[ARC085E]MUL

2020-03-11
Atcoder

题意

在序列$\{a_i\}$中可以任意次把编号为某个数倍数的全部删除,使得剩下的权值和最大

求最大值,$n\leq 100$

题解

使得去掉的负权值尽大,跑最大权闭合子图,$ 最小割=-(选的负权值-不选的正权值) $

$ 正权值和-最小割=选的正权值+选的负权值=ans $

调试记录

最大权闭合子图是关于负权值的

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#include <cstdio>
#include <queue>
#include <cstring>
#define LL long long
const int INF = 0x3f3f3f3f;
const int maxn = 105;
const int S = 0;
const int T = 104;
using namespace std;
struct E{
int to, nxt, f;
}e[maxn * 1000];
int head[maxn], tot = 1;
void addedge(int u, int v, int f){
e[++tot].to = v, e[tot].nxt = head[u], e[tot].f = f;
head[u] = tot;
}
void ins(int u, int v, int f){
addedge(u, v, f); addedge(v, u, 0);
}
int dep[maxn], now[maxn];
bool bfs(){
queue <int> q; while (!q.empty()) q.pop(); q.push(S);
memset(dep, 0, sizeof dep); dep[S] = 1;
memcpy(now, head, sizeof now);
while (!q.empty()){
int cur = q.front(); q.pop();
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (dep[v] == 0 && e[i].f > 0) dep[v] = dep[cur] + 1, q.push(v);
}
}
return dep[T] != 0;
}
LL dfs(int cur, LL Max){
if (cur == T) return Max;
int flow = 0;
for (int i = now[cur]; i; i = e[i].nxt){
int v = e[i].to; now[cur] = i;
if (dep[v] == dep[cur] + 1 && e[i].f > 0){
LL tmp = dfs(v, min(Max - flow, 1ll * e[i].f));
flow += tmp;
e[i].f -= tmp;
e[i ^ 1].f += tmp;
if (flow == Max) return flow;
}
}
return flow;
}
LL maxflow = 0;
void Dinic(){
while (bfs()) maxflow += dfs(S, INF);
}
int n, a[maxn];
LL ans = 0;
int main(){
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", a + i);
for (int i = 1; i <= n; i++)
if (a[i] > 0) ins(i, T, a[i]), ans += 1ll * a[i];
else ins(S, i, -a[i]);
for (int i = 1; i <= n; i++)
for (int j = 2; i * j <= n; j++)
ins(i, i * j, INF);
Dinic();
printf("%lld\n", ans - maxflow);
return 0;
}