牛客国庆集训派对Day7 Solution
A Relic Discovery
水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int t, n; 5 6 int main() 7 { 8 scanf("%d", &t); 9 while (t--) 10 { 11 scanf("%d", &n); 12 int res = 0; 13 for (int i = 1, a, b; i <= n; ++i) 14 { 15 scanf("%d%d", &a, &b); 16 res += a * b; 17 } 18 printf("%d\n", res); 19 } 20 return 0; 21 }
B Pocket Cube
按题意模拟即可，注意顺时针转逆时针转均可，也就是说一个面有八种旋转可能
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int t; 5 int G[10][10], tmp[10][10]; 6 7 bool ok() 8 { 9 for (int i = 1; i <= 6; ++i) 10 for (int j = 2; j <= 4; ++j) 11 if (tmp[i][j] != tmp[i][j - 1]) 12 return false; 13 return true; 14 } 15 16 void clear() 17 { 18 for (int i = 1; i <= 6; ++i) 19 for (int j = 1; j <= 4; ++j) 20 tmp[i][j] = G[i][j]; 21 } 22 23 int origin[][8][2] = 24 { 25 // top 26 {4, 3, 4, 4, 6, 1, 6, 3, 2, 2, 2, 1, 5, 4, 5, 2,}, 27 // bottom 28 {4, 1, 4, 2, 6, 1, 6, 3, 2, 4, 2, 3, 5, 3, 5, 1,}, 29 // left 30 {1, 1, 1, 3, 2, 1, 2, 3, 3, 1, 3, 3, 4, 1, 4, 3,}, 31 // right 32 {1, 4, 1, 2, 4, 4, 4, 2, 3, 4, 3, 2, 2, 4, 2, 2,}, 33 // front 34 {1, 3, 1, 4, 6, 3, 6, 4, 3, 2, 3, 1, 5, 3, 5, 4,}, 35 // back 36 {3, 3, 3, 4, 6, 2, 6, 1, 1, 2, 1, 1, 5, 2, 5, 1,}, 37 }; 38 39 int Move[][8][2] = 40 { 41 {5, 4, 5, 2, 4, 3, 4, 4, 6, 1, 6, 3, 2, 2, 2, 1,}, 42 {5, 3, 5, 1, 4, 1, 4, 2, 6, 1, 6, 3, 2, 4, 2, 3,}, 43 {4, 1, 4, 3, 1, 1, 1, 3, 2, 1, 2, 3, 3, 1, 3, 3,}, 44 {2, 4, 2, 2, 1, 4, 1, 2, 4, 4, 4, 2, 3, 4, 3, 2,}, 45 {5, 3, 5, 4, 1, 3, 1, 4, 6, 3, 6, 4, 3, 2, 3, 1,}, 46 {5, 2, 5, 1, 3, 3, 3, 4, 6, 2, 6, 1, 1, 2, 1, 1,}, 47 }; 48 49 bool work() 50 { 51 clear(); if (ok()) return true; 52 for (int i = 0; i < 6; ++i) 53 { 54 for (int j = 0; j < 8; j += 2) 55 { 56 clear(); 57 for (int k = 0; k < 8; ++k) 58 { 59 int x = (j + k) % 8; 60 tmp[origin[i][x][0]][origin[i][x][1]] = G[Move[i][x][0]][Move[i][x][1]]; 61 } 62 if (ok()) return true; 63 } 64 for (int j = 1; j < 8; j += 2) 65 { 66 clear(); 67 for (int k = 0; k > -8; --k) 68 { 69 int x = (j + k + 8) % 8; 70 int y = (x + 4) % 8; 71 tmp[origin[i][x][0]][origin[i][x][1]] = G[Move[i][y][0]][Move[i][y][1]]; 72 } 73 if (ok()) return true; 74 } 75 } 76 return false; 77 } 78 79 int main() 80 { 81 scanf("%d", &t); 82 while(t--) 83 { 84 for (int i = 1; i <= 6; ++i) 85 for (int j = 1; j <= 4; ++j) 86 scanf("%d", &G[i][j]); 87 puts(work() ? "YES" : "NO"); 88 } 89 return 0; 90 }
C Pocky
根据样例发现规律 答案为 log(l / d) + 1.0 如果 l <= d 答案为0
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const double eps = 1e-8; 5 6 int t; 7 double l, d; 8 9 int main() 10 { 11 scanf("%d", &t); 12 while (t--) 13 { 14 scanf("%lf%lf", &l, &d); 15 if (l <= d || fabs(l - d) <= eps) 16 printf("%.6f\n", 0.0); 17 else 18 printf("%.6f\n", log(l / d) + 1.0); 19 } 20 return 0; 21 }
D Lucky Coins
留坑。
E Fibonacci
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F Lambda Calculus
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G Coding Contest
题意：有n个区域，每个区域有$s_i$个人以及$b_i$个食物，有m条路，每条路最多让$c_i$个人走，其中第一个人走不产生影响，后面的每个人都有p的可能破坏 求最小的破坏可能
思路：将p取log就变成了累加最小，建立一个源点以及一个汇点，其中每块区域需要出走的人和汇点相连，可以进来的人与源点相连，跑一遍最小费用最大流即可
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const double eps = 1e-8; 6 const double EI = exp(1.0); 7 const int maxn = 10000; 8 const int maxm = 100000; 9 const int INF = 0x3f3f3f3f; 10 11 int sgn(double x) 12 { 13 if(fabs(x) < eps) return 0; 14 else return x > 0 ? 1 : -1; 15 } 16 17 struct Edge{ 18 int to, nxt, cap, flow; 19 double cost; 20 }edge[maxm]; 21 22 double ans; 23 int head[maxn], tot; 24 int pre[maxn]; 25 double dis[maxn]; 26 bool vis[maxn]; 27 int N; 28 29 void Init(int n) 30 { 31 N = n; 32 tot = 0; 33 for(int i = 0; i <= n; ++i) head[i] = -1; 34 } 35 36 void addedge(int u, int v,int cap, double cost) 37 { 38 edge[tot].to = v; 39 edge[tot].cap = cap; 40 edge[tot].cost = cost; 41 edge[tot].flow = 0; 42 edge[tot].nxt = head[u]; 43 head[u] = tot++; 44 45 edge[tot].to = u; 46 edge[tot].cap = 0; 47 edge[tot].cost = -cost; 48 edge[tot].flow = 0; 49 edge[tot].nxt = head[v]; 50 head[v] = tot++; 51 } 52 53 bool SPFA(int s,int t) 54 { 55 queue<int>q; 56 for(int i = 0; i < N; ++i) 57 { 58 dis[i] = INF; 59 vis[i] = false; 60 pre[i] = -1; 61 } 62 dis[s] = 0; 63 vis[s] = true; 64 q.push(s); 65 while(!q.empty()) 66 { 67 int u = q.front(); 68 q.pop(); 69 vis[u] = false; 70 for(int i = head[u]; ~i; i = edge[i].nxt) 71 { 72 int v = edge[i].to; 73 if(edge[i].cap > edge[i].flow && sgn(dis[v] - dis[u] - edge[i].cost) > 0) 74 { 75 dis[v] = dis[u] + edge[i].cost; 76 pre[v] = i; 77 if(!vis[v]) 78 { 79 vis[v] = true; 80 q.push(v); 81 } 82 } 83 } 84 } 85 if(pre[t] == -1) return false; 86 else return true; 87 } 88 89 int minCostMaxflow(int s, int t) 90 { 91 int flow = 0; 92 ans = 0; 93 while(SPFA(s, t)) 94 { 95 int Min = INF; 96 for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) 97 { 98 Min = min(Min, edge[i].cap - edge[i].flow); 99 } 100 for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) 101 { 102 edge[i].flow += Min; 103 edge[i ^ 1].flow -= Min; 104 ans += edge[i].cost * Min; 105 } 106 flow += Min; 107 } 108 return flow; 109 } 110 111 int n, m; 112 int s[maxn], b[maxn], tmp[maxn]; 113 114 int main() 115 { 116 int t; 117 scanf("%d", &t); 118 while(t--) 119 { 120 scanf("%d %d", &n, &m); 121 Init(n + 2); 122 for(int i = 1; i <= n; ++i) 123 { 124 scanf("%d %d", s + i, b + i); 125 if(s[i] - b[i] > 0) addedge(0, i, s[i] - b[i], 0); 126 if(s[i] - b[i] < 0) addedge(i, n + 1, b[i] - s[i], 0); 127 } 128 129 for(int i = 1; i <= m; ++i) 130 { 131 int u, v, c; 132 double p; 133 scanf("%d %d %d %lf", &u, &v, &c ,&p); 134 p = -log(1.0 - p); 135 if(c > 0) addedge(u, v, 1, 0); 136 if(c > 1) addedge(u, v, c - 1, p); 137 } 138 int flow = minCostMaxflow(0, n + 1); 139 ans = pow(EI, -ans); 140 printf("%.2f\n", 1.0 - ans); 141 } 142 return 0; 143 }
H Pattern
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I Travel Brochure
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J Cliques
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K Finding Hotels
题意：有若干酒店，以及一些旅客，每个旅客有价格接受范围，对于每个旅客要找出可接受价格以内的所有酒店中最近的
思路：KDTree 对于价格直接特判，如果超出，直接return INF 还要注意多解的情况下取id小的
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 #define DIM 10 7 #define INF 0x3f3f3f3f3f3f3f3f 8 inline ll sqr(ll x) { return x * x; } 9 namespace KD 10 { 11 int K; 12 struct Point 13 { 14 ll x[DIM]; 15 int id; 16 ll distance(const Point &b) const 17 { 18 if (x[2] < b.x[2]) return INF; 19 ll ret = 0; 20 for (int i = 0; i < K; ++i) ret += sqr(x[i] - b.x[i]); 21 return ret; 22 } 23 void input(int id) { this->id = id; for (int i = 0; i < K + 1; ++i) scanf("%lld", x + i); } 24 void output() { for (int i = 0; i < K + 1; ++i) printf("%lld%c", x[i], " \n"[i == K]); } 25 }; 26 struct cmpx 27 { 28 int div; 29 cmpx(const int &div) { this->div = div; } 30 bool operator () (const Point &a, const Point &b) 31 { 32 for (int i = 0; i < K; ++i) if (a.x[(div + i) % K] != b.x[(div + i) % K]) 33 return a.x[(div + i) % K] < b.x[(div + i) % K]; 34 return true; 35 } 36 }; 37 bool cmp(const Point &a, const Point &b, int div) { return cmpx(div)(a, b); } 38 struct node 39 { 40 Point e; 41 node *lc, *rc; 42 int div; 43 }pool[N], *tail, *root; 44 void init() { tail = pool; } 45 node* build(Point *a, int l, int r, int div) 46 { 47 if (l >= r) return NULL; 48 node *p = tail++; 49 p->div = div; 50 int mid = (l + r) >> 1; 51 nth_element(a + l, a + mid, a + r, cmpx(div)); 52 p->e = a[mid]; 53 p->lc = build(a, l, mid, (div + 1) % K); 54 p->rc = build(a, mid + 1, r, (div + 1) % K); 55 return p; 56 } 57 Point res; ll Min; 58 void search(Point p, node *x, int div) 59 { 60 if (!x) return; 61 if (cmp(p, x->e, div)) 62 { 63 search(p, x->lc, (div + 1) % K); 64 ll tmp = p.distance(x->e); 65 if (tmp < Min || tmp == Min && x->e.id < res.id) 66 { 67 Min = tmp; 68 res = x->e; 69 } 70 if (sqr(x->e.x[div] - p.x[div]) <= Min) 71 search(p, x->rc, (div + 1) % K); 72 } 73 else 74 { 75 search(p, x->rc, (div + 1) % K); 76 ll tmp = p.distance(x->e); 77 if (tmp < Min || tmp == Min && x->e.id < res.id) 78 { 79 Min = tmp; 80 res = x->e; 81 } 82 if (sqr(x->e.x[div] - p.x[div]) <= Min) 83 search(p, x->lc, (div + 1) % K); 84 } 85 } 86 void search(Point p) 87 { 88 Min = INF; 89 search(p, root, 0); 90 } 91 } 92 93 int t, n, q; 94 KD::Point p[N]; 95 96 void Run() 97 { 98 for (scanf("%d", &t); t--; ) 99 { 100 KD::K = 2; 101 scanf("%d%d", &n, &q); 102 for (int i = 0; i < n; ++i) p[i].input(i); 103 KD::init(); 104 KD::root = KD::build(p, 0, n, 0); 105 for (int i = 1; i <= q; ++i) 106 { 107 KD::Point o; o.input(0); 108 KD::search(o); 109 KD::res.output(); 110 } 111 } 112 } 113 114 int main() 115 { 116 #ifdef LOCAL 117 freopen("Test.in", "r", stdin); 118 #endif 119 120 Run(); 121 return 0; 122 123 }
L Tower Attack
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M Generator and Monitor
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