【HNOI2014】画框

题面

题解

这又是一种套路啊233

将\(\sum a_i\)和\(\sum b_i\)分别看做\(x\)和\(y\)，投射到平面直角坐标系中，于是就是找\(xy\)最小的点

于是可以先找出\(x\)最小的点\(\mathrm{A}\)和\(y\)最小的点\(\mathrm{B}\)，然后找到在\(\mathrm{AB}\)下方的最远的点\(\mathrm{C}\)

即\(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\)最小
\[ \begin{aligned} \because \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} &= (x_{\mathrm{B}} - x_{\mathrm{A}})(y_{\mathrm{C}} - y_{\mathrm{A}}) - (y_{\mathrm{B}} - y_{\mathrm{A}})(x_\mathrm{C} - x_\mathrm{A}) \\ &= (x_\mathrm B - x_\mathrm A) \times y_\mathrm C + (y_\mathrm A - y_\mathrm B) \times x_\mathrm C + y_\mathrm B x_\mathrm A - x_\mathrm B y_\mathrm A \end{aligned} \]
于是将权值改成\(\mathrm{g}[i][j] = (y_\mathrm A - y_\mathrm B) \times a[i][j] + (x_\mathrm B - x_\mathrm A)\times b[i][j]\)，然后用\(\mathrm{KM}\)找出\(\mathrm C\)。

找到\(\mathrm C\)之后用叉积判断一下\(\mathrm C\)是不是在\(\mathrm{AB}\)的下方，如果是的话，就递归处理\(\mathrm{AC, CB}\)

复杂度\(\mathrm{O}(\)能过\()\)

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<climits>
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
    int data = 0, w = 1; char ch = getchar();
    while(ch != '-' && (!isdigit(ch))) ch = getchar();
    if(ch == '-') w = -1, ch = getchar();
    while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
    return data * w;
}

const int N(75);
int T, n, a[N][N], b[N][N], ans;
struct vector { int x, y; };
inline vector operator - (const vector &lhs, const vector &rhs)
    { return (vector) {lhs.x - rhs.x, lhs.y - rhs.y}; }
inline int operator * (const vector &lhs, const vector &rhs)
    { return lhs.x * rhs.y - lhs.y * rhs.x; }
int g[N][N], lx[N], ly[N], visx[N], visy[N], match[N];
bool hungary(int x)
{
    visx[x] = 1;
    for(RG int to = 1; to <= n; to++)
        if(!visy[to] && lx[x] + ly[to] == g[x][to])
        {
            visy[to] = true;
            if(!match[to] || hungary(match[to])) return (match[to] = x, true);
        }
    return false;
}

void build(int valx, int valy)
{
    for(RG int i = 1; i <= n; i++) for(RG int j = 1; j <= n; j++)
        g[i][j] = -(valx * a[i][j] + valy * b[i][j]);
}

vector KM()
{
    for(RG int i = 1; i <= n; i++)
        ly[i] = 0, lx[i] = *std::max_element(g[i] + 1, g[i] + n + 1);
    memset(match, 0, sizeof match);
    for(RG int x = 1; x <= n; x++)
        while(1)
        {
            clear(visx, 0), clear(visy, 0);
            if(hungary(x)) break;
            int inc = INT_MAX;
            for(RG int i = 1; i <= n; i++) if(visx[i])
                for(RG int j = 1; j <= n; j++) if(!visy[j])
                    inc = std::min(inc, lx[i] + ly[j] - g[i][j]);
            for(RG int i = 1; i <= n; i++) if(visx[i]) lx[i] -= inc;
            for(RG int i = 1; i <= n; i++) if(visy[i]) ly[i] += inc;
        }
    vector ans = (vector) {0, 0};
    for(RG int i = 1; i <= n; i++)
        ans.x += a[match[i]][i], ans.y += b[match[i]][i];
    return ans;
}

void solve(const vector &A, const vector &B)
{
    build(A.y - B.y, B.x - A.x);
    vector C = KM(); ans = std::min(ans, C.x * C.y);
    if((B - A) * (C - A) >= 0) return;
    solve(A, C), solve(C, B);
}

int main()
{
    T = read();
    while(T--)
    {
        n = read();
        for(RG int i = 1; i <= n; i++)
            for(RG int j = 1; j <= n; j++)
                a[i][j] = read();
        for(RG int i = 1; i <= n; i++)
            for(RG int j = 1; j <= n; j++)
                b[i][j] = read();
        build(1, 0); vector A = KM();
        build(0, 1); vector B = KM();
        ans = std::min(A.x * A.y, B.x * B.y);
        solve(A, B); printf("%d\n", ans);
    }
    return 0;
}