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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
    
     #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
     
       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;}