BZOJ2424 [HAOI2010]订货

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 = 1061109567;
 
int n, m, maxin;
int nd[N], cost[N], head[N], tot, S, T = N - 1, 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 = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}
 
int ch(int x) {
    return ( (x + 1) ^ 1) - 1;
}
 
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, 0, -c);
}
 
int bfs() {
    memset(dis, 63, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    memset(pre, 0, sizeof(pre));
    dis[S] = 0;
    vis[S] = 1;
    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] = 1, q.push(nt);
        }
        vis[u] = 0;
    }
    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, 1, n) nd[i] = rd;
    rep(i, 1, n) cost[i] = rd;
    rep(i, 1, n) {
        add(S, i, inf, cost[i]);
        add(i, T, nd[i], 0);
        add(i, i + 1, maxin, m);
    }
    EK();
    printf("%d
", minco);
}