Luogu P4311 【士兵占领】

Description

传送门

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Solution

首先如果士兵只能给一行或一列造成贡献的答案是(sum_{i = 1}^m l_i + sum_{i = 1}^n c_i)。

但是发现有的士兵可以同时给一列和一行造成贡献。

那就算出这些士兵的个数就行了。

(S)向每一行连容量为(l_i)的有向边；每一列向(T)连容量为(c_i)的有向边。

如果坐标((i,j))不是障碍，那么第(i)行向第(j)列连容量为(1)的有向边。

跑出来的最大流就是贡献为(2)的士兵数量。

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Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int INF = 999999999;
const int N = 10050;
const int M = 200050;

int n, m, s, t, head[N], num = 1, dis[N], k, l[150], c[150], p[150][150], sum[500], ans;

struct Node
{
    int next, to, dis;
} edge[M * 2];

void Addedge(int u, int v, int w)
{
    edge[++num]= (Node){head[u], v, w};
    head[u] = num;
}

template <class T>
void Read(T &x)
{
    x = 0; int p = 0; char st = getchar();
    while (st < '0' || st > '9') p = (st == '-'), st = getchar();
    while (st >= '0' && st <= '9') x = (x << 1) + (x << 3) + st - '0', st = getchar();
    x = p ? -x : x;
    return;
}

template <class T>
void Put(T x)
{
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) Put(x / 10);
    putchar(x % 10 + '0');
    return;
}

bool Bfs()
{
    queue<int> q;
    for (int i = 0; i <= t; i++) dis[i] = 0;
    dis[s] = 1; q.push(s);
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        for (int i = head[u]; i; i = edge[i].next)
            if (!dis[edge[i].to] && edge[i].dis)
            {
                dis[edge[i].to] = dis[u] + 1;
                q.push(edge[i].to);
                if (edge[i].to == t) return 1;
            }
    }
    return 0;
}

int Dinic(int x, int flow)
{
    if (x == t || !flow) return flow;
    int rest = flow;
    for (int i = head[x]; i && rest; i = edge[i].next)
        if (edge[i].dis && dis[edge[i].to] == dis[x] + 1)
        {
            int v = edge[i].to;
            int tmp = Dinic(v, min(flow, edge[i].dis));
            rest -= tmp;
            edge[i].dis -= tmp;
            edge[i ^ 1].dis += tmp;
            if (!tmp) dis[v] = 0;
        }
    return flow - rest;
}

void Add(int u, int v, int w)
{
	Addedge(u, v, w);
	Addedge(v, u, 0);
	return;
}

int Maxflow()
{
	int maxflow = 0, tmp;
	while (Bfs())
	{
		tmp = Dinic(s, INF);
		if (tmp) maxflow += tmp;
	}
	return maxflow;
}

int main()
{
	Read(m); Read(n); Read(k);
	for (int i = 1; i <= m; i++) Read(l[i]), ans += l[i];
	for (int i = 1; i <= n; i++) Read(c[i]), ans += c[i];
	int x, y;
	for (int i = 1; i <= k; i++) Read(x), Read(y), p[x][y] = 1;
	s = 0, t = n + m + 1;
	for (int i = 1; i <= m; i++) Add(s, i, l[i]);
	for (int i = 1; i <= n; i++) Add(i + m, t, c[i]);
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= n; j++)
		{
			if (!p[i][j])
				Add(i, j + m, 1);
			sum[i] = sum[i] + !p[i][j];
			sum[j + m] = sum[j + m] + !p[i][j];
		}
	for (int i = 1; i <= m; i++) if (sum[i] < l[i]) { puts("JIONG!"); return 0; } 
	for (int j = 1; j <= n; j++) if (sum[j + m] < c[j]) { puts("JIONG!"); return 0; } 
	Put(ans - Maxflow());
    return 0;
}