一家餐厅,第$i$天需要$r_i$块餐巾,每天获取餐巾有三种途径
1、以$p$的费用买
2、以$f$的费用送到快洗部,并在$m$天后取出
3、以$s$的费用送到慢洗部,并在$n$天后取出
问满足要求时的最小费用
一道非常不错的网络流,应该不难看出是费用流。
首先进行拆点,把每个点早上和晚上,然后进行连边
从$S$向i连边$(0, r_i)$,表示到了晚上有$r_i$块脏餐巾
从$i'$向$T$连边$(0, r_i)$,表示早上有$r_i$块新餐巾
从$S$向$i'$连边$(p, INF)$,表示每天早上可以以$p$的费用无限提供餐巾
从$i$向$i'$连边$(0, INF)$,表示每天晚上的脏餐巾可以留到第二天晚上
从$i$向$i' + m$连边$(f, INF)$,表示快洗
从$i$向$i' + n$连边$(s, INF)$,表示慢洗
这样既可以保证每天的$r_i$满足要求,又能保证最小费用。so nice
#include<cstdio>
#include<cstring>
#include<queue>
#define LL long long
using namespace std;
const int MAXN = 1e5 + 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, S, T;
int r[MAXN], p, m, f, n, s;
struct Edge {
int u, v, w, f, nxt;
}E[MAXN];
int head[MAXN], num;
inline void add_edge(int x, int y, int w, int f) {
E[num] = (Edge){x, y, w, f, head[x]};
head[x] = num++;
}
inline void AddEdge(int x, int y, int w, int f) {
add_edge(x, y, w, f);
add_edge(y, x, -w, 0);
}
int dis[MAXN], vis[MAXN], Pre[MAXN];
bool SPFA() {
memset(dis, 0x3f, sizeof(dis));
memset(vis, 0, sizeof(vis));
queue<int> q; dis[S] = 0; q.push(S);
while(!q.empty()) {
int p = q.front(); q.pop(); vis[p] = 0;
for(int i = head[p]; i != -1; i = E[i].nxt) {
int to = E[i].v;
if(dis[to] > dis[p] + E[i].w && E[i].f) {
dis[to] = dis[p] + E[i].w;
Pre[to] = i;
if(!vis[to]) vis[to] = 1, q.push(to);
}
}
}
return dis[T] <= INF;
}
LL F() {
LL nowflow = INF;
for(int i = T; i != S; i = E[Pre[i]].u) nowflow = min(nowflow, (LL)E[Pre[i]].f);
for(int i = T; i != S; i = E[Pre[i]].u) E[Pre[i]].f -= nowflow, E[Pre[i] ^ 1].f += nowflow;
return nowflow * dis[T];
}
LL MCMF() {
LL ans = 0;
while(SPFA())
ans += F();
return ans;
}
int main() {
memset(head, -1, sizeof(head));
N = read();
S = 0; T = 2 * N + 1;
for(int i = 1; i <= N; i++) r[i] = read();
p = read(); m = read(); f = read(); n = read(); s = read();
for(int i = 1; i <= N; i++) AddEdge(S, i, 0, r[i]);
for(int i = 1; i <= N; i++) AddEdge(S, i + N, p, INF);
for(int i = 1; i <= N; i++) AddEdge(i + N, T, 0, r[i]);
for(int i = 1; i <= N; i++) {
if(i + m <= N) AddEdge(i, i + N + m, f, INF);
if(i + n <= N) AddEdge(i, i + N + n, s, INF);
if(i + 1 <= N) AddEdge(i, i + 1, 0, INF);
}
printf("%lld", MCMF());
}