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。

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long double LD;
 #define N 50 + 5
 #define M 100000 + 5
 #define INF 1e9
 #define eps 1e-9

 int n, m, S, T, sum, tot;
 int A[N], B[N];
 int Head[N << ], q[N << ], Dfn[N << ];
 bool Map[N][N];
 LD l = 0.0, r;

 struct Edge
 {
     int next, node;
     LD flow;
 }h[M];

 inline void addedge(int u, int v, LD fl)
 {
     h[++ tot].next = Head[u], Head[u] = tot;
     h[tot].node = v, h[tot].flow = fl;
     h[++ tot].next = Head[v], Head[v] = tot;
     h[tot].node = u, h[tot].flow = ;
 }

 inline bool BFS()
 {
     for (int i = S; i <= T; i ++)
         Dfn[i] = ;
     int l = , r = ;
     q[] = S, Dfn[S] = ;
     while (l <= r)
     {
         int z = q[l ++];
         for (int i = Head[z]; i; i = h[i].next)
         {
             int d = h[i].node;
             LD p = h[i].flow;
             if (p < eps || Dfn[d]) continue ;
             Dfn[d] = Dfn[z] + ;
             q[++ r] = d;
             if (d == T) return ;
         }
     }
     return ;
 }

 inline LD dinic(int z, LD inflow)
 {
     if (z == T || inflow < eps) return inflow;
     LD ret = inflow, flow;
     for (int i = Head[z]; i; i = h[i].next)
     {
         int d = h[i].node;
         LD p = h[i].flow;
         if (Dfn[d] != Dfn[z] + ) continue ;
         flow = dinic(d, min(p, ret));
         ret -= flow;
         h[i].flow -= flow, h[i ^ ].flow += flow;
         if (ret < eps) return inflow;
     }
     if (fabs(inflow - ret) < eps) Dfn[z] = -;
     return inflow - ret;
 }

 inline bool Judge(LD k)
 {
     tot = ;
     for (int i = S; i <= T; i ++)
         Head[i] = ;
     for (int i = ; i <= m; i ++)
         addedge(S, i, k * B[i]);
     for (int i = ; i <= n; i ++)
         addedge(i + m, T, A[i]);
     for (int i = ; i <= m; i ++)
         for (int j = ; j <= n; j ++)
             if (Map[i][j]) addedge(i, j + m, INF);
     LD res = 0.0;
     while (BFS())
         res += dinic(S, INF);
     return fabs(res - sum) < eps;
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("3993.in", "r", stdin);
         freopen("3993.out", "w", stdout);
     #endif

     scanf("%d%d", &n, &m);
     S = , T = n + m + ;
     for (int i = ; i <= n; i ++)
     {
         scanf("%d", A + i);
         r += A[i];
         sum += A[i];
     }
     for (int i = ; i <= m; i ++)
         scanf("%d", B + i);
     for (int i = ; i <= m; i ++)
         for (int j = ; j <= n; j ++)
             scanf("%d", Map[i] + j);
     while (l + 1e- < r)
     {
         LD mid = (l + r) / ;
         if (Judge(mid)) r = mid;
             else l = mid;
     }
     printf("%.4lf\n", (double) l);

     #ifndef ONLINE_JUDGE
         fclose(stdin);
         fclose(stdout);
     #endif
     return ;
 }

3993_Gromah