題意: 在X,Y坐標系中有N(N<=100)個冰塊...有些冰塊上有若幹只企鵝..每只企鵝一次最多跳M距離..一個冰塊在有Mi個企鵝離開..就會消失..問有哪些冰塊可以作為集合點..就是所有企鵝都能成功到這個冰塊上來.

這個題建圖比較有意思。

把每塊冰分成兩個點i和i+n. i表示進入i冰塊的點(可以有無數企鵝過來,所以從別的冰到i有邊,容量為INF) i+n表示從i冰塊出去的點(最多只能有Mi企鵝從這跳出去,所以從i到i+n有邊,且容量為Mi)
從源點S到i有邊(S, i, i點初始企鵝數).
從i到i+n有邊(i, i+n, Mi). 表示第i塊冰最多只有Mi個企鵝能跳走.
因為i+n表示的是第i個跳走的點,所以如果冰塊i和j之間的距離<=企鵝能跳躍的距離M,有邊(i+n, j, INF)

假設我們當前枚舉第x塊冰塊作為集合點,那麽(x分成x和x+n兩個點)x點就是匯點(不是x+n點哦),我們只要計算到x點的流量是否==企鵝總數即可.

#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#define maxn 1000
#define INF 0x3f3f3f3f

using namespace std;
struct e {
    int from, to, w, next;
}edge[1000000];

struct p {
    int x, y;
}pos[10000];
int ss, tt;
int n;
double d;
int cur[maxn];
int cont;
int head[maxn];
int map[maxn][maxn];
int dis[maxn][maxn];
int divv[maxn];
int x[maxn], y[maxn], num[maxn], m[maxn];


bool check(p a, p b) {
    double dis = sqrt(pow(fabs(a.x - b.x), 2) + pow(fabs(a.y - b.y), 2));
    if (dis <= d)
        return true;
    else
        return false;

}

void add(int u, int v, int w) {
    edge[cont].from = u;
    edge[cont].to = v;
    edge[cont].w = w;
    edge[cont].next = head[u];
    head[u] = cont++;
}

void makediv() {
    memset(divv, 0, sizeof(divv));
    divv[ss] = 1;
    queue<int> Q;
    Q.push(ss);
    while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
        for (int i = head[u]; i != -1; i = edge[i].next) {
            int w = edge[i].w;
            int v = edge[i].to;
            if (divv[v] == 0 && w) {
                divv[v] = divv[u] + 1;
                Q.push(v);
            }
        }

    }

}

int DFS(int u, int maxflow, int tt) {
    if (u == tt)
        return maxflow;
    int ret = 0;
    for (int &i = cur[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        int w = edge[i].w;
        if (divv[v] == divv[u] + 1 && w) {
            int f = DFS(v, min(maxflow - ret, w), tt);
            edge[i].w -= f;
            edge[i ^ 1].w += f;
            ret += f;
            if (ret == maxflow)
                return ret;

        }
    }
    return ret;
}

int Dinic() {
    int ans = 0;
    while (1) {
        makediv();
        if (divv[tt] == 0)
            break;
        memcpy(cur, head, sizeof(head));
        ans += DFS(ss, INF, tt);
    }
    return ans;
}

void Build() {
    for (int i = 1; i <= n; i++) {
        add(ss, i, num[i]);
        add(i, ss, 0);
        add(i, i + n, m[i]);
        add(i + n, i, 0);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == j)
                continue;
            if (check(pos[i], pos[j])) {
                add(i + n, j, INF);
                add(j, i + n, 0);
            }
        }
    }
}

int main(void) {
    int t;
    int p_n;
    scanf("%d", &t);
    while (t--) {
        bool flag = false;
        ss = 0;
        p_n = 0;
        scanf("%d%lf", &n, &d);
        memset(head, -1, sizeof(head));
        for (int i = 1; i <= n; i++) {
            scanf("%d%d%d%d", &x[i], &y[i], &num[i], &m[i]);
            if (num)
                p_n += num[i];
            pos[i].x = x[i];
            pos[i].y = y[i];
        }


        vector<int> V;
        for (int i = 1; i <= n; i++) {
            tt = i;
            cont = 0;
            memset(head, -1, sizeof(head));
            Build();
            if (Dinic() == p_n) {
                flag = true;
                V.push_back(i - 1);
            }
        }
        if (!flag)
            printf("-1\n");
        else {
            for (int i = 0; i < (int)V.size(); i++)
                if (i != V.size() - 1)
                    printf("%d ", V[i]);
                else
                    printf("%d\n", V[i]);
        }

    }
    
    return 0;
}
 

POJ 3498【最大流+拆點建圖】