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  • Label [高斯消元]

    LabelLabel


    color{red}{正解部分}

    考虑一个未知点连向两个已知点, 边权为 w1,w2w_1, w_2, 点权为 ax,a1,a2a_x, a_1, a_2,
    则现在要做的就是使得 w1(axa1)2+w2(axa2)2w_1(a_x-a_1)^2 + w_2(a_x-a_2)^2 最小,
    根据二次函数相关知识可得, 当 ax=w1a1+w2a2w1+w2a_x = frac{w_1a_1+w_2a_2}{w_1+w_2} 时, 上式取得最小值,

    推广到多个点, 当 ax=wiaiwia_x = frac{sum w_ia_i}{sum w_i} 时, 答案最小 .


    color{red}{实现部分}

    将所有 未知数 提出, 列出方程, 高斯消元 即可 .

    #include<bits/stdc++.h>
    #define reg register
    
    int read(){
            char c;
            int s = 0, flag = 1;
            while((c=getchar()) && !isdigit(c))
                    if(c == '-'){ flag = -1, c = getchar(); break ; }
            while(isdigit(c)) s = s*10 + c-'0', c = getchar();
            return s * flag;
    }
    
    const int maxn = 30005;
    const double eps = 1e-12;
    
    int N;
    int M;
    int cnt;
    int num0;
    int A[maxn];
    int B[maxn];
    int Mp[maxn];
    int head[maxn];
    
    double C[maxn];
    double K[305][305];
    
    struct Edge{ int nxt, to, w; } edge[maxn << 1];
    
    struct EDGE{ int u, v, w; } E[maxn];
    
    void Add(int from, int to, int w){ edge[++ num0] = (Edge){ head[from], to, w }; head[from] = num0; }
    
    void Gays(){
            for(reg int i = 1; i <= cnt; i ++){
                    int max_id = i;
                    for(reg int j = i+1; j <= cnt; j ++)
                            if(fabs(K[j][i]) > fabs(K[max_id][i])) max_id = j;
                    std::swap(K[max_id], K[i]);
                    for(reg int j = cnt+1; j >= i; j --) K[i][j] /= K[i][i]; // !
                    for(reg int j = i+1; j <= cnt; j ++){
                            if(fabs(K[j][i]) < eps) continue ;
                            for(reg int k = cnt+1; k >= i; k --) K[j][k] -= K[j][i] * K[i][k];
                    }
            }
            for(reg int i = cnt; i >= 1; i --){
                    for(reg int j = cnt; j > i; j --) K[i][cnt+1] -= K[i][j] * K[j][cnt+1];
                    C[B[i]] = K[i][cnt+1];
            }
    }
    
    int main(){
            N = read(), M = read();
            for(reg int i = 1; i <= M; i ++){
                    int u = read(), v = read(), w = read();
                    E[i].u = u, E[i].v = v, E[i].w = w;
                    Add(u, v, w), Add(v, u, w);
            }
            for(reg int i = 1; i <= N; i ++){
                    C[i] = A[i] = read();
                    if(A[i] == -1) B[++ cnt] = i, Mp[i] = cnt;
            }
            for(reg int i = 1; i <= N; i ++){
                    if(~A[i]) continue ;
                    double sum = 0;
                    for(reg int j = head[i]; j; j = edge[j].nxt){
                            int to = edge[j].to;
                            if(to == i) continue ;
                            sum += edge[j].w;
                            if(A[to] == -1) K[Mp[i]][Mp[to]] -= edge[j].w;
                            else K[Mp[i]][cnt+1] += 1ll * edge[j].w * A[to];
                    }
                    for(reg int j = 1; j <= cnt+1; j ++) K[Mp[i]][j] /= sum;
                    K[Mp[i]][Mp[i]] = 1;
            }
            Gays();
            double Ans = 0;
            for(reg int i = 1; i <= M; i ++) Ans += 1ll*E[i].w * (C[E[i].u] - C[E[i].v]) * (C[E[i].u] - C[E[i].v]);
            printf("%.12lf
    ", Ans);
            return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zbr162/p/11822455.html
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