AcWing 374 二分 + 二分图最大匹配

题意

传送门 Acwing 374 导弹防御塔

题解

击退入侵者的时间满足单调性，二分答案。则转化为指派问题，将入侵者作为左部节点，防御塔与发射时间的二元组作为右部节点。在二分值代表的时间内若防御塔 $T_1+k\times (T_1+T_2),k\ge 0$ 时发射的炮弹能击退入侵者，则在防御塔与入侵者之间连边。求二分图最大匹配，判断是否等于 $M$ 即可。

单次求解二分图最大匹配时间复杂度 $O(VE)$，本题中为 $O(N^5)$，显然难以胜任。观察到左部节点为 $O(N)$，右部节点为 $O(N^2)$，那么从节点数较小的左部 $DFS$ 寻找增广路，此时单次求解二分图最大匹配时间复杂度为 $O(N^4)$。总时间复杂度 $O(N^4\log [(T_1+T_2)\times M])$。

需要注意，$T_1$ 的单位为秒，$T_2$ 的单位是分。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 55, maxv = maxn + maxn * maxn, maxe = maxn * maxn * maxn;
const double eps = 1e-8;
int N, M, V, match[maxv];
int X1[maxn], Y1[maxn], X2[maxn], Y2[maxn];
int tot, head[maxv], to[maxe], nxt[maxe];
double T1, T2, T3[maxn][maxn];
bool vs[maxn];

inline void add(int x, int y)
{
    
    
    to[++tot] = y, nxt[tot] = head[x], head[x] = tot;
}

bool dfs(int x)
{
    
    
    vs[x] = 1;
    for (int i = head[x]; i; i = nxt[i])
    {
    
    
        int y = to[i], z = match[y];
        if (!z || (!vs[z] && dfs(z)))
        {
    
    
            match[y] = x;
            return 1;
        }
    }
    return 0;
}

bool judge(double x)
{
    
    
    memset(head, 0, sizeof(head));
    tot = 0;
    int t = (x - T1) / (T1 + T2) + 1;
    for (int i = 1; i <= M; ++i)
        for (int j = 1; j <= N; ++j)
        {
    
    
            double r = x - T3[i][j] - T1;
            int cnt = r / (T1 + T2) + (r >= 0);
            for (int k = 1, x, y; k <= cnt; ++k)
                x = i, y = (j - 1) * t + k + M, add(x, y), add(y, x);
        }
    memset(match, 0, sizeof(match));
    int res = 0;
    for (int i = 1; i <= M; ++i)
        memset(vs, 0, sizeof(vs)), res += dfs(i);
    return res == M;
}

int main()
{
    
    
    scanf("%d%d%lf%lf%d", &N, &M, &T1, &T2, &V);
    T1 /= 60;
    for (int i = 1; i <= M; ++i)
        scanf("%d%d", X1 + i, Y1 + i);
    for (int i = 1; i <= N; ++i)
        scanf("%d%d", X2 + i, Y2 + i);
    for (int i = 1; i <= M; ++i)
        for (int j = 1; j <= N; ++j)
        {
    
    
            int dx = X1[i] - X2[j], dy = Y1[i] - Y2[j];
            T3[i][j] = sqrt(dx * dx + dy * dy) / V;
        }
    double lb = T1, ub = T1 + 49 * (T1 + T2) + 1;
    while (ub - lb > eps)
    {
    
    
        double mid = (lb + ub) / 2;
        if (judge(mid))
            ub = mid;
        else
            lb = mid;
    }
    printf("%.6f\n", ub);
    return 0;
}