HIT 2715 Matrix3

HIT_2715

    这个题目和HIT_2543比较像，每个点有两种性质，走第一次的时候会花费w，后面再走花费就会变为0。于是不妨将一个点拆成两个点，并连两条边，一条为容量为1花费为w的边，另一条为容量为INF花费为0的边。再将其余的边连好之后做最大费用最大流即可，由于题目限制行走过程中不会出现环，所以就不必担心有正圈了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 50
#define MAXD 5010
#define MAXM 35010
#define INF 0x3f3f3f3f
#define NINF 0xc3c3c3c3
int N, K, h[MAXN][MAXN], first[MAXD], e, next[MAXM], u[MAXM], v[MAXM], flow[MAXM], cost[MAXM];
int S, T, q[MAXD], dis[MAXD], pre[MAXD], inq[MAXD];
const int Q = 2500;
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
void add(int x, int y, int f, int c)
{
    u[e] = x, v[e] = y, flow[e] = f, cost[e] = c;
    next[e] = first[x], first[x] = e ++;    
}
void init()
{
    int i, j, k, ni, nj, x, y, nx, w;
    scanf("%d%d", &N, &K);
    S = 0, T = N * N * 2 + 1;
    memset(first, -1, sizeof(first[0]) * (T + 1)), e = 0;
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= N; j ++)
        {
            x = (i - 1) * N + j, y = N * N + x;
            scanf("%d", &w);
            add(x, y, 1, w), add(y, x, 0, -w), add(x, y, INF, 0), add(y, x, 0, 0);
        }
    for(i = 1; i <= N; i ++) for(j = 1; j <= N; j ++) scanf("%d", &h[i][j]);
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= N; j ++)
        {
            x = (i - 1) * N + j, y = N * N + x;
            add(S, x, INF, 0), add(x, S, 0, 0);
            if(i == 1 || i == N || j == 1 || j == N)
                add(y, T, INF, 0), add(T, y, 0, 0);
            for(k = 0; k < 4; k ++)
            {
                ni = i + dx[k], nj = j + dy[k];
                if(ni >= 1 && ni <= N && nj >= 1 && nj <= N && h[ni][nj] < h[i][j])
                {
                    nx = (ni - 1) * N + nj;
                    add(y, nx, INF, 0), add(nx, y, 0, 0);    
                }    
            }
        }
}
int bfs()
{
    int i, x, front, rear;
    front = rear = 0;
    memset(dis, 0xc3, sizeof(dis[0]) * (T + 1));
    dis[S] = 0, pre[S] = -1, q[rear ++] = S;
    memset(inq, 0, sizeof(inq[0]) * (T + 1));
    while(front != rear)
    {
        x = q[front ++], inq[x] = 0;
        front > Q ? front = 0 : 0;
        for(i = first[x]; i != -1; i = next[i])
            if(flow[i] && dis[x] + cost[i] > dis[v[i]])
            {
                dis[v[i]] = dis[x] + cost[i], pre[v[i]] = i;
                if(!inq[v[i]])
                {
                    q[rear ++] = v[i], inq[v[i]] = 1;
                    rear > Q ? rear = 0 : 0;    
                }
            }
    }
    return dis[T] != NINF;
}
void solve()
{
    int i, j, c = 0, a, cnt = 0;
    while((++ cnt) <= K && bfs())
    {
        for(i = pre[T], a = INF; i != -1; i = pre[u[i]])
            a = std::min(a, flow[i]);
        if(a > 1)
            break;
        for(i = pre[T]; i != -1; i = pre[u[i]])
            flow[i] -= a, flow[i ^ 1] += a;
        c += dis[T];
    }
    printf("%d\n", c);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;    
}