题目描述
有N个工作，M种机器，每种机器你可以租或者买过来. 每个工作包括若干道工序，每道工序需要某种机器来完成,你可以通过购买或租用机器来完成。 现在给出这些参数，求最大利润
输入输出格式
输入格式：第一行给出 N,M(1<=N<=1200,1<=M<=1200) 下面将有N组数据。
每组数据第一行给出完成这个任务能赚到的钱(其在[1,5000])及有多少道工序
接下来若干行每行两个数，分别描述完成工序所需要的机器编号及租用它的费用(其在[1,20000]) 最后M行，每行给出购买机器的费用(其在[1,20000])
输出格式：最大利润
输入输出样例
输入样例#1：
复制
2 3 100 2 1 30 2 20 100 2 1 40 3 80 50 80 110
输出样例#1： 复制
50
建立源点st与汇点ed；
类比于 最大权闭合子图；
st 与任务连边，权值为所能赚的钱；
ed 与机器相连，表示购买的花费；
但还有一个限制就是可以租用机器；
在最大权闭合子图中，如果没有该限制，其连边应该为inf的容量；
那么考虑租用，将任务与机器的连边inf改为租金即可；
租用机器的操作只与该任务有关，而与其他无关，所以改成 moneyRent即可；
luogu上面我加了O2优化以及快读才过，不知道我这个dinic为啥会T（已加了当前弧优化还是T）
测试点信息
#1 AC 22ms/24208KB
#2 AC 14ms/26064KB
#3 AC 24ms/24096KB
#4 AC 21ms/24200KB
#5 AC 14ms/25788KB
#6 AC 24ms/24224KB
#7 AC 21ms/24184KB
#8 AC 11ms/24120KB
#9 AC 24ms/24192KB
#10 AC 24ms/24212KB
#11 AC 14ms/26096KB
#12 AC 24ms/24336KB
#13 AC 20ms/24104KB
#14 AC 12ms/24964KB
#15 AC 747ms/39924KB
#16 AC 489ms/33560KB
#17 AC 853ms/64272KB
#18 AC 716ms/39828KB
#19 AC 872ms/69156KB
#20 AC 703ms/69504KB
由几个点都接近900ms了，orz；
#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>
#pragma GCC optimize(2)
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 3000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#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 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
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++)
inline ll rd() {
ll 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);
}
ll sqr(ll 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;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int n, m;
int st, ed;
struct node {
int u, v, nxt, w;
}edge[maxn<<1];
int head[maxn], cnt;
void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w;
edge[cnt].nxt = head[u]; head[u] = cnt++;
}
int rk[maxn];
int bfs() {
queue<int>q;
ms(rk);
rk[st] = 1; q.push(st);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].v;
if (rk[to] || edge[i].w <= 0)continue;
rk[to] = rk[tmp] + 1; q.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(edge[i].w, flow - add));
if (!tmpadd) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
}
return add;
}
ll ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
int main()
{
//ios::sync_with_stdio(0);
memset(head, -1, sizeof(head));
rdint(n); rdint(m);
st = 0; ed = n + m + 1;
int sum = 0;
for (int i = 1; i <= n; i++) {
int moy, num;
moy = rd(); sum += moy;
addedge(st, i, moy); addedge(i, st, 0);
//rdint(num);
num = rd();
for (int j = 0; j < num; j++) {
int ID; ID = rd(); moy = rd();
//rdint(ID); rdint(moy);
addedge(i, n + ID, moy); addedge(n + ID, i, 0);
}
}
for (int i = 1; i <= m; i++) {
int moy;// rdint(moy);
moy = rd();
addedge(n + i, ed, moy); addedge(ed, n + i, 0);
}
dinic();
printf("%d
", sum - ans);
return 0;
}