餐巾计划问题 费用流
题目描述
一个餐厅在相继的
每天结束时,餐厅必须决定将多少块脏的餐巾送到快洗部,多少块餐巾送到慢洗部,以及多少块保存起来延期送洗。但是每天洗好的餐巾和购买的新餐巾数之和,要满足当天的需求量。
试设计一个算法为餐厅合理地安排好
输入输出格式
输入格式：
由标准输入提供输入数据。文件第 1 行有 1 个正整数
接下来的
最后一行包含5个正整数
输出格式：
将餐厅在相继的 N 天里使用餐巾的最小总花费输出
输入输出样例
输入样例#1： 复制
3 1 7 5 11 2 2 3 1
输出样例#1： 复制
134
说明
N<=2000
ri<=10000000
p,f,s<=10000
时限4s
重要是建图；
将每一天分为早上和晚上；
s向每一天晚上连容量为ri,费用为0的边，每一天早上向t连容量为ri,费用为0的边；
第 i 天晚上可以向第 i+1 天晚上连边，对于清洗来说，第 i 天晚上可以向 第 i+T1/T2 天早上连边；
最后每一天早上都可以花费 p 来购买毛巾；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
bool vis[maxn];
int n, m, s, t;
int x, y, f, z;
ll dis[maxn], pre[maxn], last[maxn], flow[maxn];
ll maxflow, mincost;
struct node {
ll to, nxt, flow, dis;
}edge[maxn << 2];
int head[maxn], cnt;
queue<int>q;
void addedge(int from, int to, int flow, int dis) {
edge[++cnt].to = to; edge[cnt].flow = flow; edge[cnt].dis = dis;
edge[cnt].nxt = head[from]; head[from] = cnt;
}
bool spfa(int s, int t) {
memset(dis, 0x7f, sizeof(dis)); memset(flow, 0x7f, sizeof(flow));
ms(vis);
q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
while (!q.empty()) {
int now = q.front(); q.pop(); vis[now] = 0;
for (int i = head[now]; i != -1; i = edge[i].nxt) {
if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
dis[edge[i].to] = edge[i].dis + dis[now];
pre[edge[i].to] = now; last[edge[i].to] = i;
flow[edge[i].to] = min(flow[now], edge[i].flow);
if (!vis[edge[i].to]) {
vis[edge[i].to] = 1; q.push(edge[i].to);
}
}
}
}
return pre[t] != -1;
}
void mincost_maxflow() {
while (spfa(s, t)) {
int now = t;
maxflow += flow[t]; mincost += flow[t] * dis[t];
while (now != s) {
edge[last[now]].flow -= flow[t];
edge[last[now] ^ 1].flow += flow[t];
now = pre[now];
}
}
}
int main()
{
//ios::sync_with_stdio(0);
memset(head, -1, sizeof(head)); cnt = 1;
n = rd(); s = 0; t = 2 * n + 2;
for (int i = 1; i <= n; i++) {
int x; x = rd();
addedge(s, i, x, 0); addedge(i, s, 0, 0);
addedge(i + n, t, x, 0); addedge(t, i + n, 0, 0);
}
int p; int m1, t1, m2, t2;
p = rd(); t1 = rd();m1 = rd(); t2 = rd(); m2 = rd();
for (int i = 1; i <= n; i++) {
if (i + 1 <= n)addedge(i, i + 1, inf, 0), addedge(i + 1, i, 0, 0);
if (i + t1 <= n)addedge(i, i + t1 + n, inf, m1), addedge(i + t1 + n, i, 0, -m1);
if (i + t2 <= n)addedge(i, i + t2 + n, inf, m2), addedge(i + t2 + n, i, 0, -m2);
addedge(s, i + n, inf, p), addedge(i + n, s, 0, -p);
}
mincost_maxflow();
printf("%lld\n", 1ll * mincost);
return 0;
}
NKDEWSM
