洛谷(餐巾计划问题)

链接:https://www.luogu.org/problemnew/show/P1251
思路:费用流题目,建图还是关键,首先很显然想到了拆点,因为有用掉和送出去两个状态,所以把每天拆为两个点,一个表示使用量,一个表示送出去的量,分别放在二分图的x和y上,建立源点汇点,对于每个点i,从源点到x上的i建立一条容量c[i],费用为0的边,表示当天使用c[i]个餐巾;从y轴上的i到汇点建立一条容量为c[i],费用为p的边,用来限制最大流。然后对于每个x上的点,向x+1建立一条容量为INF,费用为0的边,向y+m上建立一条容量为INF,费用为f的边,向y+n上建立一条容量为INF,费用为s的边,最后跑一遍费用流即可。

代码:

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#include<bits/stdc++.h>
using namespace std;

const int maxn = 5050;
const int INF = 1e9;

int t;
int c[2010];
int p,m,f,n,s;

struct edge{
int from,to,cap,flow,cost;
};

struct MCMF{
int n,m,s,t;
vector<edge> edges;
vector<int> G[maxn];
int d[maxn];
int inq[maxn];
int p[maxn];
int a[maxn];

void init(int n){
this->n = n;
for(int i=0;i<=n;i++)G[i].clear();
edges.clear();
}

void addedge(int from,int to,int cap,int cost){
edges.push_back(edge{from,to,cap,0,cost});
edges.push_back(edge{to,from,0,0,-cost});
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}

bool spfa(int s,int t,int &flow,long long &cost){
for(int i=0;i<=n;i++)d[i] = INF;
memset(inq,0,sizeof(inq));
d[s] = 0;
inq[s] = 1;
p[s] = 0;
a[s] = INF;
queue<int> q;
q.push(s);
while(!q.empty()){
int u = q.front();
q.pop();
inq[u] = 0;
for(int i=0;i<G[u].size();i++){
edge &e = edges[G[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u],e.cap-e.flow);
if(!inq[e.to]){
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if(d[t]==INF)return false;
flow+=a[t];
cost+=1LL*d[t]*a[t];
int u = t;
while(u!=s){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u = edges[p[u]].from;
}
return true;
}

long long mincost(int s,int t){
int flow = 0;
long long cost = 0;
while(spfa(s,t,flow,cost));
return cost;
}
}solver;

int main(){
scanf("%d",&t);
for(int i=1;i<=t;i++)scanf("%d",&c[i]);
scanf("%d%d%d%d%d",&p,&m,&f,&n,&s);
solver.init(2*t+1);
for(int i=1;i<=t;i++){
solver.addedge(0,i,c[i],0);
solver.addedge(i+t,2*t+1,c[i],0);
solver.addedge(0,i+t,INF,p);
if(i!=t)solver.addedge(i,i+1,INF,0);
if(i+m<=t)solver.addedge(i,i+t+m,INF,f);
if(i+n<=t)solver.addedge(i,i+t+n,INF,s);
}
long long res = solver.mincost(0,2*t+1);
printf("%lldn",res);
return 0;
}