0-1bfs

有用的链接 0-1bfs
求某一个点到其它点的最短距离,类似于dijkstra算法,不过这个图或者树它的每条边的权重都为0或1,所以有更快的
方法可以解决最短路径问题,可以使用队列去解决这个问题,如果边的权重为0,我们把点放在队首,否则放在队尾。

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using namespace std;
#define SIZE 9

struct {
int to, weight;
};
vector<Node> edges[SIZE];

void zeroOneBFS(int start){
int dist[SIZE];
for(int i = 0; i < SIZE; i++) dist[i] = INT_MAX;
deque<int> Q;
dist[start] = 0;
Q.push_back(start);
while(!Q.empty()){
int v = Q.front();
Q.pop_front();
for(int i = 0; i < edges[v].size(); i++){
if(dist[edges[v][i].to] > dist[v] + edges[v][i].weight){
dist[edges[v][i].to] = dist[v]+ edges[v][i].weight;
if(edges[v][i].weight == 0) Q.push_front(edges[v][i].to);
else Q.push_back(edges[v][i].to);
}
}
}
for(int i = 0; i < SIZE; i++) cout<<dist[i]<<" ";
}

void addEdge(int u, int v, int wt){
edges[u].push_back({v, wt});
edges[v].push_back({u, wt});
}

int main(){
addEdge(0, 1, 0);
addEdge(0, 7, 1);
addEdge(1, 7, 1);
addEdge(1, 2, 1);
addEdge(2, 3, 0);
addEdge(2, 5, 0);
addEdge(2, 8, 1);
addEdge(3, 4, 1);
addEdge(3, 5, 1);
addEdge(4, 5, 1);
addEdge(5, 6, 1);
addEdge(6, 7, 1);
addEdge(7, 8, 1);
zeroOneBFS(0);
return 0;
}