「2016山东省队集训」 分子切割 molecule

Description

Sample Input & Output

HINT

Solution

记录一下这道题，网络流建模好题！

首先不考虑1，那么答案就是每个L形两个2最大匹配是多少。

因为一个1只能用一次，因此我们把这个二分图连的边拆开。不能同时用的一对2都连到一个1的点上，然后把这个点拆开，加流量为1的限制即可。

考试的时候写了个类似于匈牙利的玩意，结果TLE+WA。越来越弱了。。。。连网络流也想不出来了。。。

然后1拆点还加了4条边，简直药丸。。。

Code

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int INF = 1009999999;const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};struct edge_type {    int to, next, r;} edge[500005];int S, T, cnte = 1, h[500005];int Q[500005];void ins(int u, int v, int f) {    ++cnte;    edge[cnte].to = v;    edge[cnte].r = f;    edge[cnte].next = h[u];    h[u] = cnte;    ++cnte;    edge[cnte].to = u;    edge[cnte].r = 0;    edge[cnte].next = h[v];    h[v] = cnte;}int tail, head, d[500005], cur[500005];bool bfs() {    tail = head = 0;    for (int i = S; i <= T; ++i) d[i] = -1;    Q[tail++] = S;    d[S] = 0;    int now;    while (head < tail) {        now = Q[head++];        for (int i = h[now]; i; i = edge[i].next) {            if (d[edge[i].to] == -1 && edge[i].r) {                d[edge[i].to] = d[now] + 1;                Q[tail++] = edge[i].to;            }        }    }    return d[T] != -1;}int dfs(int now, int a) {    if (T == now || a == 0) return a;    int ret = 0, f;    for (int &i = cur[now]; i; i = edge[i].next) {        if (d[edge[i].to] != d[now]+1) continue;        if (f = dfs(edge[i].to, min(a, edge[i].r))) {            a -= f;            ret += f;            edge[i].r -= f;            edge[i^1].r += f;            if (a == 0) break;        }    }    if (ret == 0) d[now] = -1;    return ret;}int Dinic() {    int ret = 0;    while (bfs()) {        for (int i = S; i <= T; ++i)            cur[i] = h[i];        ret += dfs(S, INF);    }    return ret;}int a[510][510], id[510][510], id2[510][510], n, m, tot;int main() {    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; ++i)        for (int j = 1; j <= m; ++j) {            scanf("%d", &a[i][j]);            id[i][j] = ++tot;            if (a[i][j] == 1)                id2[i][j] = ++tot;        }    for (int i = 1; i <= n; ++i)        for (int j = 1; j <= m; ++j)            if (a[i][j] == 1) {                ins(id[i][j], id2[i][j], 1);                for (int k = 0; k < 4; ++k) {                    int tx1 = i + dx[k], ty1 = j + dy[k],                        tx2 = i + dx[(k+1)%4], ty2 = j + dy[(k+1)%4];                    if (1 <= tx1 && tx1 <= n && 1 <= tx2 && tx2 <= n && 1 <= ty1 && ty1 <= m && 1 <= ty2 && ty2 <= m)                            if (a[tx1][ty1] == 2 && a[tx2][ty2] == 2) {                                if (tx1 & 1)                                    ins(id[tx1][ty1], id[i][j], 1),                                    ins(id2[i][j], id[tx2][ty2], 1);                                else                                     ins(id[tx2][ty2], id[i][j], 1),                                    ins(id2[i][j], id[tx1][ty1], 1);                            }                }            }    S = 0; T = ++tot;    for (int i = 1; i <= n; ++i)        for (int j = 1; j <= m; ++j) {            if (a[i][j] == 2) {                if (i & 1) ins(S, id[i][j], 1);                else ins(id[i][j], T, 1);            }        }    int ans = Dinic();    printf("%d", ans);    return 0;}