BZOJ1927: [Sdoi2010]星际竞速(最小费用最大流最小路径覆盖)-优快云博客

本文详细解析了如何利用最小费用最大流算法解决带权最小路径覆盖问题，通过构建特殊的网络流图，将二分图匹配问题转化为最小费用流问题，实现了对带权图的有效求解。

题意

题目链接

Sol

看完题不难想到最小路径覆盖，但是带权的咋做啊？qwqqq

首先冷静思考一下：最小路径覆盖 = \(n - \text{二分图最大匹配数}\)

为什么呢？首先最坏情况下是用\(n\)条路径去覆盖(就是\(n\)个点)，根据二分图的性质，每个点只能有一个和他配对，这样就保证了，每多出一个匹配，路径数就会\(-1\)

扩展到有边权的图也是同理的，\(i\)表示二分图左侧的点,\(i'\)表示二分图右侧的点，对于两点\(u, v\)，从\(u\)向\(v'\)连\((1, w_i)\)的边(前面是流量，后面是费用)

接下来从\(S\)向\(i\)连\((1, 0)\)的边，从\(i'\)向\(T\)连\((1, 0)\)的边，从\(S\)向\(i'\)连\((1, A_i)\)的边

跑最小费用最大流即可

#include<bits/stdc++.h>
#define chmin(x, y) (x = x < y ? x : y)
#define chmax(x, y) (x = x > y ? x : y)
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, S, T, dis[MAXN], vis[MAXN], Pre[MAXN], ansflow, anscost, A[MAXN];
struct Edge {
    int u, v, f, w, nxt;
}E[MAXN];
int head[MAXN], num = 0;
inline void add_edge(int x, int y, int f, int w) {
    E[num] = (Edge) {x, y, f, w, head[x]}; head[x] = num++;    
}
inline void AddEdge(int x, int y, int f, int w) {
    add_edge(x, y, f, w); add_edge(y, x, 0, -w);
}
bool SPFA() {
    queue<int> q; q.push(S);
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[S] = 0;
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = 0;
        for(int i = head[p]; ~i; i = E[i].nxt) {
            int to = E[i].v;
            if(E[i].f && dis[to] > dis[p] + E[i].w) {
                dis[to] = dis[p] + E[i].w; Pre[to] = i;
                if(!vis[to]) vis[to] = 1, q.push(to);
            }
        }
    }
    return dis[T] <= INF;
}
void F() {
    int canflow = INF;
    for(int i = T; i != S; i = E[Pre[i]].u) chmin(canflow, E[Pre[i]].f);
    for(int i = T; i != S; i = E[Pre[i]].u) E[Pre[i]].f -= canflow, E[Pre[i] ^ 1].f += canflow;
    anscost += canflow * dis[T];
}
void MCMF() {
    while(SPFA()) F();
}
int main() {   
    memset(head, -1, sizeof(head));
    N = read(); M = read(); S = N * 2 + 2, T = N * 2 + 3;
    for(int i = 1; i <= N; i++) A[i] =read(), AddEdge(S, i, 1, 0), AddEdge(i + N, T, 1, 0), AddEdge(S, i + N, 1, A[i]);
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read(), v = read();
        if(x > y) swap(x, y);
        AddEdge(x, y + N, 1, v);
    }
    MCMF();
    printf("%d\n", anscost);
    return 0;
}

转载于:https://www.cnblogs.com/zwfymqz/p/10078458.html