, j, INF), dinic.addEdge(j+n, i, INF);    }}int kcase = 1;void solve(){    scanf("%d%lf", &n, &limit);    int sumflow = 0;    for(int i = 1; i <= n; i++)        scanf("%lf%lf%d%d", &x[i], &y[i], &num[i], &can[i]), sumflow += num[i];    //memcpy(dinic.Rhead, dinic.head, sizeof(dinic.head));    //memcpy(dinic.Redge, dinic.edge, sizeof(dinic.edge));    int have = 0;    S = 0, T = 2*n+1;    printf("Case %d: ", kcase++);    for(int i = 1; i <= n; i++)    {        getMap();        dinic.addEdge(i, T, INF);        if(dinic.Maxflow(S, T) == sumflow)//满流        {            if(have)                printf(" ");            printf("%d", i-1);            have++;        }    }    if(have == 0)        printf("-1\n");    else        printf("\n");}int main(){    int t;    scanf("%d", &t);    while(t--){        solve();    }    return 0;}