bzoj4010: [HNOI2015]菜肴制作【拓扑排序】

　　想到了一个分治方法，每一次尽量放小的那个，把它依赖的放在左边，不依赖的放在右边。

　　TLE 80：

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //***********************************

 const int maxn = ;

 vector <int> Vx[maxn], Vy[maxn];
 int src[maxn], lft[maxn];

 int n, m;
 int deg[maxn], tmp[maxn];

 bool topo() {
     static int que[maxn]; int qh(), qt();
     rep(i, , n) if (!deg[i]) que[++qt] = i;
     int cnt();
     while (qh != qt) {
         int x = que[++qh];
         ++cnt;
         int len = (int) Vx[x].size();
         for (int i = ; i < len; i++) {
             if (--deg[Vx[x][i]] == ) que[++qt] = Vx[x][i];
         }
     }
     return cnt == n;
 }

 bool vis[maxn];
 void bfs(int s) {
     static int que[maxn]; int qh(), qt();
     vis[que[++qt] = s] = ;
     while (qh != qt) {
         int x = que[++qh];
         int len = (int) Vy[x].size();
         for (int i = ; i < len; i++) {
             if (!vis[Vy[x][i]]) {
                 vis[que[++qt] = Vy[x][i]] = ;
                 lft[Vy[x][i]] = ;
             }
         }
     }
 }

 void solve(int l, int r) {
     if (l >= r) return;
     int t, haha = inf;
     rep(i, l, r) if (src[i] < haha) { t = i; haha = src[i]; }
     rep(i, l, r) lft[src[i]] = vis[src[i]] = ;
     bfs(src[t]);
     int cur = l - ;
     rep(i, l, r) if (src[i] != src[t] && lft[src[i]]) tmp[++cur] = src[i];
     tmp[++cur] = src[t]; int mid = cur;
     rep(i, l, r) if (src[i] != src[t] && !lft[src[i]]) tmp[++cur] = src[i];
     memcpy(src + l, tmp + l, sizeof(int) * (r - l + ));
     solve(l, mid - );
     solve(mid + , r);
 }

 int main() {
     freopen("dishes.in", "r", stdin);
     freopen("dishes.out", "w", stdout);
     int T; scanf("%d", &T);
     while (T--) {
         scanf("%d%d", &n, &m);
         rep(i, , n) Vx[i].clear(), Vy[i].clear();
         clr(deg);
         rep(i, , m) {
             int x, y; scanf("%d%d", &x, &y);
             Vx[x].pb(y);
             Vy[y].pb(x);
             deg[y]++;
         }
         if (!topo()) { puts("Impossible!"); continue; }
         rep(i, , n) src[i] = i;
         solve(, n);
         rep(i, , n) printf(i == n ? "%d\n" : "%d ", src[i]);
     }
     return ;
 }

　　正解：

　　　　有依赖关系，我们建立反图，加入我们的到了这个反图的一个拓扑序，那么怎样对应到最优的答案呢？应该是尽量让编号小的出现在队列后方，所以最大字典序的拓扑排序即可。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //***********************************

 const int maxn = ;

 vector <int> Vx[maxn], Vy[maxn];
 priority_queue<int> Q;
 int deg[maxn], ans[maxn], n, m;

 bool topo() {
     rep(i, , n) if (!deg[i]) Q.push(i);
     int cnt();
     while (!Q.empty()) {
         int x = Q.top(); Q.pop();
         ans[++cnt] = x;
         int len = (int) Vy[x].size();
         for (int i = ; i < len; i++) {
             if (--deg[Vy[x][i]] == ) Q.push(Vy[x][i]);
         }
     }
     return cnt == n;
 }

 int main() {
     freopen("dishes.in", "r", stdin);
     freopen("dishes.out", "w", stdout);
     int T; scanf("%d", &T);
     while (T--) {
         scanf("%d%d", &n, &m);
         rep(i, , n) Vx[i].clear(), Vy[i].clear();
         clr(deg);
         rep(i, , m) {
             int x, y; scanf("%d%d", &x, &y);
             Vx[x].pb(y);
             Vy[y].pb(x);
             deg[x]++;
         }
         if (!topo()) puts("Impossible!");
         else { drep(i, n, ) printf("%d ", ans[i]); puts(""); }
     }
     return ;
 }