BZOJ 3993 [SDOI 2015] 星际战争 解题报告
首先我们可以二分答案。
假设当前二分出来的答案是 $Ans$ ，那么我们考虑用网络流检验：
设武器为 $X$，第 $i$ 个武器的攻击力为 $B_i$；
设机器人为 $Y$，第 $i$ 个机器人的装甲为 $A_i$；
设 $Map[i][j]$ 表示第 $i$ 个机器人是否能攻击第 $j$ 号机器人。
设源为 $S$，汇为 $T$，现在考虑连边：
- $S\rightarrow X_i$，容量为 $Ans * B_i$；
- $Y_i\rightarrow T$，容量为 $A_i$；
- $\forall (i,j),Map[i][j]=1:X_i\rightarrow Y_j$，容量为 $\infty$
然后跑网络流，假设最大流为 $M$，那么看是否有：$M=\sum A_i$。
如果是，那么说明当前答案是满足的，更新上界，否则更新下界。
毕竟 Gromah 太弱，只会做水题。
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 typedef long double LD; 8 #define N 50 + 5 9 #define M 100000 + 5 10 #define INF 1e9 11 #define eps 1e-9 12 13 int n, m, S, T, sum, tot; 14 int A[N], B[N]; 15 int Head[N << 1], q[N << 1], Dfn[N << 1]; 16 bool Map[N][N]; 17 LD l = 0.0, r; 18 19 struct Edge 20 { 21 int next, node; 22 LD flow; 23 }h[M]; 24 25 inline void addedge(int u, int v, LD fl) 26 { 27 h[++ tot].next = Head[u], Head[u] = tot; 28 h[tot].node = v, h[tot].flow = fl; 29 h[++ tot].next = Head[v], Head[v] = tot; 30 h[tot].node = u, h[tot].flow = 0; 31 } 32 33 inline bool BFS() 34 { 35 for (int i = S; i <= T; i ++) 36 Dfn[i] = 0; 37 int l = 1, r = 1; 38 q[1] = S, Dfn[S] = 1; 39 while (l <= r) 40 { 41 int z = q[l ++]; 42 for (int i = Head[z]; i; i = h[i].next) 43 { 44 int d = h[i].node; 45 LD p = h[i].flow; 46 if (p < eps || Dfn[d]) continue ; 47 Dfn[d] = Dfn[z] + 1; 48 q[++ r] = d; 49 if (d == T) return 1; 50 } 51 } 52 return 0; 53 } 54 55 inline LD dinic(int z, LD inflow) 56 { 57 if (z == T || inflow < eps) return inflow; 58 LD ret = inflow, flow; 59 for (int i = Head[z]; i; i = h[i].next) 60 { 61 int d = h[i].node; 62 LD p = h[i].flow; 63 if (Dfn[d] != Dfn[z] + 1) continue ; 64 flow = dinic(d, min(p, ret)); 65 ret -= flow; 66 h[i].flow -= flow, h[i ^ 1].flow += flow; 67 if (ret < eps) return inflow; 68 } 69 if (fabs(inflow - ret) < eps) Dfn[z] = -1; 70 return inflow - ret; 71 } 72 73 inline bool Judge(LD k) 74 { 75 tot = 1; 76 for (int i = S; i <= T; i ++) 77 Head[i] = 0; 78 for (int i = 1; i <= m; i ++) 79 addedge(S, i, k * B[i]); 80 for (int i = 1; i <= n; i ++) 81 addedge(i + m, T, A[i]); 82 for (int i = 1; i <= m; i ++) 83 for (int j = 1; j <= n; j ++) 84 if (Map[i][j]) addedge(i, j + m, INF); 85 LD res = 0.0; 86 while (BFS()) 87 res += dinic(S, INF); 88 return fabs(res - sum) < eps; 89 } 90 91 int main() 92 { 93 #ifndef ONLINE_JUDGE 94 freopen("3993.in", "r", stdin); 95 freopen("3993.out", "w", stdout); 96 #endif 97 98 scanf("%d%d", &n, &m); 99 S = 0, T = n + m + 1; 100 for (int i = 1; i <= n; i ++) 101 { 102 scanf("%d", A + i); 103 r += A[i]; 104 sum += A[i]; 105 } 106 for (int i = 1; i <= m; i ++) 107 scanf("%d", B + i); 108 for (int i = 1; i <= m; i ++) 109 for (int j = 1; j <= n; j ++) 110 scanf("%d", Map[i] + j); 111 while (l + 1e-4 < r) 112 { 113 LD mid = (l + r) / 2; 114 if (Judge(mid)) r = mid; 115 else l = mid; 116 } 117 printf("%.4lf\n", (double) l); 118 119 #ifndef ONLINE_JUDGE 120 fclose(stdin); 121 fclose(stdout); 122 #endif 123 return 0; 124 }
