有向图
1.邻接矩阵,稠密图
2.邻接表 (常用)单链表,每一个点都有一个单链表 ,插入一般在头的地方插,
图的邻接表的存储方式
特殊的深度优先搜索,难点是如何实现,一条道走到黑
const int N=100010,M=n*2;
int h[N],e[N],ne[N],idx;
bool st[N];//记录状态
void add(int a,int b)
{
e[idx]=b;
ne[idx]=h[a];
h[a]=idx++;
}
void dfs(int u)
{
st[u]=true;
for(i=h[u];i!=-1;i=ne[i])
{
int j=e[i];//当前节点对应的图的值;
if(!st[j])dfs(j);
}
}
int main()
{
memset(h,-1,sizeof(h));
return 0;
}
例题:图的层序搜索
#include
#include
#include
#include
#include
using namespace std;
const int N=100010;
int n,m;
int d[N];
int e[N],h[N],idx,ne[N];
void add(int a,int b)
{
e[idx]=b;
ne[idx]=h[a];
h[a]=idx++;
}
void bfs()
{memset(d,-1,sizeof d);
queue q;
d[1]=0;
q.push(1);
while(q.size())
{
auto t=q.front();
q.pop();
for(int i=h[t];i!=-1;i=ne[i])
{
int j=e[i];
if(d[j]==-1)
{
d[j]=d[t]+1;
q.push(j);
}
}
}
printf("%d",d[n]);
}
int main()
{
cin>>n>>m;
memset(h,-1,sizeof h);
for(int i=0;i>a>>b;
add(a,b);
}
bfs();
return 0;
}
例题 :有向图的拓扑序列
#include
#include
#include
using namespace std;
const int N = 100010;
int n, m;
int h[N], e[N], ne[N], idx;
int d[N];
int q[N];
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}
bool topsort()
{
int hh = 0, tt = -1;
for (int i = 1; i <= n; i ++ )
if (!d[i])
q[ ++ tt] = i;
while (hh <= tt)
{
int t = q[hh ++ ];
for (int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if (-- d[j] == 0)
q[ ++ tt] = j;
}
}
return tt == n - 1;
}
int main()
{
scanf("%d%d", &n, &m);
memset(h, -1, sizeof h);
for (int i = 0; i < m; i ++ )
{
int a, b;
scanf("%d%d", &a, &b);
add(a, b);
d[b] ++ ;
}
if (!topsort()) puts("-1");
else
{
for (int i = 0; i < n; i ++ ) printf("%d ", q[i]);
puts("");
}
return 0;
}
#include
#include
#include
#include
using namespace std;
const int a1=510;
int n,m;
int g[a1][a1];
int dist[a1];
bool st[a1];
int dijk()
{
memset(dist,0x3f,sizeof dist);
dist[1]=0;
for(int i=0;idist[j]))t=j;
}
for(int j=1;j<=n;j++)
dist[j]=min(dist[j],dist[t]+g[t][j]);
st[t]=true;
}
if(dist[n]==0x3f3f3f3f)return -1;
return dist[n];
}
int main()
{
cin>>n>>m;
memset(g,0x3f,sizeof g);
while(m--)
{
int a,b,c;
cin>>a>>b>>c;
g[a][b]=min(g[a][b],c);
}
cout<
#include
#include
#include
#include
#include
using namespace std;
typedef pair pii;
const int N =1e6 + 10;
int n,m,a,b,c;
int h[N],e[N],ne[N],w[N],idx;
int dist[N];
bool st[N];
void add(int a,int b,int c)
{
e[idx]=b,w[idx]=c,ne[idx]=h[a],h[a]=idx++;
}
int dijk()
{
memset(dist,0x3f3f3f3f,sizeof dist);
dist[1]=0;
priority_queue, greater> heap;
heap.push({0,1});
while(heap.size())
{
auto t=heap.top();
heap.pop();
int ver=t.second,distance=t.first;
if(st[ver])continue;
st[ver]=true;
for(int i=h[ver];i!=-1;i=ne[i])
{
int j=e[i];
if(dist[j]>dist[ver]+w[i])
{
dist[j]=dist[ver]+w[i];
heap.push({dist[j],j});
}
}
}
if(dist[n]==0x3f3f3f3f)return -1;
return dist[n];
}
int main()
{
cin>>n>>m;
memset(h,-1,sizeof h);
while(m--)
{
cin>>a>>b>>c;
add(a,b,c);
}
cout<