BZOJ1927: [Sdoi2010]星际竞速(最小费用最大流 最小路径覆盖)

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;
}