牛客国庆集训派对Day6 Solution
A Birthday
思路：设置一个源点，一个汇点，每次对$源点对a_i, b_i , a_i 对 b_i 连一条流为1，费用为0的边$
每个点都再连一条 1, 3, 5, 7, ....的边到汇点之间，因为每多加一个流的费用，然后就是最小费用最大流
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const int INF = 0x3f3f3f3f; 6 7 const int maxn = 1010; 8 const int maxm = 10010; 9 10 struct Edge{ 11 int to, nxt, cap, flow, cost; 12 }edge[maxm]; 13 14 int head[maxn], tot; 15 int pre[maxn], dis[maxn]; 16 bool vis[maxn]; 17 int N; 18 19 void Init(int n) 20 { 21 N = n; 22 tot = 0; 23 memset(head, -1, sizeof head); 24 } 25 26 void addedge(int u, int v, int cap, int cost) 27 { 28 edge[tot].to = v; 29 edge[tot].cap = cap; 30 edge[tot].cost = cost; 31 edge[tot].flow = 0; 32 edge[tot].nxt = head[u]; 33 head[u] = tot++; 34 35 edge[tot].to = u; 36 edge[tot].cap = 0; 37 edge[tot].cost = -cost; 38 edge[tot].flow = 0; 39 edge[tot].nxt = head[v]; 40 head[v] = tot++; 41 } 42 43 bool SPFA(int s, int t) 44 { 45 queue<int>q; 46 for(int i = 0; i < N; ++i) 47 { 48 dis[i] = INF; 49 vis[i] = false; 50 pre[i] = -1; 51 } 52 dis[s] = 0; 53 vis[s] = true; 54 q.push(s); 55 while(!q.empty()) 56 { 57 int u = q.front(); 58 q.pop(); 59 vis[u] = false; 60 for(int i = head[u]; ~i; i = edge[i].nxt) 61 { 62 int v = edge[i].to; 63 if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost) 64 { 65 dis[v] = dis[u] + edge[i].cost; 66 pre[v] = i; 67 if(!vis[v]) 68 { 69 vis[v] = true; 70 q.push(v); 71 } 72 } 73 } 74 } 75 if(pre[t] == -1) return false; 76 else return true; 77 } 78 79 int solve(int s,int t) 80 { 81 int flow = 0 ; 82 int cost = 0; 83 while(SPFA(s, t)) 84 { 85 int Min = INF; 86 for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) 87 { 88 Min = min(Min, edge[i].cap - edge[i].flow); 89 } 90 for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) 91 { 92 edge[i].flow += Min; 93 edge[i ^ 1].flow -= Min; 94 cost += edge[i].cost * Min; 95 } 96 flow += Min; 97 } 98 return cost; 99 } 100 101 int n, m; 102 103 int main() 104 { 105 while(~scanf("%d %d", &n, &m)) 106 { 107 Init(n + m + 2); 108 int s = 0, t = n + m + 1; 109 for(int i = 1; i <= n; ++i) 110 { 111 int u, v; 112 scanf("%d %d", &u, &v); 113 addedge(0, i, 1, 0); 114 addedge(i, u + n, 1, 0); 115 addedge(i, v + n, 1, 0); 116 } 117 for(int i = n + 1; i <= n + m; ++i) 118 { 119 for(int j = 1; j < 100; j += 2) 120 { 121 addedge(i, t, 1, j); 122 } 123 } 124 int cost = solve(0, n + m + 1); 125 printf("%d\n", cost); 126 } 127 return 0; 128 }
B Board
思路一：考虑 $a[x], b[x]$ 分别表示第x行加了多少， 第x列加了多少 那么 $arr[x][y]$ 就可以表示成 $a[x] + b[y]$ 那么 就可以通过找一个点来解，比如说找一个点$x_1, y_1$ 那么 $a[x] + b[y] = a[x] + b[y_1] + a[x_1] + b[y] - a[x_1] - b[x_1] = arr[x][y_1] + arr[x_1][y] - arr[x_1][y_1]$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 6 int n; 7 int arr[N][N]; 8 9 int Move[4][2] = 10 { 11 -1, -1, 12 1, 1, 13 -1, 1, 14 1, -1, 15 }; 16 17 bool ok(int x, int y) 18 { 19 if (x < 1 || x > n || y < 1 || y > n) return false; 20 return true; 21 } 22 23 int main() 24 { 25 while (scanf("%d", &n) != EOF) 26 { 27 int x, y; 28 for (int i = 1; i <= n; ++i) 29 for (int j = 1; j <= n; ++j) 30 { 31 scanf("%d", &arr[i][j]); 32 if (arr[i][j] == -1) 33 x = i, y = j; 34 } 35 if (n == 1) 36 printf("%d\n", arr[1][1] == -1 ? 0 : arr[1][1]); 37 else 38 { 39 for (int i = 0; i < 4; ++i) 40 { 41 int nx = x + Move[i][0]; 42 int ny = y + Move[i][1]; 43 if (ok(nx, ny)) 44 { 45 printf("%d\n", arr[x][ny] + arr[nx][y] - arr[nx][ny]); 46 break; 47 } 48 } 49 } 50 } 51 return 0; 52 }
思路二：考虑分成n块，包含n行n列，那么每次对每一列加上一个数或者对每一行加上一个数，相当于对n块都加上了，也就是说，这n个块最后相加的数是相同的，然后就可以求解
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 1010 6 7 int n; 8 int arr[N][N]; 9 int sum[N]; 10 int x, y; 11 12 int main() 13 { 14 while(~scanf("%d", &n)) 15 { 16 memset(sum, 0, sizeof sum); 17 for(int i = 1; i <= n; ++i) 18 { 19 for(int j = 1; j <= n; ++j) 20 { 21 scanf("%d", &arr[i][j]); 22 if(arr[i][j] != -1) sum[(i + j) % n] += arr[i][j]; 23 else x = i, y = j; 24 } 25 } 26 int tmp = (x + y) % n; 27 int ans = 0; 28 if(tmp != 0) 29 { 30 ans = sum[0] - sum[tmp]; 31 } 32 else 33 { 34 ans = sum[1] - sum[tmp]; 35 } 36 printf("%d\n", ans); 37 } 38 return 0; 39 }
C Circle
水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int n; 5 6 int main() 7 { 8 while (scanf("%d", &n) != EOF) 9 printf("%d\n", n); 10 return 0; 11 }
D 土龙弟弟
留坑。
E Growth
思路：考虑将$x, y 分别离散 dp[x][y]$ 表示 $a_i, b_i 到达 x, y $的状态 的时候可以拿到多少奖励了
$value[x][y] 表示的是 到达x并且到达y 的奖励一共有多少$
$dp[i][j] = max(dp[i][j - 1] + value[i][j - 1] *(yy[j] - yy[j - 1]), dp[i - 1][j] + value[i - 1][j] * (xx[i] - xx[i - 1]));$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 #define ll long long 6 7 int n, mx, my; 8 ll m; 9 int xx[N], yy[N], zz[N]; 10 ll dp[N][N], value[N][N]; 11 12 struct node 13 { 14 int x, y, z; 15 void scan(int i) 16 { 17 scanf("%d%d%d", &x, &y, &z); 18 xx[i] = x; yy[i] = y; 19 } 20 }arr[N]; 21 22 void Init() 23 { 24 sort(xx + 1, xx + 1 + n); 25 sort(yy + 1, yy + 1 + n); 26 mx = unique(xx + 1, xx + 1 + n) - xx - 1; 27 my = unique(yy + 1, yy + 1 + n) - yy - 1; 28 memset(dp, 0, sizeof dp); 29 memset(value, 0, sizeof value); 30 for (int i = 1; i <= n; ++i) 31 { 32 arr[i].x = lower_bound(xx + 1, xx + 1 + mx, arr[i].x) - xx; 33 arr[i].y = lower_bound(yy + 1, yy + 1 + my, arr[i].y) - yy; 34 value[arr[i].x][arr[i].y] += arr[i].z; 35 } 36 } 37 38 int main() 39 { 40 while (scanf("%d%lld", &n, &m) != EOF) 41 { 42 for (int i = 1; i <= n; ++i) arr[i].scan(i); Init(); 43 for (int i = 1; i <= mx; ++i) 44 { 45 for (int j = 1; j <= my; ++j) 46 { 47 value[i][j] += value[i - 1][j] + value[i][j - 1] - value[i - 1][j - 1]; 48 } 49 } 50 for (int i = 1; i <= mx; ++i) 51 { 52 for (int j = 1; j <= my; ++j) 53 { 54 dp[i][j] = max(dp[i][j], dp[i - 1][j] + value[i - 1][j] * (xx[i] - xx[i - 1])); 55 dp[i][j] = max(dp[i][j], dp[i][j - 1] + value[i][j - 1] * (yy[j] - yy[j - 1])); 56 } 57 } 58 ll ans = 0; 59 for (int i = 1; i <= mx; ++i) 60 { 61 for (int j = 1; j <= my; ++j) 62 { 63 if(xx[i] + yy[j] <= m) ans = max(ans, dp[i][j] + value[i][j] * (m - xx[i] - yy[j] + 1)); 64 } 65 } 66 printf("%lld\n", ans); 67 } 68 return 0; 69 }
F kingdom
留坑。
G Matrix
留坑。
H Mountain
水。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 1010 6 7 int n; 8 int arr[N]; 9 10 int main() 11 { 12 while(~scanf("%d", &n)) 13 { 14 for(int i = 1; i <= n; ++i) scanf("%d", arr + i); 15 int ans = 0; 16 for(int i = 1; i <= n; ++i) 17 { 18 ans = max(ans, arr[i]); 19 } 20 ans = ans * 2; 21 printf("%d\n", ans); 22 } 23 return 0; 24 }
I 清明梦超能力者黄YY
思路：树链剖分维护u->v 路径上染色次数，倒着操作，每次找有没有第k次染色的点，有的话拿出来更新的答案，并且把那个点的然后次数置位-INF, 每个点最多拿出来一次
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define INF 0x3f3f3f3f 6 7 int n, m, k; 8 vector <int> G[N]; 9 int ans[N]; 10 11 struct QUE 12 { 13 int u, v, c; 14 void scan() 15 { 16 scanf("%d%d%d", &u, &v, &c); 17 } 18 }que[N]; 19 20 int top[N], fa[N], deep[N], num[N], p[N], fp[N],son[N], pos; 21 22 void Init() 23 { 24 memset(son, -1, sizeof son); 25 pos = 0; 26 fa[1] = 1, deep[1] = 0; 27 } 28 29 void DFS(int u) 30 { 31 num[u] = 1; 32 for (auto v : G[u]) if (v != fa[u]) 33 { 34 fa[v] = u; deep[v] = deep[u] + 1; 35 DFS(v); num[u] += num[v]; 36 if (son[u] == -1 || num[v] > num[son[u]]) 37 son[u] = v; 38 } 39 } 40 41 void getpos(int u, int sp) 42 { 43 top[u] = sp; 44 p[u] = ++pos; 45 fp[pos] = u; 46 if (son[u] == -1) return; 47 getpos(son[u], sp); 48 for (auto v : G[u]) if (v != son[u] && v != fa[u]) 49 getpos(v, v); 50 } 51 52 struct SEG 53 { 54 struct node 55 { 56 int l, r; 57 int lazy, Max, pos; 58 node () {} 59 node (int _l, int _r) 60 { 61 l = _l, r = _r; 62 Max = 0; 63 pos = _l; 64 lazy = 0; 65 } 66 }a[N << 2]; 67 68 void pushup(int id) 69 { 70 if (a[id << 1].Max > a[id << 1 | 1].Max) 71 { 72 a[id].Max = a[id << 1].Max; 73 a[id].pos = a[id << 1].pos; 74 } 75 else 76 { 77 a[id].Max = a[id << 1 | 1].Max; 78 a[id].pos = a[id << 1 | 1].pos; 79 } 80 } 81 82 void pushdown(int id) 83 { 84 if (a[id].l >= a[id].r) return; 85 if (a[id].lazy) 86 { 87 int lazy = a[id].lazy; a[id].lazy = 0; 88 a[id << 1].lazy += lazy; 89 a[id << 1 | 1].lazy += lazy; 90 a[id << 1].Max += lazy; 91 a[id << 1 | 1].Max += lazy; 92 } 93 } 94 95 void build(int id, int l, int r) 96 { 97 a[id] = node(l, r); 98 if (l == r) return; 99 int mid = (l + r) >> 1; 100 build(id << 1, l, mid); 101 build(id << 1 | 1, mid + 1, r); 102 pushup(id); 103 } 104 105 void update(int id, int l, int r, int val) 106 { 107 if (a[id].l >= l && a[id].r <= r) 108 { 109 a[id].Max += val; 110 a[id].lazy += val; 111 return; 112 } 113 pushdown(id); 114 int mid = (a[id].l + a[id].r) >> 1; 115 if (l <= mid) update(id << 1, l, r, val); 116 if (r > mid) update(id << 1 | 1, l, r, val); 117 pushup(id); 118 } 119 }segtree; 120 121 void change(int u, int v, int val) 122 { 123 int fu = top[u], fv = top[v]; 124 while (fu != fv) 125 { 126 if (deep[fu] < deep[fv]) 127 { 128 swap(fu, fv); 129 swap(u, v); 130 } 131 int l = p[fu], r = p[u]; 132 segtree.update(1, l, r, 1); 133 while (1) 134 { 135 // puts("bug"); 136 if (segtree.a[1].Max < k) break; 137 int pos = segtree.a[1].pos; 138 // cout << pos << " " << segtree.a[1].Max << endl; 139 ans[fp[pos]] = val; 140 segtree.update(1, pos, pos, -INF); 141 // cout << pos << " " << segtree.a[1].Max << endl; 142 } 143 u = fa[fu], fu = top[u]; 144 } 145 if (deep[u] > deep[v]) swap(u, v); 146 int l = p[u], r = p[v]; 147 // printf("%d %d\n", l, r); 148 segtree.update(1, l, r, 1); 149 while (1) 150 { 151 if (segtree.a[1].Max < k) break; 152 int pos = segtree.a[1].pos; 153 // cout << pos << endl; 154 ans[fp[pos]] = val; 155 // cout << segtree.a[1].pos << " " << segtree.a[1].Max << endl; 156 segtree.update(1, pos, pos, -INF); 157 // cout << segtree.a[1].pos << " " << segtree.a[1].Max << endl; 158 // puts("..................."); 159 } 160 } 161 162 int main() 163 { 164 while (scanf("%d%d%d", &n, &m, &k) != EOF) 165 { 166 Init(); 167 for (int i = 1, u, v; i < n; ++i) 168 { 169 scanf("%d%d", &u, &v); 170 G[u].push_back(v); 171 G[v].push_back(u); 172 } DFS(1), getpos(1, 1); 173 // for (int i = 1; i <= n; ++i) printf("%d %d %d\n", i, p[i], fp[i]); 174 for (int i = 1; i <= m; ++i) que[i].scan(); 175 memset(ans, 0, sizeof ans); 176 segtree.build(1, 1, n); 177 for (int i = m; i >= 1; --i) change(que[i].u, que[i].v, que[i].c); 178 for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]); 179 } 180 return 0; 181 }
J 最短路
思路一（应该不是正解）：对于环内的点，拿出来，分别跑Dijkstra, 这样的点数应该在100左右，然后每次询问的时候更新答案。但是有一点奇怪的是，用并查集找出环内的点，就不会T，标记访问直接DFS找就会T。
1 #include <bits/stdc++.h> 2 using namespace std; 3 using pii = pair <int, int>; 4 5 #define N 101010 6 7 const int INF = 0x3f3f3f3f; 8 9 template <class T> 10 inline bool scan_d(T &ret) 11 { 12 char c; int sgn; 13 if (c = getchar(), c == EOF) return 0; 14 while (c != '-' && (c < '0' || c > '9')) c = getchar(); 15 sgn = (c == '-') ? -1 : 1; 16 ret = (c == '-') ? 0 : (c - '0'); 17 while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0'); 18 ret *= sgn; 19 return 1; 20 } 21 22 inline void out(int x) 23 { 24 if (x / 10) out(x / 10); 25 putchar(x % 10 + '0'); 26 } 27 28 vector <int> G[N]; 29 int vis[N], deep[N], stk[N], top; 30 31 struct ST 32 { 33 int mm[N << 1]; 34 int dp[N << 1][20]; 35 int rmq[N << 1]; 36 int F[N << 1], P[N], cnt; 37 38 void Init(int n) 39 { 40 mm[0] = -1; 41 for (int i = 1; i <= n; ++i) 42 { 43 mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1]; 44 dp[i][0] = i; 45 } 46 for (int j = 1; j <= mm[n]; ++j) 47 for (int i = 1; i + (1 << j) - 1 <= n; ++i) 48 dp[i][j] = rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1]; 49 } 50 51 int query(int a, int b) 52 { 53 if (a > b) swap(a, b); 54 int k = mm[b - a + 1]; 55 return rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? dp[a][k] : dp[b - (1 << k) + 1][k]; 56 } 57 58 void DFS(int u, int fa) 59 { 60 F[++cnt] = u; 61 rmq[cnt] = deep[u]; 62 P[u] = cnt; 63 vis[u] = 1; 64 for (auto v : G[u]) if (v != fa) 65 { 66 if (vis[v]) 67 { 68 stk[++top] = v; 69 continue; 70 } 71 deep[v] = deep[u] + 1; 72 DFS(v, u); 73 F[++cnt] = u; 74 rmq[cnt] = deep[u]; 75 } 76 } 77 78 void init(int root, int node_num) 79 { 80 cnt = 0; 81 deep[1] = 0; 82 DFS(root, root); 83 Init(2 * node_num - 1); 84 } 85 86 int query_lca(int u, int v) 87 { 88 return F[query(P[u], P[v])]; 89 } 90 }st; 91 92 int n, m; 93 int dis[210][N]; 94 queue <int> q; 95 96 void Dijkstra(int it, int st) 97 { 98 for (int i = 1; i <= n; ++i) 99 { 100 vis[i] = 0; 101 dis[it][i] = INF; 102 } 103 dis[it][st] = 0; 104 q.push(st); 105 while (!q.empty()) 106 { 107 int u = q.front(); q.pop(); 108 if (vis[u]) continue; 109 vis[u] = 1; 110 for (auto v : G[u]) 111 { 112 if (dis[it][v] > dis[it][u] + 1) 113 { 114 dis[it][v] = dis[it][u] + 1; 115 q.push(v); 116 } 117 } 118 } 119 } 120 121 vector <pii> tmp; 122 123 int pre[N]; 124 125 int find(int x) 126 { 127 if (x != pre[x]) 128 pre[x] = find(pre[x]); 129 return pre[x]; 130 } 131 132 void Run() 133 { 134 while (scan_d(n), scan_d(m)) 135 { 136 for (int i = 1; i <= n; ++i) pre[i] = i; top = 0; 137 for (int i = 1, u, v; i <= m; ++i) 138 { 139 scan_d(u), scan_d(v); 140 int fu = find(u), fv = find(v); 141 if (fu == fv) 142 { 143 tmp.emplace_back(u, v); 144 stk[++top] = u; 145 } 146 else 147 { 148 G[u].push_back(v); 149 G[v].push_back(u); 150 pre[fu] = fv; 151 } 152 } 153 st.init(1, n); 154 sort(stk + 1, stk + 1 + top); 155 int nn = unique(stk + 1, stk + 1 + top) - stk - 1; 156 for (auto it : tmp) 157 { 158 G[it.first].push_back(it.second); 159 G[it.second].push_back(it.first); 160 } 161 for (int i = 1; i <= nn; ++i) Dijkstra(i, stk[i]); 162 int q; 163 scan_d(q); 164 while (q--) 165 { 166 int u, v; 167 scan_d(u), scan_d(v); 168 int root = st.query_lca(u, v); 169 int ans = deep[u] + deep[v] - 2 * deep[root]; 170 for (int i = 1; i <= nn; ++i) 171 ans = min(ans, dis[i][u] + dis[i][v]); 172 out(ans), putchar('\n'); 173 } 174 } 175 } 176 int main() 177 { 178 #ifdef LOCAL 179 freopen("Test.in", "r", stdin); 180 #endif 181 182 Run(); 183 return 0; 184 }
K 排序
留坑。
