牛客练习赛45 F Magic Slab

  这道题，可以这样理解，有N* N个点，他们都能产生正贡献，有2N个点，他们都会产生负贡献，还有M个点也是有正贡献的，现在就是连边构成一个最大权闭合子图来求解这个问题了，于是，这题变成了一个模板题了。

最大权闭合子图

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 3e3 + 7, maxM = 1.6e4 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to; ll flow;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
    int deep[maxN], cur[maxN], S, T, node;
    queue<int> Q;
    inline bool bfs()
    {
        while(!Q.empty()) Q.pop();
        for(int i=0; i<=node; i++) deep[i] = 0;
        deep[S] = 1;
        Q.push(S);
        int u;
        while(!Q.empty())
        {
            u = Q.front(); Q.pop();
            ll f;
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to; f = edge[i].flow;
                if(f && !deep[v])
                {
                    deep[v] = deep[u] + 1;
                    Q.push(v);
                }
            }
        }
        return deep[T];
    }
    ll dfs(int u, ll dist)
    {
        if(u == T) return dist;
        ll f, di;
        for(int &i=cur[u], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to; f = edge[i].flow;
            if(f && deep[v] == deep[u] + 1)
            {
                di = dfs(v, min(dist, f));
                if(di)
                {
                    edge[i].flow -= di; edge[i ^ 1].flow += di;
                    return di;
                }
            }
        }
        return 0;
    }
    inline ll Dinic()
    {
        ll ans = 0, tmp;
        while(bfs())
        {
            for(int i=0; i<=node; i++) cur[i] = head[i];
            while((tmp = dfs(S, INF))) ans += tmp;
        }
        return ans;
    }
} mf;
inline void init()
{
    cnt = 0; mf.S = 0; mf.T = N * N + 2 * N + M + 1; mf.node = N * N + 2 * N + M + 2;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
}
inline int get_id(int x, int y) { return (x - 1) * N + y; }
int main()
{
    scanf("%d%d", &N, &M);
    init(); ll sum = 0;
    for(int i=1, cij, id; i<=N; i++)
    {
        for(int j=1; j<=N; j++)
        {
            scanf("%d", &cij); sum += cij;
            _add(mf.S, id = get_id(i, j), cij);
            _add(id, N * N + N + i, INF);
            _add(id, N * N + j, INF);
        }
    }
    for(int i=1, cost; i<=N; i++)
    {
        scanf("%d", &cost);
        _add(N * N + N + i, mf.T, cost);
    }
    for(int i=1, cost; i<=N; i++)
    {
        scanf("%d", &cost);
        _add(N * N + i, mf.T, cost);
    }
    for(int i=1, x1, y1, x2, y2, ki, u, v, id; i<=M; i++)
    {
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &ki); sum += ki;
        u = get_id(x1, y1); v = get_id(x2, y2); id = N * N + 2 * N + i;
        _add(mf.S, id, ki); _add(id, u, INF); _add(id, v, INF);
    }
    printf("%lld\n", sum - mf.Dinic());
    return 0;
}