图（领接矩阵和vector）的dfs和bfs
//Uva572 图的dfs求连通块
#include
#include
#include
using namespace std;
int m,n;
int dir[8][2] = {{-1,0},{-1,1},{0,1},{1,1},
{ 1,0},{1,-1},{0,-1},{-1,-1}};
bool vis[105][105];
char map[105][105];
void dfs(int x, int y){
for(int i=0; i<8; ++i){
int dx=x+dir[i][0], dy=y+dir[i][1];
if(dx>=0 && dx
dfs(dx, dy);
}
}
}
int main(){
#ifdef LOCAL
freopen("input.txt","r",stdin);
#endif
int i,j;
while(scanf("%d %d%*c",&m,&n) && m && n){
for(i=0; i
memset(vis, 0, sizeof(vis));
int cnt=0;
for(i=0; i
++cnt;
dfs(i, j);
}
}
}
printf("%d\n",cnt);
}
return 0;
}
//Uva1103
#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXH = 210;
const int MAXW = 100;
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
char* bin[]= {"0000","0001","0010","0011","0100","0101","0110","0111",
"1000","1001","1010","1011","1100","1101","1110","1111"};
char code[] = "WAKJSD";
int W,H;
char pic1[MAXH][MAXW];
char pic[MAXH][MAXW<<2];
int color[MAXH][MAXW<<2];
void dfs(int x, int y, int col)
{
color[x][y] = col;
for(int i = 0; i < 4; ++i){
int xx = x + dx[i], yy = y + dy[i];
if(xx >= 0 && xx < H && yy >= 0 && yy < W
&& pic[x][y] == pic[xx][yy] && !color[xx][yy]){
dfs(xx,yy,col);
}
}
}
int main(void)
{
//freopen("input.txt","r",stdin);
int cas = 1;
while(scanf("%d %d", &H, &W), H || W){
memset(pic,0,sizeof(pic));
memset(color,0,sizeof(color));
for(int i = 0; i < H; ++i)
scanf("%s",pic1[i]);
for(int i = 0 ; i < H; ++i){
for(int j = 0; j < W; ++j){
if(isdigit(pic1[i][j]))
pic1[i][j] -= '0';
else
pic1[i][j] -= 'a' - 10;
for(int k = 0; k < 4; ++k)
pic[i+1][1 + 4 * j + k] = bin[pic1[i][j]][k] - '0';
}
}
H += 2;
W = 4 * W + 2;
int cnt = 0;
vector
for(int i = 0; i < H; ++i){
for(int j = 0; j < W; ++j){
if(!color[i][j]){
dfs(i,j,++cnt);
if(pic[i][j] == 1) cc.push_back(cnt);
}
}
}
vector
for(int i = 0; i < H; ++i){
for(int j = 0; j < W; ++j){
if(pic[i][j]){
for(int k = 0; k < 4; ++k){
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < H && y >= 0 && y < W &&
pic[x][y] == 0 && color[x][y] != 1)
neigh[color[i][j]].insert(color[x][y]);
}
}
}
}
vector
for(int i = 0, sz = cc.size(); i < sz; ++i)
ans.push_back(code[neigh[cc[i]].size()]);
sort(ans.begin(),ans.end());
printf("Case %d: ",cas++);
for(int i = 0, sz = ans.size(); i < sz; ++i)
printf("%c",ans[i]);
puts("");
}
return 0;
}
//图的bfs求最短路
#include int cnt=0;
for(int i=vec.size()-1;i>=0;i--)
{
cnt++;
if(cnt==1) printf(" ");
printf("(%d,%d)",vec[i].r,vec[i].c);
if(cnt!=10&&i!=0) printf(" ");
if(cnt==10&&i!=0) {printf("\n");cnt=0;}
}
printf("\n");
}
void solve()
{
queue printf(" No Solution Possible\n");
}
int main()
{
while(cin>>s1)
{
if(s1=="END") break;
memset(has_edge,0,sizeof(has_edge));
memset(d,-1,sizeof(d));
memset(p,0,sizeof(p));
int a,count=0;
int a1,a2,dir1,turn1;
while(cin>>a)
{
if(a==0) break;
count++;
if(count==1)
{
r0=a;
cin>>c0>>s>>r2>>c2;
for(int i=0;i<4;i++)
{
if(s[0]==dirs[i])
{
dir=i;
break;
}
}
r1=r0+dr[dir];
c1=c0+dc[dir];
}
else
{
a1=a;
cin>>a2;
while(cin>>s)
{
if(s[0]=='*') break;
for(int i=0;i<4;i++)
{
if(s[0]==dirs[i])
{
dir1=i;
break;
}
}
for(int i=1;i {
for(int j=0;j<3;j++)
{
if(s[i]==turns[j])
{
turn1=j;
break;
}
}
has_edge[a1][a2][dir1][turn1]=1;
}
}
}
}
// for(int i=1;i<=9;i++)
// {
// for(int j=1;j<=9;j++)
// {
// for(int k=0;k<4;k++)
// {
// for(int l=0;l<3;l++)
// {
// if(has_edge[i][j][k][l])
// printf("[%d][%d][%d][%d]\n",i,j,k,l);
// }
// }
// }
// }
solve();
}
return 0;
} //图的拓扑排序DAG（有向图的节点排序问题）排序失败说明不是DAG
//有向图的拓扑排序
//WA..有一个陷阱就是n!=0 &&m==0 的情况
#include
#include
#define MAX 100
int c[MAX];
int topo[MAX],t;
int n,m;
int g[MAX][MAX];
bool dfs(int u)
{
c[u]=-1;
for(int v=0;v if(g[u][v]) { if(c[v]<0)return false; else if(!c[v] && !dfs(v))returnfalse; } c[u]=1; topo[--t]=u; return true; } bool toposort() { t=n; memset(c,0,sizeof(c)); for(int u=0;u if(!c[u] && !dfs(u))returnfalse; return true; } int main() { int i,j; while(scanf("%d%d",&n,&m),n) { memset(g,0,sizeof(g)); while(m--) { scanf("%d%d",&i,&j); i--; j--; g[i][j]=1; } toposort(); for(i=0;i { if(i!=0)printf(" "); printf("%d",topo[i]+1); } printf("\n"); } return 0; }
//图的欧拉回路
#include
#include
int indgr[30],outdgr[30],p[30];
char b[1010];
int find(int x)
{
return p[x]==x?x:(p[x]=find(p[x]));
}
int main()
{
int i,j,first,last,t,n,num,ok;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(indgr,0,sizeof(indgr));
memset(outdgr,0,sizeof(outdgr));
for(i=0;i<26;i++)
p[i]=i;
for(i=0;i
scanf("%s",b);
first=b[0]-'a';
last=b[strlen(b)-1]-'a';
outdgr[first]++;
indgr[last]++;
if(find(first)!=find(last))
p[find(first)]=find(last);
}
num=0;
ok=1;
for(i=0;!indgr[i]&&!outdgr[i];i++);
for(j=i+1;j<26;j++)
if((indgr[j]||outdgr[j])&&find(i)!=find(j))
{
ok=0;
break;
}
if(ok)
for(i=0;i<26;i++)
{
if(indgr[i]>outdgr[i])
{
if(indgr[i]-outdgr[i]>1)
{
ok=0;
break;
}
else
num++;
}
else if(outdgr[i]>indgr[i])
{
if(outdgr[i]-indgr[i]>1)
{
ok=0;
break;
}
else
num++;
}
}
if(!ok||num>2)
printf("The door cannot be opened.\n");
else
printf("Ordering is possible.\n");
}
return 0;
}