글 읽기 - 골목길

dmon 6년 전 0

코드가 별로 볼만하지가 못한데.

그래도 어디가 잘못됐는지 한번만 봐주세요.

제법 실행은 됩니다.

Bellman-Ford 알고리즘 입니다.

근데 정답이 한명도 없는걸로 봐서

채점에 오류가 있는거 아닐까요?

// A C / C++ program for Bellman-Ford's single source 
// shortest path algorithm.

#include <climits>
#include <iostream>
#include <string>
using namespace std;
 
// a structure to represent a weighted edge in graph
struct Edge
{
		int src, dest, weight;
};
 
// a structure to represent a connected, directed and 
// weighted graph
struct Graph
{
    // V-> Number of vertices, E-> Number of edges
    int V, E;
 
    // graph is represented as an array of edges.
    struct Edge* edge;
};
 
// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E);
void printArr(string path[], int n);
void BellmanFord(struct Graph* graph, int src, int dest);
 
// Driver program to test above functions
int main()
{
    /* Let us create the graph given in above example */
    int V = 2;  // Number of vertices in graph	(2<=V<=100)
    int E = 1;  // Number of edges in graph			(1<=E<=20,000)
    
    cin >> V >> E;
//    if (V < 2) V = 2; if (V > 100) V = 100;
//    if (E < 1) E = 1; if (E > 20000) E = 20000;
    
    struct Graph* graph = createGraph(V, E);

    for ( int i = 0; i < E; i++)
    {
            cin >> graph->edge[i].src 
                >> graph->edge[i].dest 
                >> graph->edge[i].weight;						// (0 ≤ |w| ≤ 1,000)
		}
		
    BellmanFord(graph, 0, V-1);
    
    return 0;
}


// The main function that finds shortest distances from src to
// all other vertices using Bellman-Ford algorithm.  The function
// also detects negative weight cycle
void BellmanFord(struct Graph* graph, int src, int dest)
{
    int V = graph->V;
    int E = graph->E;
    int dist[V];
    string path[V];
    
    // Step 1: Initialize distances from src to all other vertices
    // as INFINITE
    for (int i = 0; i < V; i++)
    {
        dist[i]   = INT_MIN;
        path[i]		= "1";
    }
    dist[src] = 0;
    
    // Step 2: Relax all edges |V| - 1 times. A simple shortest 
    // path from src to any other vertex can have at-most |V| - 1 
    // edges
    for (int i = 1; i <= V-1; i++)
    {
        for (int j = 0; j < E; j++)
        {
            int u = graph->edge[j].src - 1;
            int v = graph->edge[j].dest - 1;
            int weight = graph->edge[j].weight;
            
            if (dist[u] != INT_MIN && dist[u] + weight > dist[v])
            {
                dist[v] = dist[u] + weight;
                path[v] = path[u] + " " + to_string(v+1);
            }
        }
    }
    
    // Step 3: check for negative-weight cycles.  The above step 
    // guarantees shortest distances if graph doesn't contain 
    // negative weight cycle.  If we get a shorter path, then there
    // is a cycle.
    for (int i = 0; i < E; i++)
    {
        int u = graph->edge[i].src - 1;
        int v = graph->edge[i].dest - 1;
        int weight = graph->edge[i].weight;
        
        if (dist[u] != INT_MIN && dist[u] + weight > dist[v]) {
	        	const string u1 = to_string(u + 1);
	        	const string v1 = to_string(v + 1);
	        	if (path[V-1].find(u1, 0)+1 && path[V-1].find(v1, 0)+1) {
		            cout << -1;
		            exit(EXIT_SUCCESS);
	          }
        }
    }
    
    printArr(path, V-1);
    
    return;
}

// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E)
{
    struct Graph* graph = (struct Graph*) malloc( sizeof(struct Graph) );
    graph->V = V;
    graph->E = E;
 
    graph->edge = (struct Edge*) malloc( graph->E * sizeof( struct Edge ) );
 
    return graph;
}

// A utility function used to print the solution
void printArr(string path[], int n)
{
		if (path[n] == "1")	{
				cout << -1;
		}
		else {
				cout << path[n];
		}
}

댓글을 작성하려면 로그인해야 합니다.

문제

문제

출처

ICPC

골목길

dmon 6년 전 0

Baekjoon Online Judge

채점 현황

문제

유저 대회

고등학교 대회

출처

대학교 대회

도움말