HDU 3338 Kakuro Extension

HDU_3338

    至于具体怎么建图在其他博客里可以找得到（网络流的题解实在写起来比较费劲，所以这次就偷懒一下了……），不过值得一提的是，不必像大多数博客说的那样每个空白格子既要向管辖行的run连条边又要向管辖列的run连条边，实际上只要把对应的管辖行的run和对应的管辖列的run连条边就可以了，这条边就代表了这个空白格子，流过的流量就是这个格子要填的数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 20010
#define MAXN 110
#define MAXM 60010
#define INF 0x3f3f3f3f
int N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM], id[MAXN][MAXN];
int S, T, d[MAXD], work[MAXD], q[MAXD];
int R, C, g[MAXN][MAXN], right[MAXN][MAXN], down[MAXN][MAXN], rid[MAXN][MAXN], cid[MAXN][MAXN];
char b[10];
void init()
{
    int i, j, k;
    memset(g, 0, sizeof(g));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
        {
            scanf("%s", b);
            if(b[0] == '.') g[i][j] = 1;
            else if(b[3] != 'X')
            {
                b[3] = '\0';
                g[i][j] = down[i][j] = right[i][j] = -1;
                if(b[0] != 'X') sscanf(b, "%d", &down[i][j]);
                if(b[4] != 'X') sscanf(b + 4, "%d", &right[i][j]);
            }
        }
    R = C = 0;
    memset(rid, 0, sizeof(rid));
    memset(cid, 0, sizeof(cid));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(g[i][j] == 1)
            {
                if(!rid[i][j])
                {
                    ++ R;
                    for(k = j; k <= M && g[i][k] == 1; k ++) rid[i][k] = R;
                }
                right[i][j - 1] -= k - j;
                if(!cid[i][j])
                {
                    ++ C;
                    for(k = i; k <= N && g[k][j] == 1; k ++) cid[k][j] = C;
                }
                down[i - 1][j] -= k - i;
            }
}
void add(int x, int y, int z)
{
    v[e] = y, flow[e] = z;
    next[e] = first[x], first[x] = e ++;
}
void build()
{
    int i, j, k, p;
    S = 0, T = R + C + 1;
    memset(first, -1, sizeof(first[0]) * (T + 1)), e = 0;
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(g[i][j] == 1)
            {
                id[i][j] = e;
                add(rid[i][j], R + cid[i][j], 8), add(R + cid[i][j], rid[i][j], 0);
            }
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(g[i][j] == -1)
            {
                if(right[i][j] != -1)
                    p = rid[i][j + 1], add(S, p, right[i][j]), add(p, S, 0);
                if(down[i][j] != -1)
                    p = cid[i + 1][j], add(R + p, T, down[i][j]), add(T, R + p, 0);
            }
}
int bfs()
{
    int i, j, rear = 0;
    memset(d, -1, sizeof(d[0]) * (T + 1));
    d[S] = 0, q[rear ++] = S;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(flow[j] && d[v[j]] == -1)
            {
                d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
                if(v[j] == T) return 1;
            }
    return 0;
}
int dfs(int cur, int a)
{
    if(cur == T) return a;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i], std::min(a, flow[i])))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic()
{
    int ans = 0, t;
    while(bfs())
    {
        memcpy(work, first, sizeof(first[0]) * (T + 1));
        while(t = dfs(S, INF)) ans += t;
    }
    return ans;
}
void solve()
{
    int i, j;
    build();
    dinic();
    for(i = 1; i <= N; i ++)
    {
        for(j = 1; j <= M; j ++)
        {
            if(j != 1) printf(" ");
            if(g[i][j] == 1) printf("%d", flow[id[i][j] ^ 1] + 1);
            else printf("_");
        }
        printf("\n");
    }
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();
    }
    return 0;
}