HDU 1269 迷宫城堡

Problem Description

为了训练小希的方向感，Gardon建立了一座大城堡，里面有N个房间(N<=10000)和M条通道(M<=100000)，每个通道都是单向的，就是说若称某通道连通了A房间和B房间，只说明可以通过这个通道由A房间到达B房间，但并不说明通过它可以由B房间到达A房间。Gardon需要请你写个程序确认一下是否任意两个房间都是相互连通的，即：对于任意的i和j，至少存在一条路径可以从房间i到房间j，也存在一条路径可以从房间j到房间i。

Input

输入包含多组数据，输入的第一行有两个数：N和M，接下来的M行每行有两个数a和b，表示了一条通道可以从A房间来到B房间。文件最后以两个0结束。

Output

对于输入的每组数据，如果任意两个房间都是相互连接的，输出"Yes"，否则输出"No"。

Sample Input

3 31 22 33 13 31 22 33 20 0

Sample Output

YesNo

判断图是否是完全连通的。

以任何一个点为根正负边各自建一个图，然后遍历记录即可。

#include<stdio.h>int f[10001][300],z[10001][300];int ff[10001],zz[10001];int ss1(int x){int i;for (i=1;i<=f[x][0];i++) if(ff[f[x][i]]) {ff[f[x][i]]=0;ss1(f[x][i]);}return 0;}int ss2(int x){int i;for (i=1;i<=z[x][0];i++) if(zz[z[x][i]]) {zz[z[x][i]]=0;ss2(z[x][i]);}return 0;}int main(){int i,n,m,a,b,w;while (1){scanf("%d%d",&n,&m);if (n==0&&m==0) break;for (i=1;i<=n;i++) {f[i][0]=0;z[i][0]=0;ff[i]=1; zz[i]=1;}for (i=1;i<=m;i++){scanf("%d%d",&a,&b);f[a][++f[a][0]]=b;z[b][++z[b][0]]=a;}ff[1]=0; zz[1]=0; ss1(1);  ss2(1);for (w=0,i=1;i<=n;i++) if (ff[i]||zz[i]) w=1;if (w==0) printf("Yes\n"); else printf("No\n");}return 0;}

用vector更加简便。

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;vector<int> f[2][10005];int c[10005];void dfs(int x,int u){c[x]=1;for (int i=0;i<f[u][x].size();i++)if (!c[f[u][x][i]]) dfs(f[u][x][i],u);}int main(){int n,m;while (scanf("%d%d",&n,&m),n+m){for (int i=1;i<=n;i++) f[0][i].clear(),f[1][i].clear();for (int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);f[0][x].push_back(y);f[1][y].push_back(x);}int ff=1;memset(c,0,sizeof(c));dfs(1,0);for (int i=1;i<=n;i++) if (c[i]!=1) {ff=0;break;}memset(c,0,sizeof(c));dfs(1,1);for (int i=1;i<=n;i++) if (c[i]!=1) {ff=0;break;}if (ff) printf("Yes\n"); else printf("No\n");}return 0;}

强连通的tarjan算法。

#include<stack>#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<vector>using namespace std;const int maxn = 10005;int ff, t, tot, n, m, x, y, dfn[maxn], low[maxn], ins[maxn], fa[maxn];vector<int> tree[maxn];stack<int> p;void dfs(int x){int j;low[x] = dfn[x] = ++tot;ins[x] = 1;p.push(x);for (int i = 0; i < tree[x].size(); i++){j = tree[x][i];if (!dfn[j]){dfs(j);low[x] = min(low[x], low[j]);}else if (ins[j]) low[x] = min(dfn[j], low[x]);}if (dfn[x] == low[x]){t++;do{j = p.top();p.pop();ins[j] = 0;fa[j] = t;} while (j != x);}}int main(){while (~scanf("%d%d", &n, &m), n + m){for (int i = 1; i <= n; i++) tree[i].clear();while (m--){scanf("%d%d", &x, &y);tree[x].push_back(y);}memset(dfn, 0, sizeof(dfn));memset(fa, 0, sizeof(fa));memset(ins, 0, sizeof(ins));tot = t = 0; dfs(1);for (int i = ff = 1; i <= n; i++) if (fa[i] != fa[1]) { ff = 0; break; }if (ff) printf("Yes\n"); else printf("No\n");}return 0;}

还可以这样写

#include<stack>#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<vector>using namespace std;const int maxn = 10005;int t, tot, n, m, x, y, dfn[maxn], low[maxn], ins[maxn];vector<int> tree[maxn];stack<int> p;void dfs(int x){low[x] = dfn[x] = ++tot;ins[x] = 1;p.push(x);for (int i = 0; i < tree[x].size(); i++){int y = tree[x][i], j;if (!ins[y]){dfs(y);low[x] = min(low[x], low[y]);if (dfn[x] < low[y]){t++;do{j = p.top();p.pop();} while (j != y);}}else low[x] = min(dfn[y], low[x]);}}int main(){while (~scanf("%d%d", &n, &m), n + m){while (!p.empty()) p.pop();for (int i = 1; i <= n; i++) tree[i].clear();while (m--){scanf("%d%d", &x, &y);tree[x].push_back(y);}memset(ins, 0, sizeof(ins));tot = t = 0; dfs(1);for (int i = 1; i <= n; i++) if (!ins[i]) t++;if (!t) printf("Yes\n"); else printf("No\n");}return 0;}