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）。。。。

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。

 

1 /*zhen hao*/ 2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 using namespace std; 9 10 typedef pair
            
              pii;11 const int maxn = 1e3 + 10;12 int cnt[maxn], father[maxn];13 int n, m, k;14 vector
             
               tree[maxn];15 16 struct Node {17 int id, d, a;18 Node(int id=0, int d=0, int a=0) : id(id), d(d), a(a) {}19 bool operator < (const Node& o) const {20 return d > o.d;21 }22 };23 24 priority_queue
              
                pq;25 vector
               
                 mon[maxn];26 27 void dfs(int u, int fa) {28 father[u] = fa;29 if (!mon[u].empty()) {30 for (int i = 0; i < (int)mon[u].size(); i++) pq.push(mon[u][i]);31 return;32 }33 for (int i = 0; i < (int)tree[u].size(); i++) {34 int v = tree[u][i];35 if (v == fa) continue;36 dfs(v, u);37 }38 }39 40 int main() {41 // freopen("case.in", "r", stdin);42 int T;43 cin >> T;44 while (T--) {45 scanf("%d%d", &n, &m);46 for (int i = 0; i < n; i++) tree[i].clear(), mon[i].clear();47 for (int i = 0; i < n - 1; i++) {48 int u, v;49 scanf("%d%d", &u, &v);50 tree[u].push_back(v);51 tree[v].push_back(u);52 }53 scanf("%d", &k);54 for (int i = 0; i < m; i++) {55 int x, y, z;56 scanf("%d%d%d", &x, &y, &z);57 mon[x].push_back(Node(x, y, z));58 }59 60 while (!pq.empty()) pq.pop();61 62 dfs(0, -1);63 memset(cnt, 0, sizeof cnt);64 65 while (!pq.empty()) {66 Node p = pq.top(); pq.pop();67 // cout << p.id << " " << p.d << " " << p.a << " " << pq.size() << endl;68 if (k <= p.d) { k = -1; break; }69 k += p.a;70 int u = p.id;71 if (++cnt[u] >= (int)mon[u].size()) {72 for (int i = 0; i < (int)tree[u].size(); i++) {73 int v = tree[u][i];74 if (v == father[u]) continue;75 dfs(v, u);76 }77 }78 }79 // cout << k << endl;80 if (k != -1) puts("Oh yes.");81 else puts("Good Good Study,Day Day Up.");82 }83 }
               
              
             
            
           
          
         
        
       
      

 

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；

1 /*zhen hao*/ 2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 #include 
            
              9 using namespace std;10 11 struct Node {12 int mon, atk, id, buff;13 };14 15 Node A[110], B[110], C[110], D[110];16 int dp1[10010], dp2[10010];17 18 int main() {19 // freopen("case.in", "r", stdin);20 int T;21 cin >> T;22 while (T--) {23 int m, a, b, c, d;24 cin >> m >> a >> b >> c >> d;25 memset(dp1, 0, sizeof dp1);26 memset(dp2, 0, sizeof dp2);27 28 for (int i = 0; i < a; i++) {29 scanf("%d%d", &A[i].mon, &A[i].atk);30 dp1[A[i].mon] = max(dp1[A[i].mon], A[i].atk);31 }32 for (int i = 0; i < b; i++) {33 scanf("%d%d%d%d", &B[i].mon, &B[i].atk, &B[i].id, &B[i].buff);34 dp1[B[i].mon] = max(dp1[B[i].mon], B[i].atk);35 }36 for (int i = 0; i < c; i++) {37 scanf("%d%d", &C[i].mon, &C[i].atk);38 dp2[C[i].mon] = max(dp2[C[i].mon], C[i].atk);39 }40 for (int i = 0; i < d; i++) {41 scanf("%d%d", &D[i].mon, &D[i].atk);42 dp2[D[i].mon] = max(dp2[D[i].mon], D[i].atk);43 }44 45 for (int i = 0; i < a; i++)46 for (int j = 0; j < b; j++) {47 int temp;48 if (B[j].buff != -1 && B[j].id == i) temp = A[i].atk + B[j].atk + B[j].buff;49 else temp = A[i].atk + B[j].atk;50 dp1[A[i].mon + B[j].mon] = max(dp1[A[i].mon + B[j].mon], temp);51 }52 53 for (int i = 0; i < c; i++)54 for (int j = 0; j < i; j++)55 dp2[C[i].mon + C[j].mon] = max(dp2[C[i].mon + C[j].mon], C[i].atk + C[j].atk);56 57 for (int i = 1; i <= m; i++) {58 dp1[i] = max(dp1[i], dp1[i - 1]);59 dp2[i] = max(dp2[i], dp2[i - 1]);60 }61 int ans = 0;62 for (int i = 0; i <= m; i++) {63 ans = max(ans, dp1[i] + dp2[m - i]);64 ans = max(ans, dp1[i]);65 ans = max(ans, dp2[i]);66 }67 cout << ans << endl;68 }69 return 0;70 }
            
           
          
         
        
       
      

 

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。

 

1 /*zhen hao*/  2 #include 
      
         3 #include 
       
          4 #include 
        
           5 #include 
         
            6 #include 
          
             7 #include 
           
             8 #include 
            
              9 using namespace std; 10 11 const int maxn = 1e2 + 10, maxp = 1e4 + 10, inf = 1 << 30, mod = 1e9 + 9; 12 const int dx[] = { 0, 0, 1, -1}, dy[] = { 1, -1, 0, 0}; 13 int head[maxp], dist[maxp], done[maxp], dp[maxp]; 14 int n, src, target, tot; 15 int val[maxn][maxn], id[maxn][maxn]; 16 17 struct Edge { 18 int v, w, next; 19 Edge(int v = 0, int w = 0, int next = 0) : v(v), w(w), next(next) {} 20 } edges[maxp * 100]; 21 22 struct Node { 23 int d, u; 24 Node(int d = 0, int u = 0) : d(d), u(u) {} 25 bool operator < (const Node& o) const { 26 return d > o.d; 27 } 28 }; 29 30 void init() { 31 memset(head, -1, sizeof head); 32 tot = 0; 33 } 34 35 void add_edge(int u, int v, int w) { 36 edges[tot] = Edge(v, w, head[u]); 37 head[u] = tot++; 38 } 39 40 void dijkstra() { 41 priority_queue
             
               pq; 42 memset(done, 0, sizeof done); 43 for (int i = src; i <= target; i++) dist[i] = inf; 44 dist[src] = 0; 45 pq.push(Node(0, src)); 46 while (!pq.empty()) { 47 Node r = pq.top(); pq.pop(); 48 int u = r.u; 49 if (done[u]) continue; 50 done[u] = true; 51 for (int i = head[u]; ~i; i = edges[i].next) { 52 Edge e = edges[i]; 53 if (dist[e.v] > dist[u] + e.w) { 54 dist[e.v] = dist[u] + e.w; 55 pq.push(Node(dist[e.v], e.v)); 56 } 57 } 58 } 59 } 60 61 void print() { 62 for (int i = src; i <= target; i++) { 63 printf("%d: ", i); 64 for (int j = head[i]; ~j; j = edges[j].next) { 65 printf("(%d %d) ", edges[j].v, edges[j].w); 66 } 67 puts(""); 68 } 69 } 70 71 int walk(int u) { 72 if (u == target) return 1; 73 if (dp[u] != -1) return dp[u]; 74 int& ret = dp[u]; 75 ret = 0; 76 for (int i = head[u]; ~i; i = edges[i].next) { 77 Edge e = edges[i]; 78 if (dist[u] + e.w == dist[e.v]) { 79 ret += walk(e.v); 80 if (ret >= mod) ret -= mod; 81 } 82 } 83 return ret; 84 } 85 86 int main() { 87 // freopen("case.in", "r", stdin); 88 int T; 89 cin >> T; 90 while (T--) { 91 scanf("%d", &n); 92 for (int i = 0; i < n; i++) 93 for (int j = 0; j < n; j++) { 94 scanf("%d", &val[i][j]); 95 } 96 97 for (int i = 0, k = n - 1; i < n; i++, k--) 98 for (int j = 0, l = n - 1; j < n - i - 1; j++, l--) 99 val[i][j] += val[l][k];100 101 src = 0, target = 0;102 init();103 for (int i = 0; i < n; i++)104 for (int j = 0; j < n - i; j++)105 id[i][j] = target++;106 107 for (int i = 0; i < n; i++)108 for (int j = 0; j < n - i; j++) {109 int u = id[i][j];110 // cout << i << " " << j << " " << id[i][j] << targetl;111 for (int k = 0; k < 4; k++) {112 int ii = i + dx[k], jj = j + dy[k];113 if (ii >= 0 && ii < n && jj >= 0 && jj < n - ii) {114 int v = id[ii][jj];115 add_edge(u, v, val[ii][jj]);116 }117 }118 }119 120 for (int i = 0; i < n; i++) {121 int x = id[i][n - i - 1];122 add_edge(x, target, 0);123 add_edge(target, x, 0);124 }125 if (target >= maxp || tot > maxp * 100) {126 while (true);127 }128 129 // print();130 dijkstra();131 memset(dp, -1, sizeof dp);132 printf("%d\n", walk(src));133 }134 return 0;135 }
             
            
           
          
         
        
       
      

 

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。

 

1 /*zhen hao*/ 2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 using namespace std; 7   8 typedef long long ll; 9 const int maxn = 60, con = 1e9;10 int v[maxn], nv[maxn];11 int n, cnt, ans;12  13 int gcd(int a, int b) {14   if (!b) return a;15   return gcd(b, a % b);16 }17  18 void dfs(int now, int cur, ll LCM) {19   if (cur >= cnt) {20     if (now) ans += (con / LCM) * (now & 1 ? -1 : 1);21     return;22   }23   if (LCM > con) return;24   dfs(now, cur + 1, LCM);25   if (now) {26     if (LCM % nv[cur]) LCM *= nv[cur] / gcd(nv[cur], LCM);27     dfs(now + 1, cur + 1, LCM);28   } else {29     dfs(now + 1, cur + 1, nv[cur]);30   }31 }32  33 int main() {34 //  freopen("case.in", "r", stdin);35   int T;36   scanf("%d", &T);37   while (T--) {38     scanf("%d", &n);39     for (int i = 0; i < n; i++) scanf("%d", v + i);40     cnt = 0;41     sort(v, v + n);42  43     if (v[0] == 1) {44       puts("0"); continue;45     }46  47     for (int i = 0; i < n; i++) {48       bool ok = true;49       for (int j = 0; j < i; j++) if (v[i] % v[j] == 0) ok = false;50       if (ok) nv[cnt++] = v[i];51     }52     ans = con;53     dfs(0, 0, 1);54     printf("%lld\n", ans);55   }56   return 0;57 }