POJ 1149 PIGS

Posted 2020-12-12 ignorance

原题连接 ： http://poj.org/problem?id=1149

题目大意：有m个养猪场，n个顾客，每个顾客都有若干把养猪场的钥匙，最开始养猪场是全部锁上的，每个顾客来的时候可以用钥匙打开固定的养猪场并进行购买，每个顾客都有一个购买上限；

购买完后可以把打开的猪场里的猪任意的调整到别的打开的猪场；求购买数最大是多少；

解题思路：有一个源点s和汇点t，然后把所有的养猪场的第一个顾客与s连一条边，边权位这些养猪场的总猪数；将每个养猪场的后续顾客和前一个顾客连边权值为无穷；

将所有顾客和t连边，权值为这个顾客的购买上限；跑一遍最大流算法；

ac代码 ：

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <functional>
#include <map>
#include <set>
#include <stack>
#define FT(a, b) memset(a, b, sizeof(a))
#define FAT(a) memset(a, 0, sizeof(a))
using namespace std;
typedef long long ll;
const int M = 2e4 + 100;
const int N = 1e2 + 100;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m, k;
int h[N], e[M], ne[M], w[M], idx;
int deepth[N], gap[N];
int pig[N * 10];
int s = 0, t = 1;
int min_cost, max_flow;
void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
void f_add(int a, int b, int c)
{
    add(a, b, c), add(b, a, 0);
}

void bfs()
{
    FT(deepth, -1);
    FAT(gap);
    queue<int> q;
    q.push(t);
    deepth[t] = 0;
    gap[deepth[t]]++;
    while (q.size())
    {
        int a = q.front();
        q.pop();
        for (int i = h[a]; ~i; i = ne[i])
        {
            int j = e[i];
            if (deepth[j] == -1)
            {
                deepth[j] = deepth[a] + 1;
                gap[deepth[j]]++;
                q.push(j);
            }
        }
    }
}
int dfs(int now, int flow)
{
    if (now == t)
        return flow;
    int nowflow = 0;
    for (int i = h[now]; ~i; i = ne[i])
    {
        int j = e[i];
        if (w[i] && deepth[j] == deepth[now] - 1)
        {
            int k = dfs(j, min(flow - nowflow, w[i]));
            w[i] -= k, w[i ^ 1] += k, nowflow += k;
            if (flow == nowflow)
                return nowflow;
        }
    }
    --gap[deepth[now]];
    if (!gap[deepth[now]])
        gap[s] = n + 2;
    ++deepth[now];
    ++gap[deepth[now]];
    return nowflow;
}
int ISAP()
{
    int ans = 0;
    bfs();
    while (gap[s] < n)
        ans += dfs(s, INF);
    return ans;
}
vector<int> v[N * 10];
int main()
{

#ifdef ONLINE_JUDGE
#else
    freopen("/home/wjy/code/in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // scanf("%d", &T);
    while (T--)
    {
        idx = 0;
        FT(h, -1);
        cin >> m >> n;
        s = 0, t = n + 1;
        for (int i = 1; i <= m; i++)
            cin >> pig[i];
        for (int i = 1; i <= n; i++)
        {
            int a, b, k;
            cin >> a;
            int val = 0;
            while (a--)
            {
                int key;
                cin >> key;
                if (v[key].empty())
                    val += pig[key];
                v[key].push_back(i);
            }
            cin >> b;
            if (val)
                f_add(s, i, val);
            f_add(i, t, b);
        }
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j < v[i].size(); j++)
            {
                f_add(v[i][j - 1], v[i][j], INF);
            }
        }
        printf("%d
", ISAP());
    }
    return 0;
}