2016北京集训测试赛（六）Problem B: 矩阵

Solution

最小割.
参考BZOJ 3144切糕
在那道题的基础上将建图方法稍作变形: 我们对格子进行黑白染色, 对于两个格子之和\(\le k\)的限制, 就可以确定其中一个是白色格子, 一个是黑色格子. 我们让黑色格子和白色格子的点的顺序相反, 就可以表示限制了.
目前的代码还是WA的.

#include <cstdio>
#include <cctype>
#include <vector>
#include <deque>
#include <algorithm>

namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1;
        char c;
        while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = 50, M = 50, INF = (int)1e9;
int D[N][M], R[N][M];
struct graph
{
    struct node;
    struct edge
    {
        node *v; int cap; edge *op;
    };
    struct node
    {
        std::vector<edge*> edg;
        int dep;
        inline node()
        {
            edg.clear();
        }
    }nd[N][M][11], *S, *T;
    inline graph()
    {
        S = new node; T = new node;
    }
    inline void addEdge(node *u, node *v, int cap)
    {
        edge *a = new edge, *b = new edge;
        a->v = v; a->cap = cap; a->op = b;
        b->v = u; b->cap = 0; b->op = a;
        u->edg.push_back(a); v->edg.push_back(b);
    }
    int cnt;
    void clear(node *u)
    {
        u->dep = - cnt;
        for(auto edg : u->edg) if(edg->v->dep != -cnt) clear(edg->v);
    }
    inline int BFS()
    {
        ++ cnt; clear(S); S->dep = 0;
        static std::deque<node*> que; que.clear();
        que.push_back(S);
        for(; ! que.empty(); que.pop_front())
        {
            node *u = que.front();
            for(auto edg : u->edg) if(edg->cap && edg->v->dep == - cnt) edg->v->dep = u->dep + 1, que.push_back(edg->v);
        }
        return T->dep != - cnt;
    }
    int DFS(node *u, int flw)
    {
        if(u == T) return flw;
        int flowSum = 0;
        for(auto edg : u->edg) if(edg->cap && edg->v->dep == u->dep + 1)
        {
            int currentFlow = DFS(edg->v, std::min(edg->cap, flw - flowSum));
            flowSum += currentFlow;
            edg->cap -= currentFlow; edg->op->cap += currentFlow;
            if(flowSum == flw) return flowSum;
        }
        if(! flowSum) u->dep = -1;
        return flowSum;
    }
    inline int dinic()
    {
        cnt = 0;
        int res = 0;
        while(BFS())
            res += DFS(S, INF);
        return res;
    }
}G;
int main()
{

    #ifndef ONLINE_JUDGE

    freopen("matrix.in", "r", stdin);
    freopen("matrix.out", "w", stdout);

    #endif

    using namespace Zeonfai;
    int n = getInt(), m = getInt();
    for(int i = 0; i < n - 1; ++ i) for(int j = 0; j < m; ++ j) D[i][j] = getInt();
    for(int i = 0; i < n; ++ i) for(int j = 0; j < m - 1; ++ j) R[i][j] = getInt();
    for(int i = 0; i < n; ++ i) for(int j = 0; j < m; ++ j)
    {
        if(i ^ j & 1)
        {
            G.addEdge(G.S, &G.nd[i][j][9], INF);
            for(int k = 9; k >= 1; -- k) G.addEdge(&G.nd[i][j][k], &G.nd[i][j][k - 1], 10 - k);
            G.addEdge(&G.nd[i][j][0], G.T, INF);
            for(int k = 0; k <= 9; ++ k)
            {
                if(i + 1 < n && D[i][j] - k <= 10 && D[i][j] - k >= 1) G.addEdge(&G.nd[i + 1][j][D[i][j] - k], &G.nd[i][j][k], INF);
                if(j + 1 < m && R[i][j] - k <= 10 && R[i][j] - k >= 1) G.addEdge(&G.nd[i][j + 1][R[i][j] - k], &G.nd[i][j][k], INF);
            }
        }
        else
        {
            G.addEdge(G.S, &G.nd[i][j][1], INF);
            for(int k = 1; k <= 9; ++ k) G.addEdge(&G.nd[i][j][k], &G.nd[i][j][k + 1], 10 - k);
            G.addEdge(&G.nd[i][j][10], G.T, INF);
            for(int k = 1; k <= 10; ++ k)
            {
                if(i + 1 < n && D[i][j] - k <= 9 && D[i][j] - k >= 0) G.addEdge(&G.nd[i][j][k], &G.nd[i + 1][j][D[i][j] - k], INF);
                if(j + 1 < m && R[i][j] - k <= 9 && R[i][j] - k >= 0) G.addEdge(&G.nd[i][j][k], &G.nd[i][j + 1][R[i][j] - k], INF);
            }
        }
    }
    printf("%d\n", n * m * 10 - G.dinic());
}