2018宁夏邀请赛F Clever King（最大权闭合子图）

Clever King

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].

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

链接

https://nanti.jisuanke.com/t/26172

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 200010;
int N, M;
const int inf = 0x3f3f3f3f;
struct Edge
{
int u, v, cap;
Edge() {}
Edge(int u, int v, int cap): u(u), v(v), cap(cap) {}
} es[maxn];
int R, S, T;
vector<int> tab[maxn]; // 边集
int dis[maxn];
int current[maxn];
void addedge(int u, int v, int cap)
{
tab[u].push_back(R);
es[R++] = Edge(u, v, cap); // 正向边
tab[v].push_back(R);
es[R++] = Edge(v, u, 0); // 反向边容量为0
// 正向边下标通过异或就得到反向边下标, 2 ^ 1 == 3 ; 3 ^ 1 == 2
}
int BFS()
{
queue<int> q;
q.push(S);
memset(dis, 0x3f, sizeof(dis));
dis[S] = 0;
while (!q.empty())
{
int h = q.front();
q.pop();
for (int i = 0; i < tab[h].size(); i++)
{
Edge &e = es[tab[h][i]];
if (e.cap > 0 && dis[e.v] == 0x3f3f3f3f)
{
dis[e.v] = dis[h] + 1;
q.push(e.v);
}
}
}
return dis[T] < 0x3f3f3f3f; // 返回是否能够到达汇点
}
int dinic(int x, int maxflow)
{
if (x == T)
return maxflow;
// i = current[x] 当前弧优化
for (int i = current[x]; i < tab[x].size(); i++)
{
current[x] = i;
Edge &e = es[tab[x][i]];
if (dis[e.v] == dis[x] + 1 && e.cap > 0)
{
int flow = dinic(e.v, min(maxflow, e.cap));
if (flow)
{
e.cap -= flow; // 正向边流量降低
es[tab[x][i] ^ 1].cap += flow; // 反向边流量增加
return flow;
}
}
}
return 0; // 找不到增广路 退出
}
long long DINIC()
{
long long ans = 0;

while (BFS()) // 建立分层图
{
int flow;
memset(current, 0, sizeof(current)); // BFS后应当清空当前弧数组
while (flow = dinic(S, 0x3f3f3f3f)) // 一次BFS可以进行多次增广
ans += flow;
}
return ans;
}
int main()
{
int kase;
scanf("%d", &kase);
while (kase--)
{
scanf("%d%d", &N, &M);
long long sum = 0;
R = 0;
S = 0, T = N + M + 1;
for (int i = 0; i <= T; i++)
tab[i].clear();
for (int i = 1; i <= N; i++)
{
int cap;
scanf("%d", &cap);
sum += cap;
}
for (int i = 1; i <= M; ++i)
{
int cap;
scanf("%d", &cap);
addedge(N + i, N + M + 1, cap);
}
for (int i = 1; i <= N; i++)
{
int u, n1, n2;
scanf("%d%d", &n1, &n2);
while (n1--)
{
scanf("%d", &u);
}
while (n2--)
{
scanf("%d", &u);
}
}
printf("%lld\n", sum - DINIC());
}
return 0;
}

The end.
2018-05-02 星期三