BZOJ 3144: [Hnoi2013]切糕

3144: [Hnoi2013]切糕

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 1564  Solved: 839
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Description

Input

第一行是三个正整数P,Q,R，表示切糕的长P、 宽Q、高R。第二行有一个非负整数D，表示光滑性要求。接下来是R个P行Q列的矩阵，第z个 矩阵的第x行第y列是v(x,y,z) (1≤x≤P, 1≤y≤Q, 1≤z≤R)。 
100%的数据满足P,Q,R≤40，0≤D≤R，且给出的所有的不和谐值不超过1000。

Output

仅包含一个整数，表示在合法基础上最小的总不和谐值。

Sample Input

2 2 2
1
6 1
6 1
2 6
2 6

Sample Output

6

HINT

最佳切面的f为f(1,1)=f(2,1)=2,f(1,2)=f(2,2)=1

Source

 

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OTZ 最小割

如果用$(x,y,z)$表示网络中的点，那么有两类边：

　　1. $<(x,y,z),(x,y,z+1),V[x][y][z]>$

　　2. $<(x,y,z),(x',y',z-D),INF>$

求最小割即可，注意细节。

#include <cstdio>
#include <cstring>

inline char Char(void)
{
    static const short siz = 1024;
    
    static char buf[siz];
    static char *hd = buf + siz;
    static char *tl = buf + siz;
    
    if (hd == tl)
        fread(hd = buf, 1, siz, stdin);
    
    return *hd++;
}

inline int Int(void)
{
    int ret = 0, neg = 0, c = Char();
    
    for (; c < 48; c = Char())
        if (c == '-')neg ^= true;
    
    for (; c > 47; c = Char())
        ret = ret * 10 + c - '0';
    
    return neg ? -ret : ret;
}

const int mxn = 45;
const int siz = 2000005;
const int inf = 1000000007;

int s, t;
int hd[siz];
int to[siz];
int fl[siz];
int nt[siz];

inline void add(int u, int v, int f)
{
    static int tot = 0, init = 1;
    
    if (init)memset(hd, -1, sizeof(hd)), init = 0;
    
    nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; hd[u] = tot++;
    nt[tot] = hd[v]; to[tot] = u; fl[tot] = 0; hd[v] = tot++;
}

int dep[siz];

inline bool bfs(void)
{
    static int que[siz], head, tail;
    memset(dep, 0, sizeof(dep));
    head = 0, tail = 0;
    que[tail++] = s;
    dep[s] = 1;
    
    while (head != tail)
    {
        int u = que[head++], v; 
        
        for (int i = hd[u]; ~i; i = nt[i])
            if (!dep[v = to[i]] && fl[i])
                dep[que[tail++] = v] = dep[u] + 1;
    }
    
    return dep[t];
}

int cur[siz];

inline int min(int a, int b)
{
    return a < b ? a : b;
}

int dfs(int u, int f)
{
    if (u == t || !f)
        return f;
    
    int used = 0, flow, v;
    
    for (int i = hd[u]; ~i; i = nt[i])
        if (fl[i] && dep[v = to[i]] == dep[u] + 1)
        {
            flow = dfs(v, min(fl[i], f - used));
            
            used += flow;
            fl[i] -= flow;
            fl[i^1] += flow;
            
            if (fl[i])
                cur[u] = i;
            
            if (used == f)
                return f;
        }
        
    if (!used)
        dep[u] = 0;
        
    return used;
}

inline int minCut(void)
{
    int maxFlow = 0, newFlow;
    
    while (bfs())
    {
        memcpy(cur, hd, sizeof(cur));
            
        while (newFlow = dfs(s, inf))
            maxFlow += newFlow;
    }
    
    return maxFlow;
}

int P, Q, R, D, V[mxn][mxn][mxn];

inline int pos(int i, int j, int k)
{
    if (k < 1)return s;
    
    return (i - 1) * (Q * R) + (j - 1) * R + k;
}

const int mv[4][2] = 
{
    {0, +1},
    {0, -1},
    {-1, 0},
    {+1, 0}
};

signed main(void)
{
    P = Int();
    Q = Int();
    R = Int();
    D = Int();
    
    for (int k = 1; k <= R; ++k)
        for (int i = 1; i <= P; ++i)
            for (int j = 1; j <= Q; ++j)
                V[i][j][k] = Int();
                
    s = 0, t = P * Q * R + 1;
    
    for (int i = 1; i <= P; ++i)
        for (int j = 1; j <= Q; ++j)
        {
            for (int k = 1; k <= R; ++k)
            {
                add(pos(i, j, k - 1), pos(i, j, k), V[i][j][k]);
                
                for (int l = 0; l < 4; ++l)
                {
                    int x = i + mv[l][0];
                    int y = j + mv[l][1];
                    
                    if (x < 1 || x > P)continue;
                    if (y < 1 || y > Q)continue;
                    
                    if (k - D > 0)
                        add(pos(i, j, k), pos(x, y, k - D), inf);
                }
            }
            
            add(pos(i, j, R), t, inf);
        }
            
    printf("%d
", minCut());
}

@Author: YouSiki