Submission #68605

#	Time	Username	Problem	Language	Result	Execution time	Memory
68605		renatsj	Min-max tree (BOI18_minmaxtree)	C++14	Compilation error	0 ms	0 KiB

minmaxtree

// C++ program for implementation of Ford Fulkerson algorithm
#include "homecoming.h"
#include <iostream>
#include <limits.h>
#include <string.h>
#include <queue>
using namespace std;

// Number of vertices in given graph
#define V 100

/* Returns true if there is a path from source 's' to sink 't' in
  residual graph. Also fills parent[] to store the path */
bool bfs(int rGraph[V][V], int s, int t, int parent[])
{
    // Create a visited array and mark all vertices as not visited
    bool visited[V];
    memset(visited, 0, sizeof(visited));

    // Create a queue, enqueue source vertex and mark source vertex
    // as visited
    queue <int> q;
    q.push(s);
    visited[s] = true;
    parent[s] = -1;

    // Standard BFS Loop
    while (!q.empty())
    {
        int u = q.front();
        q.pop();

        for (int v=0; v<V; v++)
        {
            if (visited[v]==false && rGraph[u][v] > 0)
            {
                q.push(v);
                parent[v] = u;
                visited[v] = true;
            }
        }
    }

    // If we reached sink in BFS starting from source, then return
    // true, else false
    return (visited[t] == true);
}

// Returns the maximum flow from s to t in the given graph
int fordFulkerson(long long graph[V][V], int s, int t)
{
    int u, v;

    // Create a residual graph and fill the residual graph with
    // given capacities in the original graph as residual capacities
    // in residual graph
    int rGraph[V][V]; // Residual graph where rGraph[i][j] indicates
                     // residual capacity of edge from i to j (if there
                     // is an edge. If rGraph[i][j] is 0, then there is not)
    for (u = 0; u < V; u++)
        for (v = 0; v < V; v++)
             rGraph[u][v] = graph[u][v];

    int parent[V];  // This array is filled by BFS and to store path

    int max_flow = 0;  // There is no flow initially

    // Augment the flow while tere is path from source to sink
    while (bfs(rGraph, s, t, parent))
    {
        // Find minimum residual capacity of the edges along the
        // path filled by BFS. Or we can say find the maximum flow
        // through the path found.
        int path_flow = INT_MAX;
        for (v=t; v!=s; v=parent[v])
        {
            u = parent[v];
            path_flow = min(path_flow, rGraph[u][v]);
        }

        // update residual capacities of the edges and reverse edges
        // along the path
        for (v=t; v != s; v=parent[v])
        {
            u = parent[v];
            rGraph[u][v] -= path_flow;
            rGraph[v][u] += path_flow;
        }

        // Add path flow to overall flow
        max_flow += path_flow;
    }

    // Return the overall flow
    return max_flow;
}
long long i,j;
long long solve(int N, int K, int *A, int *B)
{
    long long rez=0;
    long long graph[V][V];
    i=0;
    while (i<V)
    {
        j=0;
        while (j<V)
        {
            graph[i][j]=0;
            j++;
        }
        i++;
    }
    i=1;
    while (i<=N)
    {
        rez+=A[i-1];
        rez-=B[i-1];
        graph[0][i]=A[i-1];
        i++;
    }
    i=1;
    while (i<=N)
    {
        j=0;
        while (j<K&&j<N)
        {
            graph[i][(i+j-1)%N+1+N]=1000000000000000000;
            j++;
        }
        i++;
    }
    i=N+1;
    while (i<=2*N)
    {
        graph[i][2*N+1]=B[i-N-1];
        i++;
    }
    return rez-fordFulkerson(graph, 0, 2*N+1);
}
// Driver program to test above functions
/*int main()
{
    // Let us create a graph shown in the above example
    /*int graph[V][V] = { {0, 16, 13, 0, 0, 0},
                        {0, 0, 10, 12, 0, 0},
                        {0, 4, 0, 0, 14, 0},
                        {0, 0, 9, 0, 0, 20},
                        {0, 0, 0, 7, 0, 4},
                        {0, 0, 0, 0, 0, 0}
                      };

    int a[5],b[5];
    a[0]=40;
    a[1]=80;
    a[2]=100;
    b[0]=140;
    b[1]=0;
    b[2]=20;
    cout << i << "\n\n\n";
    cout << solve(3,2,a,b);
    //cout << "The maximum possible flow is ";

    return 0;
}*/

Compilation message (stderr)

minmaxtree.cpp:2:10: fatal error: homecoming.h: No such file or directory
 #include "homecoming.h"
          ^~~~~~~~~~~~~~
compilation terminated.