Clever King

题目传送门:https://nanti.jisuanke.com/t/26172

Description:

In order to increase the happiness index of people’s lives, King Y has decided to develop the manufacturing industry vigorously. There are total n kinds of products that King can choose to produce, different products can improve the happiness index of poeple’s lives in different degrees, of course, the production of goods needs raw materials, different products need different ore or other products as raw materials. There are total m mines, and each mine can exploit different ore, Therefore, there are m types of ores, the cost of each mining for each mine is different, king Y want to maximize the income, the calculation method of income is:∑increased happiness index - ∑mining costs.
If you choose to exploit a mine, there will be an unlimited number of this kind of ore. What’s more, if you produce one product, the happiness index will definitely increase, no matter how many you produce.

Input:

The first line of the input has an integer T(1<=T<=50), which represents the number of test cases.
In each test case, the first line of the input contains two integers n(1<=n<=200)–the number of the products and m(1<=m<=200)–the number of mines. The second line contains n integers, val[i] indicates the happiness index that number i product can increase. The third line contains m integers, cost[i] indicates the mining cost of number i mine. The next n lines, each line describes the type of raw material needed for the number i product, in each line, the first two integers n1(1<=n1<=m)–the number of ores that this product needs, n2(1<=n2<=n)–the number of products that this product needs, the next n1 + n2 integers indicate the id of ore and product that this product needs. it guarantees that ∑n1+∑n2<=2000.

Output:

Each test case output an integer that indicates the maximum value ∑val[i]-∑cost[i].
忽略每行输出的末尾多余空格

样例输入

2
3 3
600 200 400
100 200 300
1 2 1 2 3
1 0 2
1 0 3
3 4
600 400 200
100 200 300 1000
2 1 1 2 3
1 0 1
1 0 1

样例输出

600
900

题目来源

2018 ACM-ICPC 中国大学生程序设计竞赛线上赛

学习……

最大权闭合子图
大佬POJ2987的题解
HiHoCoder 119周 最大权闭合子图

Code

#include <bits/stdc++.h>
using namespace std;
const int mn=5555;
const int mm=222222;
const int oo=1e9;
int node, st, sd, edge;
int ver[mm], flow[mm], Next[mm];
int head[mn], work[mn], dis[mn], q[mn], visit[mn];

inline void init(int _node, int _st, int _sd) {
    node=_node, st=_st, sd=_sd;
    for(int i=0; i<node; i++)
        head[i]=-1;
    edge=0;
}

void addedge(int u, int v, int c1, int c2) {
    ver[edge]=v, flow[edge]=c1, Next[edge]=head[u],head[u]=edge++;
    ver[edge]=u, flow[edge]=c2, Next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs() {
    int i,u,v,l,r=0;
    for(i=0; i<node; ++i)dis[i]=-1;
    dis[q[r++]=st]=0;
    for(l=0; l<r; ++l)
        for(i=head[u=q[l]]; i>=0; i=Next[i])
            if(flow[i]&&dis[v=ver[i]]<0) {
                dis[q[r++]=v]=dis[u]+1;
                if(v==sd)return 1;
            }
    return 0;
}
long long Dinic_dfs(int u, int exp) {
    if(u==sd) return exp;
    for(int &i=work[u]; i>=0; i=Next[i]) {
        int v=ver[i], tp;
        if(flow[i]&&dis[v]==dis[u]+1&&(tp=Dinic_dfs(v,min(flow[i],exp)))>0) {
            flow[i]-=tp;
            flow[i^1]+=tp;
            return tp;
        }
    }
    return 0;
}
long long Dinic_flow() {
    int i,delta;
    long long ret=0;
    while(Dinic_bfs()) {
        for(i=0; i<node; ++i)work[i]=head[i];
        while(delta=Dinic_dfs(st,oo))ret+=delta;
    }
    return ret;
}

void DFS(int u) {
    visit[u]=1;
    for(int i=head[u], v; i>=0; i=Next[i])
        if(flow[i]>0&&!visit[v=ver[i]]) DFS(v);
}
int T,n,m,w,v,u,n1,n2;
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        init(n+m+2,0,n+m+1);
        long long sum=0;
        for(int i=1; i<=n; i++) {
            scanf("%d",&w);
            sum+=w;
            addedge(st,i,w,0);
        }
        for (int i=n+1;i<=n+m;++i) {
            scanf("%d",&w);
            addedge(i,sd,w,0);
        }

        for (int i=1;i<=n;++i) {
            scanf("%d%d",&n1,&n2);
            while (n1--) {
                scanf("%d",&u);
                addedge(i,u+n,oo,0);
            }
            while (n2--) {
                scanf("%d",&u);
                addedge(i,u,oo,0);
            }
        }

        long long maxflow=Dinic_flow();
        memset(visit,0,sizeof(visit));

        cout<<sum-maxflow<<endl;
    }
    return 0;
}
内容概要:本文档定义了一个名为 `xxx_SCustSuplier_info` 的视图,用于整合和展示客户(Customer)和供应商(Supplier)的相关信息。视图通过连接多个表来获取组织单位、客户账户、站点使用、位置、财务代码组合等数据。对于客户部分,视图选择了与账单相关的记录,并提取了账单客户ID、账单站点ID、客户名称、账户名称、站点代码、状态、付款条款等信息;对于供应商部分,视图选择了有效的供应商及其站点信息,包括供应商ID、供应商名称、供应商编号、状态、付款条款、财务代码组合等。视图还通过外连接确保即使某些字段为空也能显示相关信息。 适合人群:熟悉Oracle ERP系统,尤其是应付账款(AP)和应收账款(AR)模块的数据库管理员或开发人员;需要查询和管理客户及供应商信息的业务分析师。 使用场景及目标:① 数据库管理员可以通过此视图快速查询客户和供应商的基本信息,包括账单信息、财务代码组合等;② 开发人员可以利用此视图进行报表开发或数据迁移;③ 业务分析师可以使用此视图进行数据分析,如信用评估、付款周期分析等。 阅读建议:由于该视图涉及多个表的复杂连接,建议读者先熟悉各个表的结构和关系,特别是 `hz_parties`、`hz_cust_accounts`、`ap_suppliers` 等核心表。此外,注意视图中使用的外连接(如 `gl_code_combinations_kfv` 表的连接),这可能会影响查询结果的完整性。
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