1227 方格取数 2

题目描述 Description

给出一个n*n的矩阵,每一格有一个非负整数Aij,(Aij <= 1000)现在从(1,1)出发,可以往右或者往下走,最后到达(n,n),每达到一格,把该格子的数取出来,该格子的数就变成0,这样一共走K次,现在要求K次所达到的方格的数的和最大

输入描述 Input Description

第一行两个数n,k（1<=n<=50, 0<=k<=10）

接下来n行,每行n个数,分别表示矩阵的每个格子的数

输出描述 Output Description

一个数,为最大和

样例输入 Sample Input

3 1

1 2 3

0 2 1

1 4 2

样例输出 Sample Output

11

数据范围及提示 Data Size & Hint

1<=n<=50, 0<=k<=10

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<functional>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

/*
每个点只能计算一次（不是只能过一次）
于是拆点，把一个点拆成入点和出点，入点和出点之间加两条边，一条是容量为1，费用为权值，一条是容量为INF，费用为0
把每个点向下方和右方加边
*/

const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to, next, cap, flow, cost;
}edge[MAXM];
int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s, int t)
{
    queue<int>q;
    for (int i = 0; i < N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
                if (edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost)
                {
                    dis[v] = dis[u] + edge[i].cost;
                    pre[v] = i;
                    if (!vis[v])
                    {
                        vis[v] = true;
                        q.push(v);
                    }
                }
        }
    }
    if (pre[t] == -1)return false;
    else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost)
{
    int flow = 0;
    cost = 0;
    while (spfa(s, t))
    {
        int Min = INF;
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
        {
            if (Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
        {
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}
int g[55][55];
int n, k;
inline int index(int i, int j)
{
    return (i - 1)*n + j;
}
int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> k;
    init(2*n*n + 2);
    for (int i = 1; i <= n; i++)
    {
        for(int j = 1;j<=n;j++)
        {
            cin >> g[i][j];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            int i1 = index(i, j);
            addedge(i1, i1 + n*n, 1, -g[i][j]);
            addedge(i1, i1 + n*n, INF, 0);
            if (i != n)
            {
                addedge(i1 + n*n, index(i + 1, j), INF, 0);
            }
            if (j != n)
            {
                addedge(i1 + n*n, index(i, j + 1), INF, 0);
            }
        }
    }
    addedge(0, 1, k, 0);
    addedge(n*n * 2, 2 * n * n + 1, INF, 0);
    int ans, tmp;
    tmp = minCostMaxflow(0, 2 * n*n + 1, ans);
    cout << -ans << endl;

}