【费用流】Kaka's Matrix Travels

最新推荐文章于 2018-09-12 19:44:14 发布

Whjpji

最新推荐文章于 2018-09-12 19:44:14 发布

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分类专栏： OI 文章标签： matrix numbers each file output input

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本文探讨了在N×N棋盘上，通过特定规则移动骑士以最大化累计得分的方法，并利用费用流算法解决该问题。重点在于理解如何通过拆分点、设定超级源和求解最大费用流来找到最优路径。

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Description
On an N × N chessboard with a non-negative number in each grid, Kaka starts his matrix travels
with SUM = 0. For each travel, Kaka moves one rook from the left-upper grid to the right-bottom
one, taking care that the rook moves only to the right or down. Kaka adds the number to SUM in
each grid the rook visited, and replaces it with zero. It is not difficult to know the maximum
SUM Kaka can obtain for his first travel. Now Kaka is wondering what is the maximum SUM he can
obtain after his Kth travel. Note the SUM is accumulative during the K travels.

Input
The first line contains two integers N and K (1 ≤ N ≤ 50, 0 ≤ K ≤ 10) described above. The
following N lines represents the matrix. You can assume the numbers in the matrix are no more
than 1000.

Output
The maximum SUM Kaka can obtain after his Kth travel.

Sample Input
3 2
1 2 3
0 2 1
1 4 2

Sample Output
15

此题考察费用流的应用。把每个点拆成上下两个点，加个超级源，流量设为K，求一次最大费用流即可。
Accode：

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <cstring>
#define pos(i, j, x) (((i) * n + (j) << 1) + 1 + (x))

const char fi[] = "number.in";
const char fo[] = "number.out";
const int maxN = 5010, SIZE = 0xffff;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int u, v, f, d; Edge *next, *back;
    Edge(int u, int v, int f, int d, Edge *next):
        u(u), v(v), f(f), d(d), next(next) {}
} *edge[maxN], *pre[maxN]; bool marked[maxN];
int dist[maxN], q[SIZE + 1], N, n, S, T;

void init_file()
{
    freopen(fi, "r", stdin);
    freopen(fo, "w", stdout);
    return;
}

inline int getint()
{
    int res = 0; char tmp;
    while (!isdigit(tmp = getchar()));
    do res = (res << 3) + (res << 1) + tmp - '0';
    while (isdigit(tmp = getchar()));
    return res;
}

inline void Ins(int u, int v, int f, int d)
{
    edge[u] = new Edge(u, v, f, d, edge[u]);
    edge[v] = new Edge(v, u, 0, -d, edge[v]);
    edge[u] -> back = edge[v];
    edge[v] -> back = edge[u];
    return;
}

void readdata()
{
    n = getint(); N = getint();
    T = pos(n - 1, n - 1, 1); S = T + 1;
    for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
    {
        int x = getint();
        if (i < n - 1)
            Ins(pos(i, j, 1), pos(i + 1, j, 0), INF, 0);
        if (j < n - 1)
            Ins(pos(i, j, 1), pos(i, j + 1, 0), INF, 0);
        Ins(pos(i, j, 0), pos(i, j, 1), 1, x);
        Ins(pos(i, j, 0), pos(i, j, 1), INF, 0);
    }
    Ins(S, 1, N, 0);
    return;
}

inline bool Spfa()
{
    memset(dist, ~0x3f, sizeof dist);
    memset(marked, 0, sizeof marked);
    memset(pre, 0, sizeof pre); int f = 0, r = 0;
    dist[q[r++] = S] = 0; marked[S] = 1;
    while (f - r)
    {
        int u = q[f++]; marked[u] = 0; f &= SIZE; //
        for (Edge *p = edge[u]; p; p = p -> next)
        if (p -> f > 0)
        {
            int v = p -> v;
            if (dist[u] + p -> d > dist[v])
            {
                dist[v] = dist[u] + p -> d;
                pre[v] = p;
                if (!marked[v])
                    marked[q[r++] = v] = 1, r &= SIZE; //
            }
        }
    }
    return (bool)pre[T];
}

void work()
{
    int ans = 0;
    while (Spfa())
    {
        int Max_flow = INF;
        for (Edge *p = pre[T]; p; p = pre[p -> u])
            Max_flow = std::min(Max_flow, p -> f);
        for (Edge *p = pre[T]; p; p = pre[p -> u])
        {
            p -> f -= Max_flow;
            p -> back -> f += Max_flow;
        }
        ans += dist[T];
    }
    printf("%d\n", ans);
    return;
}

int main()
{
    init_file();
    readdata();
    work();
    return 0;
}