牛客练习赛19 托米去购物
标签: 网络流
题目链接
https://www.nowcoder.com/acm/contest/111/D
分析
- 因为总金额有限,所以要求现金使用最小也就意味着优惠券使用最多
- 源点连优惠券,流量为优惠券金额;优惠券连物品,流量为INF;物品连汇点,流量为物品价格
代码
// 感谢会长优秀板子
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e5+50;
const int mod = 1e9+7;
struct Edge{
int from,to,cap,flow;
Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic{
int s,t;
vector<Edge>edges;
vector<int> G[N];
bool vis[N];
int d[N];
int cur[N];
void init(){
for (int i=0;i<=t+1;i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap){
edges.push_back(Edge(from,to,cap,0));
edges.push_back(Edge(to,from,0,0));
int mm=edges.size();
G[from].push_back(mm-2);
G[to].push_back(mm-1);
}
bool BFS(){
memset(vis,0,sizeof(vis));
queue<int>q;
q.push(s);
d[s]=0;
vis[s]=1;
while (!q.empty()){
int x = q.front();q.pop();
for (int i = 0;i<G[x].size();i++){
Edge &e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow){
vis[e.to]=1;
d[e.to] = d[x]+1;
q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a){
if (x==t || a==0)
return a;
int flow = 0,f;
for(int &i=cur[x];i<G[x].size();i++){
Edge &e = edges[G[x][i]];
if (d[x]+1 == d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0){
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if (a==0)
break;
}
}
return flow;
}
int Maxflow(int s,int t){
this->s=s;
this->t=t;
int flow = 0;
while (BFS()){
memset(cur,0,sizeof(cur));
flow+=DFS(s,inf);
}
return flow;
}
}dc;
int main(){
int T;
scanf("%d", &T);
while(T--){
int n,m;
scanf("%d%d",&n,&m);
int s=n+m+1,t=n+m+2;
dc.init();
int sum=0;
for (int i = 1; i <= n; ++i)
{
int x;
scanf("%d", &x);
sum+=x;
dc.AddEdge(i,t,x);
}
for (int i = n+1; i <= n+m; ++i)
{
int x;
scanf("%d", &x);
dc.AddEdge(s,i,x);
}
for (int i = n+1; i <= n+m; ++i)
{
int k;
scanf("%d", &k);
while(k--){
int x;
scanf("%d", &x);
dc.AddEdge(i,x,inf);
}
}
printf("%d
", sum-dc.Maxflow(s,t));
}
return 0;
}