BZOJ3118 Orz the MST 【單純形 + 生成樹】

題目連接

BZOJ3118

題解

少有的單純形好題啊
咱們先抽離出生成樹
生成樹中的邊只可能減，其它邊只可能加
對於不在生成樹的邊，其權值必定要比生成樹中其端點之間的路徑上全部的邊都大
而後就是一個最小化的線性規劃
爲了防止限制過多
咱們只需對原先生成樹中的比該邊大的邊創建限制便可
而後就是單純形 + 對偶

雙倍經驗

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 1005,maxm = 10005,NN = 1005;
const double eps = 1e-10,INF = 1e15;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt,id;}ed[maxn << 1];
inline void build(int u,int v,int id){
    ed[++ne] = (EDGE){v,h[u],id}; h[u] = ne;
    ed[++ne] = (EDGE){u,h[v],id}; h[v] = ne;
}
int n,m,N,M;
int dep[maxn],fa[maxn],id[maxn];
int U[NN],V[NN],W[NN],F[NN],A[NN],B[NN];
double a[maxn][maxm];
void dfs(int u){
    Redge(u) if ((to = ed[k].to) != fa[u]){
        fa[to] = u; dep[to] = dep[u] + 1; id[to] = ed[k].id;
        dfs(to);
    }
}
void Pivot(int l,int e){
    double t = a[l][e]; a[l][e] = 1;
    for (int j = 0; j <= n; j++) a[l][j] /= t;
    for (int i = 0; i <= m; i++) if (i != l && fabs(a[i][e]) > 0){
        t = a[i][e]; a[i][e] = 0;
        for (int j = 0; j <= n; j++)
            a[i][j] -= a[l][j] * t;
    }
}
void simplex(){
    while (true){
        int l = 0,e = 0; double mn = INF;
        for (int j = 1; j <= n; j++) if (a[0][j] > eps){e = j; break;}
        if (!e) break;
        for (int i = 1; i <= m; i++) if (a[i][e] > eps && a[i][0] / a[i][e] < mn)
            mn = a[i][0] / a[i][e],l = i;
        Pivot(l,e);
    }
}
int main(){
    N = read(); M = read();
    REP(i,M){
        U[i] = read(); V[i] = read(); W[i] = read();
        F[i] = read(); A[i] = read(); B[i] = read();
        if (F[i]) build(U[i],V[i],i);
    }
    dfs(1);
    n = M;
    REP(i,M){
        if (F[i]) a[i][0] = B[i];
        else{
            a[i][0] = A[i];
            int u = U[i],v = V[i],x;
            while (u != v){
                if (dep[u] < dep[v]) swap(u,v);
                x = id[u];
                if (W[x] > W[i]){
                    m++; 
                    a[x][m] = a[i][m] = 1;
                    a[0][m] = W[x] - W[i];
                }
                u = fa[u];
            }
        }
    }
    swap(n,m);
    simplex();
    printf("%.0lf\n",-a[0][0]);
    return 0;
}