HDU-4971-a simple brute force problem. (Max weight closure diagram)

Question: n (n <= 20) projects, M (M <= 50) technical issues, and profit (<= 1000) after a project is completed ), after a project is completed, the corresponding technical problems must be solved, and cost (<= 1000) is required to solve a technical problem. The technical problems are dependent on each other to maximize the benefits.

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4971

--> The project must solve the corresponding technical problems, and there are dependencies between the technical problems, corresponding to the closed diagram, and the maximum benefit corresponds to the maximum right .. As a result, the largest right closed chart, the smallest cut, and the largest stream playing ..

Graph creation:

1) Super source S = N + M, super sink t = n + m + 1

2) for each project I: S-> I (profit [I])

3) for each technical question I: I + N-> T (cost [I])

4) technical issues that must be addressed by project I j: I-> J + N (INF)

5) the technical problem J must solve first I: I + N-> J + N (INF) (here I think it should be: J + N-> I + N (INF). This understanding is correct. However, if you are not familiar with the sample, you can submit wa ..)

Then, dinic goes on stage ..

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using std::queue;using std::min;const int MAXN = 20 + 50 + 10;const int MAXM = 20 + 1000 + 2500 + 50 + 10;const int INF = 0x3f3f3f3f;int n, m, sum;int hed[MAXN];int cur[MAXN], h[MAXN];int ecnt;int S, T;struct EDGE{    int to;    int cap;    int flow;    int nxt;} edges[MAXM << 1];void Init(){    ecnt = 0;    memset(hed, -1, sizeof(hed));    sum = 0;}void AddEdge(int u, int v, int cap){    edges[ecnt].to = v;    edges[ecnt].cap = cap;    edges[ecnt].flow = 0;    edges[ecnt].nxt = hed[u];    hed[u] = ecnt++;    edges[ecnt].to = u;    edges[ecnt].cap = 0;    edges[ecnt].flow = 0;    edges[ecnt].nxt = hed[v];    hed[v] = ecnt++;}void Read(){    int profit, cost, pc, tp;    scanf("%d%d", &n, &m);    S = n + m;    T = n + m + 3;    for (int i = 0; i < n; ++i)    {        scanf("%d", &profit);        AddEdge(S, i, profit);        sum += profit;    }    for (int i = 0; i < m; ++i)    {        scanf("%d", &cost);        AddEdge(i + n, T, cost);    }    for (int i = 0; i < n; ++i)    {        scanf("%d", &pc);        for (int j = 0; j < pc; ++j)        {            scanf("%d", &tp);            AddEdge(i, tp + n, INF);        }    }    for (int i = 0; i < m; ++i)    {        for (int j = 0; j < m; ++j)        {            scanf("%d", &tp);            if (tp)            {                AddEdge(i + n, j + n, INF);            }        }    }}bool Bfs(){    memset(h, -1, sizeof(h));    queue<int> qu;    qu.push(S);    h[S] = 0;    while (!qu.empty())    {        int u = qu.front();        qu.pop();        for (int e = hed[u]; e != -1; e = edges[e].nxt)        {            int v = edges[e].to;            if (h[v] == -1 && edges[e].cap > edges[e].flow)            {                h[v] = h[u] + 1;                qu.push(v);            }        }    }    return h[T] != -1;}int Dfs(int u, int cap){    if (u == T || cap == 0) return cap;    int flow = 0, subFlow;    for (int e = cur[u]; e != -1; e = edges[e].nxt)    {        cur[u] = e;        int v = edges[e].to;        if (h[v] == h[u] + 1 && (subFlow = Dfs(v, min(cap, edges[e].cap - edges[e].flow))) > 0)        {            flow += subFlow;            edges[e].flow += subFlow;            edges[e ^ 1].flow -= subFlow;            cap -= subFlow;            if (cap == 0) break;        }    }    return flow;}int Dinic(){    int maxFlow = 0;    while (Bfs())    {        memcpy(cur, hed, sizeof(hed));        maxFlow += Dfs(S, INF);    }    return maxFlow;}int main(){    int t, kase = 0;    scanf("%d", &t);    while (t--)    {        Init();        Read();        printf("Case #%d: %d\n", ++kase, sum - Dinic());    }    return 0;}

HDU-4971-a simple brute force problem. (Max weight closure diagram)