记一次坑爹的广工校赛

6/6

4-10，广工校赛，一场坑爹的邂逅。。。。。。

 

首先，路程坑爹，问了好久路才到什么GUI工学馆。。。。

其次，天气坑爹，怎么老是下雨。。。。

最后，比赛坑爹，本垃圾被坑得不要不要的，为什么题A1小时前和一小时后看的是不一样的，改题面，改样例，恩，你倒不如新增一道题算了；为什么题C无解啊！广工：大家注意，没有这种数据（没有你放出来干*啊）。。。。

传送门

Problem A: Krito的讨伐

题意：略

题解：实质上是贪心模拟题，先从根节点往下走，把根节点的怪都消灭掉之后，在把它的儿子都加进来，最后看有没有打不过的就可以了。

代码怎么写呢？先用dfs从根出发，然后扫到一个有怪物的结点，加进去之后就退出来，不在dfs，从优先队列拿出元素之后，记录这个点有没有全部拿完怪兽，有的dfs这个点就好了。

 

/*zhen hao*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
 
typedef pair<int,int> pii;
const int maxn = 1e3 + 10;
int cnt[maxn], father[maxn];
int n, m, k;
vector<int> tree[maxn];
 
struct Node {
  int id, d, a;
  Node(int id=0, int d=0, int a=0) : id(id), d(d), a(a) {}
  bool operator < (const Node& o) const {
    return d > o.d;
  }
};
 
priority_queue<Node> pq;
vector<Node> mon[maxn];
 
void dfs(int u, int fa) {
  father[u] = fa;
  if (!mon[u].empty()) {
    for (int i = 0; i < (int)mon[u].size(); i++)  pq.push(mon[u][i]);
    return;
  }
  for (int i = 0; i < (int)tree[u].size(); i++) {
    int v = tree[u][i];
    if (v == fa) continue;
    dfs(v, u);
  }
}
 
int main() {
//  freopen("case.in", "r", stdin);
  int T;
  cin >> T;
  while (T--) {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) tree[i].clear(), mon[i].clear();
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      tree[u].push_back(v);
      tree[v].push_back(u);
    }
    scanf("%d", &k);
    for (int i = 0; i < m; i++) {
      int x, y, z;
      scanf("%d%d%d", &x, &y, &z);
      mon[x].push_back(Node(x, y, z));
    }
 
    while (!pq.empty()) pq.pop();
 
    dfs(0, -1);
    memset(cnt, 0, sizeof cnt);
 
    while (!pq.empty()) {
      Node p = pq.top(); pq.pop();
//      cout << p.id << " " << p.d << " " << p.a << " " << pq.size() << endl;
      if (k <= p.d) { k = -1; break; }
      k += p.a;
      int u = p.id;
      if (++cnt[u] >= (int)mon[u].size()) {
        for (int i = 0; i < (int)tree[u].size(); i++) {
          int v = tree[u][i];
          if (v == father[u]) continue;
          dfs(v, u);
        }
      }
    }
//    cout << k << endl;
    if (k != -1) puts("Oh yes.");
    else puts("Good Good Study,Day Day Up.");
  }
}

Problem B: Sward Art Online

题意：略

题解：暴力水题，枚举首饰和头盔，枚举单手剑即可，dp1[i]表示用了i元钱能够买到最大首饰盒头盔的攻击力，dp2[i]表示用了i元钱能够买到最大的剑的攻击力，然后总结一下最大值即可。

注意两个坑点：第一id == -1 buff == -1buff无效，第二剑只有一把；

/*zhen hao*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;
 
struct Node {
  int mon, atk, id, buff;
};
 
Node A[110], B[110], C[110], D[110];
int dp1[10010], dp2[10010];
 
int main() {
//  freopen("case.in", "r", stdin);
  int T;
  cin >> T;
  while (T--) {
    int m, a, b, c, d;
    cin >> m >> a >> b >> c >> d;
    memset(dp1, 0, sizeof dp1);
    memset(dp2, 0, sizeof dp2);
 
    for (int i = 0; i < a; i++) {
      scanf("%d%d", &A[i].mon, &A[i].atk);
      dp1[A[i].mon] = max(dp1[A[i].mon], A[i].atk);
    }
    for (int i = 0; i < b; i++) {
      scanf("%d%d%d%d", &B[i].mon, &B[i].atk, &B[i].id, &B[i].buff);
      dp1[B[i].mon] = max(dp1[B[i].mon], B[i].atk);
    }
    for (int i = 0; i < c; i++) {
      scanf("%d%d", &C[i].mon, &C[i].atk);
      dp2[C[i].mon] = max(dp2[C[i].mon], C[i].atk);
    }
    for (int i = 0; i < d; i++) {
      scanf("%d%d", &D[i].mon, &D[i].atk);
      dp2[D[i].mon] = max(dp2[D[i].mon], D[i].atk);
    }
 
    for (int i = 0; i < a; i++)
      for (int j = 0; j < b; j++) {
        int temp;
        if (B[j].buff != -1 && B[j].id == i) temp = A[i].atk + B[j].atk + B[j].buff;
        else temp = A[i].atk + B[j].atk;
        dp1[A[i].mon + B[j].mon] = max(dp1[A[i].mon + B[j].mon], temp);
      }
 
    for (int i = 0; i < c; i++)
      for (int j = 0; j < i; j++)
        dp2[C[i].mon + C[j].mon] = max(dp2[C[i].mon + C[j].mon], C[i].atk + C[j].atk);
 
    for (int i = 1; i <= m; i++) {
      dp1[i] = max(dp1[i], dp1[i - 1]);
      dp2[i] = max(dp2[i], dp2[i - 1]);
    }
    int ans = 0;
    for (int i = 0; i <= m; i++) {
      ans = max(ans, dp1[i] + dp2[m - i]);
      ans = max(ans, dp1[i]);
      ans = max(ans, dp2[i]);
    }
    cout << ans << endl;
  }
  return 0;
}

Problem C: wintermelon的魔界寻路之旅

题意：略

题解：连成一个有向图，然后跑dij之后，看最短路有多少条即可。

/*zhen hao*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;
 
const int maxn = 1e2 + 10, maxp = 1e4 + 10, inf = 1 << 30, mod = 1e9 + 9;
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
int head[maxp], dist[maxp], done[maxp], dp[maxp];
int n, src, target, tot;
int val[maxn][maxn], id[maxn][maxn];
 
struct Edge {
  int v, w, next;
  Edge(int v = 0, int w = 0, int next = 0) : v(v), w(w), next(next) {}
} edges[maxp * 100];
 
struct Node {
  int d, u;
  Node(int d = 0, int u = 0) : d(d), u(u) {}
  bool operator < (const Node& o) const {
    return d > o.d;
  }
};
 
void init() {
  memset(head, -1, sizeof head);
  tot = 0;
}
 
void add_edge(int u, int v, int w) {
  edges[tot] = Edge(v, w, head[u]);
  head[u] = tot++;
}
 
void dijkstra() {
  priority_queue<Node> pq;
  memset(done, 0, sizeof done);
  for (int i = src; i <= target; i++) dist[i] = inf;
  dist[src] = 0;
  pq.push(Node(0, src));
  while (!pq.empty()) {
    Node r = pq.top(); pq.pop();
    int u = r.u;
    if (done[u]) continue;
    done[u] = true;
    for (int i = head[u]; ~i; i = edges[i].next) {
      Edge e = edges[i];
      if (dist[e.v] > dist[u] + e.w) {
        dist[e.v] = dist[u] + e.w;
        pq.push(Node(dist[e.v], e.v));
      }
    }
  }
}
 
void print() {
  for (int i = src; i <= target; i++) {
    printf("%d: ", i);
    for (int j = head[i]; ~j; j = edges[j].next) {
      printf("(%d %d) ", edges[j].v, edges[j].w);
    }
    puts("");
  }
}
 
int walk(int u) {
  if (u == target) return 1;
  if (dp[u] != -1) return dp[u];
  int& ret = dp[u];
  ret = 0;
  for (int i = head[u]; ~i; i = edges[i].next) {
    Edge e = edges[i];
    if (dist[u] + e.w == dist[e.v]) {
      ret += walk(e.v);
      if (ret >= mod) ret -= mod;
    }
  }
  return ret;
}
 
int main() {
//  freopen("case.in", "r", stdin);
  int T;
  cin >> T;
  while (T--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        scanf("%d", &val[i][j]);
      }
 
    for (int i = 0, k = n - 1; i < n; i++, k--)
      for (int j = 0, l = n - 1; j < n - i - 1; j++, l--)
        val[i][j] += val[l][k];
 
    src = 0, target = 0;
    init();
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n - i; j++)
        id[i][j] = target++;
 
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n - i; j++) {
        int u = id[i][j];
//        cout <<  i << " " << j << " " << id[i][j] << targetl;
        for (int k = 0; k < 4; k++) {
          int ii = i + dx[k], jj = j + dy[k];
          if (ii >= 0 && ii < n && jj >= 0 && jj < n - ii) {
            int v = id[ii][jj];
            add_edge(u, v, val[ii][jj]);
          }
        }
      }
 
    for (int i = 0; i < n; i++) {
      int x = id[i][n - i - 1];
      add_edge(x, target, 0);
      add_edge(target, x, 0);
    }
    if (target >= maxp || tot > maxp * 100) {
      while (true);
    }
 
//    print();
    dijkstra();
    memset(dp, -1, sizeof dp);
    printf("%d
", walk(src));
  }
  return 0;
}

Problem F: 我是好人4

题意：略

题解：容斥+dfs枚举子集。不能用位来枚举，因为有个关键的剪枝就是：如果这个数大于le9就不用再继续枚举下去了，这样想想，好像真的很小的复杂度，因为还有个去重，就是如果aj是某个ai的备受，那么就不用管这个aj；挺好的一道题。

/*zhen hao*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long ll;
const int maxn = 60, con = 1e9;
int v[maxn], nv[maxn];
int n, cnt, ans;
 
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
 
void dfs(int now, int cur, ll LCM) {
  if (cur >= cnt) {
    if (now) ans += (con / LCM) * (now & 1 ? -1 : 1);
    return;
  }
  if (LCM > con) return;
  dfs(now, cur + 1, LCM);
  if (now) {
    if (LCM % nv[cur]) LCM *= nv[cur] / gcd(nv[cur], LCM);
    dfs(now + 1, cur + 1, LCM);
  } else {
    dfs(now + 1, cur + 1, nv[cur]);
  }
}
 
int main() {
//  freopen("case.in", "r", stdin);
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", v + i);
    cnt = 0;
    sort(v, v + n);
 
    if (v[0] == 1) {
      puts("0"); continue;
    }
 
    for (int i = 0; i < n; i++) {
      bool ok = true;
      for (int j = 0; j < i; j++) if (v[i] % v[j] == 0) ok = false;
      if (ok) nv[cnt++] = v[i];
    }
    ans = con;
    dfs(0, 0, 1);
    printf("%lld
", ans);
  }
  return 0;
}