BZOJ3417 Poi2013 Tales of seafaring

一眼，什么神题。。。

然后发现对每个点无脑bfs一下是O(mn)的。。。额好吧Σ( ° △ °|||)

觉得clrs的写法很好呢，学习了！

/**************************************************************
    Problem: 3417
    User: rausen
    Language: C++
    Result: Accepted
    Time:7648 ms
    Memory:12916 kb
****************************************************************/
 
#include <cstdio>
 
using namespace std;
const int N = 10010;
const int K = 1000010;
 
struct edge {
  int next, to;
  edge() {}
  edge(int _n, int _t) : next(_n), to(_t) {}
} e[N];
 
struct Edge {
  int next, to, v;
  Edge() {}
  Edge(int _n, int _t, int _v) : next(_n), to(_t), v(_v) {}
} E[K];
 
int first[N], First[N], tot, Tot;
 
struct queue {
  int x, y;
  queue() {}
  queue(int _x, int _y) : x(_x), y(_y) {}
} q[N];
 
int n, m, Q, l, r;
int pos[N][2], dis[N][2];
 
int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch)
    ch = getchar();
  while ('0' <= ch && ch <= '9')
    (x *= 10) += ch - '0', ch = getchar();
  return x;
}
 
inline void add_edges(int x, int y) {
  e[++tot] = edge(first[x], y), first[x] = tot;
  e[++tot] = edge(first[y], x), first[y] = tot;
}
 
inline void Add_Edge(int x, int y, int z) {
  E[++Tot] = Edge(First[x], y, z), First[x] = Tot;
}
 
inline void update(int x, int y, int z, int i) {
  if (pos[x][y] < i)
    pos[x][y] = i, dis[x][y] = z, q[++r] = queue(x, y);
}
 
int main() {
  int i, j, x, y, z, X, Y;
  n = read(), m = read(), Q = read();
  for (i = 1; i <= m; ++i)
    add_edges(read(), read());
  for (i = 1; i <= Q; ++i) {
    x = read(), y = read(), z = read();
    Add_Edge(x, y, z);
  }
  for (i = 1; i <= n; ++i) if (First[i]) {
      l = 1, r = 0, update(i, 0, 0, i);
      while (l <= r) {
    X = q[l].x, Y = q[l].y, ++l;
    for (x = first[X]; x; x = e[x].next)
      update(e[x].to, Y ^ 1, dis[X][Y] + 1, i);
      }
      for (x = First[i]; x; x = E[x].next)
    if ((E[x].to != i || first[i]) && pos[E[x].to][E[x].v & 1] == i && dis[E[x].to][E[x].v & 1] <= E[x].v) E[x].to = 0;
    }
  for (i = 1; i <= Tot; ++i)
    puts(E[i].to ? "NIE" : "TAK");
  return 0;
}