[CEOI2008]order BZOJ1391 网络流

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

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\n", sum - ans);

    return 0;

}