luogu P4553 80人环游世界

https://www.luogu.com.cn/problem/P4553

容易想到拆点建二分图，跑最小费用路径覆盖，不过要强制每个点都流满

于是把$i->i+n，c$表示这个点必须要被经过c次

然后跑最小费用可行流即可

code:

#include<bits/stdc++.h>
#define N 805
#define ll long long
using namespace std;
struct edge {
    int v, c, nxt;
    ll w;
} e[N * N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void insert(int u, int v, int c, ll w) {
    e[eid].v = v;
    e[eid].c = c;
    e[eid].w = w;
    e[eid].nxt = p[u];
    p[u] = eid ++;
}
void add(int u, int v, int c, ll w) { //printf("%d --> %d  %d %d\n", u, v, c, w);
    insert(u, v, c, w), insert(v, u, 0, -w);
}
int n, m, S, T, pre[N], vis[N];
ll dis[N];
queue<int> q;
const ll INF = 1e18;
const int inf = 1e4;
int bfs() {
    for(int i = 0; i <= T; i ++) dis[i] = INF, vis[i] = 0, pre[i] = -1;
    dis[S] = 0; q.push(S);
    while(q.size()) {
        int u = q.front(); q.pop();
        vis[u] = 0;
        for(int i = p[u]; i + 1; i = e[i].nxt) {
            int v = e[i].v; //printf("*%d %d %d %d  %d %d\n", u, v, dis[u], dis[v], e[i].c, e[i].w);
            if(e[i].c && dis[u] + e[i].w < dis[v]) {
                dis[v] = dis[u] + e[i].w;
                pre[v] = i;
                if(!vis[v]) vis[v] = 1, q.push(v);
            }
        }
    }
//    for(int i = 1; i <= T; i ++) printf("%lld ", dis[i]); printf("\n");
    return pre[T] != -1;
}
ll mcfc() {
    ll ret = 0;
    for(; bfs() ;) {
        ll flow = INF;
        for(int i = T; i != S; i = e[pre[i] ^ 1].v) flow = min(flow, (ll)e[pre[i]].c);
        for(int i = T; i != S; i = e[pre[i] ^ 1].v) {
            ret += flow * e[pre[i]].w;
            e[pre[i]].c -= flow, e[pre[i] ^ 1].c += flow;
        }
       // printf("*%d*", flow);
       // break;
    }
    return ret;
}
ll sum[N], totc;
void addl(int u, int v, int l, int r, int c) { //printf("%d-->%d %d %d %d\n", u, v, l, r, c);
    if(r - l > 0) add(u, v, r - l, c);
    sum[u] -= l, sum[v] += l; totc += 1ll * c * l;
}
int SS, TT;
void build() {
    for(int i = 1; i <= TT; i ++) {
        if(sum[i] > 0) add(S, i, sum[i], 0);
        if(sum[i] < 0) add(i, T, - sum[i], 0);
    }
  //  for(int i = 1; i <= TT; i ++) printf(" %lld ", sum[i]); printf("\n");
    add(TT, SS, m, 0);
}
int main() {
    init();
    scanf("%d%d", &n, &m);
    SS = 2 * n + 1, TT = SS + 1, S = TT + 1, T = S + 1;
    for(int i = 1, c; i <= n; i ++) {
        scanf("%d", &c);
        addl(SS, i, 0, inf, 0);
        addl(i, i + n, c, c, 0);
        addl(i + n, TT, 0, inf, 0);
    }
    for(int i = 1; i <= n; i ++)
        for(int j = i + 1, c; j <= n; j ++) {
            scanf("%d", &c);
            if(c >= 0) addl(i + n, j, 0, inf, c);
        }
    build();
   // printf("%d %d\n", S, T);
    printf("%lld", mcfc() + totc);
    return 0;
}