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洛谷P1251 餐巾计划问题(最小费用最大流)

2018-07-27 06:10:24来源：博客园阅读 ()

题意

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

Sol

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());
}