UVA - 11902 （有向图的关节点）

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链接：

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18888

题意：一个有向图，如果从0点出发到达某一点必须经过某些点题目就是求出这些点。

点数不多可以删点然后dfs搜索，之前能搜到的点但是删了该点之后搜不到了那么这个删点就是从起始点搜不到的必经之路。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <stack>#include <cmath>#include <queue>#include <map>#include <set>#define lson o<<1,l,m#define rson o<<1|1,m+1,r#define mem(a) memset(a,0,sizeof(a))typedef long long ll;const int N = 105;const int M = 10005;const ll mod = 1000000009;using namespace std;int n, T;int g[N][N], a[N][N];int v1[N], v2[N];int ans[N][N];char out[N*3][N*3];void dfs(int u) {if(v1[u]) return;v1[u] = 1;for(int i = 1; i <= n; i++) {if(g[u][i] && v1[i] == 0) dfs(i);}}void del(int u) {mem(g[u]);}void rec(int u) {memcpy(g[u], a[u], sizeof(a[u]));}void solve(int u) {mem(v1);dfs(1);for(int i = 1; i <= n; i++) {if(v1[i] == 0 && v2[i] == 1) {ans[u][i] = 1;}}}int main() {//freopen("in.txt", "r", stdin);cin >> T;int ca = 1;while(T--) {cin >> n;for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {scanf("%d", &g[i][j]);a[i][j] = g[i][j];}}mem(v1);dfs(1);mem(ans);for(int i = 1; i <= n; ++i) if(v1[i]) {ans[1][i] = 1;ans[i][i] = 1;}memcpy(v2, v1, sizeof(v1));for(int i = 2; i <= n; i++) {del(i);solve(i);rec(i);}mem(out);for(int i = 1; i <= n+n+1; i++) {if(i & 1) {out[i][1] = out[i][n+n+1] = '+';for(int j = 2; j <= n+n; j++) {out[i][j] = '-';}} else {for(int j = 1; j <= n+n+1; j++) {if(j & 1) {out[i][j] = '|';} else {if(ans[i/2][j/2]) {out[i][j] = 'Y';} else out[i][j] = 'N';}}}}printf("Case %d:\n", ca++);for(int i = 1; i <= n+n+1; i++) {for(int j = 1; j <= n+n+1; j++) {printf("%c", out[i][j]);}puts("");}//for(int i = 1; i <= n; i++) {//for(int j = 1; j <= n; j++) {//printf("%d ", ans[i][j]);//}//puts("");//}puts("");}return 0;}

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