USACO4.3.2 Street Race (race3)

第一问n-2次spfa，枚举每一个不可避免的路口k，spfa时遇到k时continue，如果从起点无法走到终点则k为不可避免的路口。

第二问，显然中间路口一定是一个不可避免路口，所以先枚举每一个不可避免路口k，以k为起点spfa，如果从比k编号小的节点回到了k，则k一定不是中间路口。否则k一定是中间路口。

/*
ID:shijiey1
PROG:race3
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

#define INF 0x3f3f3f3f

using namespace std;

int V = 0;
vector<int> edges[55];
vector<int> t_edges[55];
void addEdge(int a, int b) {
edges[a].push_back(b);
t_edges[a].push_back(b);
t_edges[b].push_back(a);
}

int done[55];
int dist[55];
bool spfa(int k) {
for (int i = 0; i <= V; i++) {
dist[i] = INF;
}
dist[0] = 0;
memset(done, 0, sizeof(done));
queue<int> q;
done[0] = 1;
q.push(0);
while (!q.empty()) {
int current = q.front();
q.pop();
for (int i = 0; i < edges[current].size(); i++) {
int v = edges[current][i];
if (v == k) continue;
if (dist[v] > dist[current] + 1) {
dist[v] = dist[current] + 1;
if (!done[v]) {
done[v] = 1;
q.push(v);
}
}
}
done[current] = 0;
}
return dist[V - 1] == INF;
}

bool spfa2(int s) {
for (int i = 0; i <= V; i++) {
dist[i] = INF;
}
dist[s] = 0;
memset(done, 0, sizeof(done));
queue<int> q;
done[s] = 1;
q.push(s);
while (!q.empty()) {
int current = q.front();
q.pop();
for (int i = 0; i < edges[current].size(); i++) {
int v = edges[current][i];
if (v == s && current < s) return false;
if (dist[v] > dist[current] + 1) {
dist[v] = dist[current] + 1;
if (!done[v]) {
done[v] = 1;
q.push(v);
}
}
}
done[current] = 0;
}
return true;
}

vector<int> ans1;
vector<int> ans2;
int main() {
freopen("race3.in", "r", stdin);
freopen("race3.out", "w", stdout);
while (true) {
int t;
while (scanf("%d", &t)) {
if (t < 0) break;
addEdge(V, t);
}
if (t == -1) break;
V++;
}

for (int k = 1; k <= V - 2; k++) {
if (spfa(k)) {
ans1.push_back(k);
}
}
int size = ans1.size();
printf("%d", size);
for (int i = 0; i < size; i++) {
int t = ans1[i];
printf(" %d", t);
if (spfa2(t)) {
ans2.push_back(t);
}
}
size = ans2.size();
printf("\n%d", size);
for (int i = 0; i < size; i++) {
printf(" %d", ans2[i]);
}
printf("\n");
return 0;
}