校内模拟赛 Label

题意：

　　n个点m条边的无向图，有些点有权值，有些没有。边权都为正。给剩下的点标上数字，使得$\sum\limits_{(u,v)\in E}len(u,v) \times (w[u] - w[v]) ^ 2$最小。

分析：

　　$$\begin{aligned}\sum_{x\to v}(w_v-w_x)^2\cdot len_v&=\sum_{x\to v}(w_v^2-2w_vw_x+w_x^2)\cdot len_v\\&=\left(\sum len_v\right)w_x^2+\left(\sum-2len_vw_v\right)w_x+\sum len_vw_v^2\end{aligned}$$

　　这是一个二次函数，可以知道取最小值的时候：$$w_x=\frac{\sum w_vlen_v}{\sum len_v}$$

　　根据这个可以列出很多个方程，然后高斯消元即可。将已确定的值放到右边，未确定的放到左边。

　　只与为什么一定有解，可以感性的理解一下，或者戳这

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<bitset>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 505;
const double eps = 1e-9;
struct Edge { int to, nxt, w; } e[100005];
double A[N][N], w[N];
int a[N], b[N], c[N], head[N], En;
bool B[N];

inline void add_edge(int u,int v,int w) { 
    ++En; e[En].to = v, e[En].w = w, e[En].nxt = head[u]; head[u] = En;
    ++En; e[En].to = u, e[En].w = w, e[En].nxt = head[v]; head[v] = En;
}
inline double sqr(double x) { return x * x; }
void Gauss(int n) {
    for (int k = 1; k <= n; ++k) {
        if (!B[k]) k ++;
        int r = k;
        for (int i = k + 1; i <= n; ++i) if (fabs(A[i][k]) > fabs(A[r][k])) r = k;
        if (r != k) for (int j = 1; j <= n + 1; ++j) swap(A[k][j], A[r][j]);
        if (fabs(A[k][k]) < eps) continue;
        for (int i = k + 1; i <= n; ++i) {
            if (fabs(A[i][k]) > eps) {
                double t = A[i][k] / A[k][k];
                for (int j = 1; j <= n + 1; ++j) A[i][j] -= t * A[k][j];
            }
        }
    }
    for (int i = n; i; --i) {
        if (!B[i]) continue;
        for (int j = i + 1; j <= n + 1; ++j) A[i][n + 1] -= A[i][j] * w[j];
        w[i] = A[i][n + 1] / A[i][i];
    }
}
int main() {
    int n = read(), m = read();
    for (int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        if (u != v) add_edge(u, v, w);
    }
    for (int i = 1; i <= n; ++i) w[i] = read();
    for (int i = 1; i <= n; ++i) {
        if (w[i] < 0) {
            B[i] = 1;
            for (int j = head[i]; j; j = e[j].nxt) {
                int v = e[j].to;
                if (w[v] < 0) A[i][v] -= e[j].w;
                else A[i][n + 1] += 1.0 * w[v] * e[j].w;
                A[i][i] += e[j].w;
            }
        }
    }
    Gauss(n);
    double ans = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = head[i]; j; j = e[j].nxt)
            if (e[j].to > i) ans += sqr(w[i] - w[e[j].to]) * e[j].w;
    printf("%.10lf\n", ans);
    return 0;
}