CSU 1319 CX‘s dreams 最大权闭合图 求最多的正点权个数

标签：http   io   ar   os   sp   for   on   ad   ef   

题目链接：点击打开链接

思路：

显然就是问最大权闭合图 和 能取最多的正点权个数

1、首先对于正权值的付出，直接取，而对于梦想也忽略正权值的付出，这样就转成一个裸的最大权闭合图了。

2、计算此时的正点权个数：把所有点权*大数C，然后把正点权值+1，跑出来流量就是 flow / C, 最多的正点权个数就是 正点权点集-flow%C.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
//点标 [0,n]
typedef long long ll;
const int N = 3010;
const int M = 500010;
const ll INF = 1e18;
template<class T>
struct Max_Flow {
    int n;
    int Q[N], sign;
    int head[N], level[N], cur[N], pre[N];
    int nxt[M], pnt[M], E;
    T cap[M];
    void Init(int n) {
        this->n = n+1;
        E = 0;
        std::fill(head, head + this->n, -1);
    }
    //有向rw 就= 0
    void add(int from, int to, T c, T rw) {
        pnt[E] = to;
        cap[E] = c;
        nxt[E] = head[from];
        head[from] = E++;

        pnt[E] = from;
        cap[E] = rw;
        nxt[E] = head[to];
        head[to] = E++;
    }
    bool Bfs(int s, int t) {
        sign = t;
        std::fill(level, level + n, -1);
        int *front = Q, *tail = Q;
        *tail++ = t; level[t] = 0;
        while(front < tail && level[s] == -1) {
            int u = *front++;
            for(int e = head[u]; e != -1; e = nxt[e]) {
                if(cap[e ^ 1] > 0 && level[pnt[e]] < 0) {
                    level[pnt[e]] = level[u] + 1;
                    *tail ++ = pnt[e];
                }
            }
        }
        return level[s] != -1;
    }
    void Push(int t, T &flow) {
        T mi = INF;
        int p = pre[t];
        for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
            mi = std::min(mi, cap[p]);
        }
        for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
            cap[p] -= mi;
            if(!cap[p]) {
                sign = pnt[p ^ 1];
            }
            cap[p ^ 1] += mi;
        }
        flow += mi;
    }
    void Dfs(int u, int t, T &flow) {
        if(u == t) {
            Push(t, flow);
            return ;
        }
        for(int &e = cur[u]; e != -1; e = nxt[e]) {
            if(cap[e] > 0 && level[u] - 1 == level[pnt[e]]) {
                pre[pnt[e]] = e;
                Dfs(pnt[e], t, flow);
                if(level[sign] > level[u]) {
                    return ;
                }
                sign = t;
            }
        }
    }
    T Dinic(int s, int t) {
        pre[s] = -1;
        T flow = 0;
        while(Bfs(s, t)) {
            std::copy(head, head + n, cur);
            Dfs(s, t, flow);
        }
        return flow;
    }
};
Max_Flow <ll>F;

ll dream[N], work[N], ans;
int from, to;
int n, m;
const ll C = 1e6;
void input(){
	ans = 0;
	from = 0; to = n+m+1;
	F.Init(to);
	for(int i = 1; i <= n; i++)
	{
		scanf("%lld", &dream[i]);
		F.add(from, i, dream[i]*C+1LL, 0);
		ans += dream[i];
	}
	for(int i = 1; i <= m; i++)
	{
		scanf("%lld", &work[i]);
		if(work[i] >= 0) ans += work[i];
		else
			F.add(n +i, to, -work[i]*C, 0);
	}
	for(int i = 1, siz, u; i <= n; i++)
	{
		scanf("%d", &siz);
		while(siz--){
			scanf("%d", &u);
			if(work[u] >= 0)continue;
			F.add(i, n+u, INF, 0);
		}
	}
}
int main() {
	while(~scanf("%d %d", &n, &m)){
		input();
		ll flow = F.Dinic(from,to);
	//	cout<<"FLOW:"<<flow<<endl;
		cout<< ans - flow / C << " " << n-flow % C <<endl;
	}
	return 0;
}
/*
3 3
2 3 5
-2 -3 -5
2 1 2
2 1 2
1 3

3 3
0 0 0
-1 -10 -10
2 1 2
2 1 2
3 3 2 1


*/

CSU 1319 CX‘s dreams 最大权闭合图 求最多的正点权个数

标签：http   io   ar   os   sp   for   on   ad   ef   

原文地址：http://blog.csdn.net/qq574857122/article/details/41495073