BZOJ2424 [HAOI2010]订货 - 费用流

题解

（非常裸的费用流

题意有一点表明不清： 该月卖出的商品可以不用算进仓库里面。

然后套上费用流模板

代码

 #include<cstring>
 #include<queue>
 #include<cstdio>
 #include<algorithm>
 #define rd read()
 #define rep(i,a,b) for(int i = (a); i <= (b); ++i)
 using namespace std;

 const int N = 1e4;
 const int inf = ;

 int n, m, maxin;
 int nd[N], cost[N], head[N], tot, S, T = N - , maxflow, minco;
 int dis[N], pre[N], vis[N];

 queue<int> q;

 struct edge {
     int nxt, val, to, c;
 }e[N];

 int read() {
     int X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 int ch(int x) {
     return ( (x + ) ^ ) - ;
 }

 void added(int fr, int to, int val, int c) {
     e[++tot].to = to;
     e[tot].val = val;
     e[tot].c = c;
     e[tot].nxt = head[fr];
     head[fr] = tot;
 }

 void add(int fr, int to, int val, int c) {
     added(fr, to, val, c);
     added(to, fr, , -c);
 }

 int bfs() {
     memset(dis, , sizeof(dis));
     memset(vis, , sizeof(vis));
     memset(pre, , sizeof(pre));
     dis[S] = ;
     vis[S] = ;
     q.push(S);
     for(int u, nt; !q.empty(); ) {
         u = q.front(); q.pop();
         for(int i = head[u]; i; i = e[i].nxt ) {
             nt = e[i].to;
             if(dis[nt] <= dis[u] + e[i].c || !e[i].val) continue;
             dis[nt] = dis[u] + e[i].c;
             pre[nt] = i;
             if(!vis[nt]) vis[nt] = , q.push(nt);
         }
         vis[u] = ;
     }
     return dis[T];
 }

 void EK() {
     for(; bfs() != inf; ) {
         int tmp = inf;
         for(int i = pre[T]; i; i = pre[e[ch(i)].to]) tmp = min(tmp, e[i].val);
         for(int i = pre[T]; i; i = pre[e[ch(i)].to]) e[i].val -= tmp, e[ch(i)].val += tmp;
         maxflow += tmp;
         minco += tmp * dis[T];
     }
 }

 int main()
 {
     n = rd; m = rd; maxin = rd;
     rep(i, , n) nd[i] = rd;
     rep(i, , n) cost[i] = rd;
     rep(i, , n) {
         add(S, i, inf, cost[i]);
         add(i, T, nd[i], );
         add(i, i + , maxin, m);
     }
     EK();
     printf("%d\n", minco);
 }