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;

}