UVa1104/LA5131 Chips Challenge

题目链接

  本题是2011年icpc世界总决赛的D题

题意

  在一个N×N（N≤40）网格里放芯片。其中一些格子已经放了芯片（用C表示）,有些格子不能放（用/表示），有些空格子可以放或者不放芯片（用.表示）。还要放一些芯片（用W表示），使得第i行的总芯片个数（C和W之后）等于第i列。为了保证散热，任意行/列的芯片（包括C和W）不能超过总芯片数的A/B。

分析

  先不考虑任意行/列的芯片不能超过总部件数的A/B，则是上下界循环费用流问题：将行$i$看成结点$x_i$，列$j$看成结点$y_j$，已经放了芯片的格子$(i,j)$对应上下界均为1费用为0的有向边$i\to j$，可以放或者不放的格子$(i,j)$对应容量为1费用为0的有向边$x_i\to y_j$；$y_i$向$x_i$连容量为$inf$费用为-1的有向边来满足第i行的总芯片个数等于第i列的约束。

  再来看任意行/列的芯片不能超过总部件数的A/B这一点怎么处理，如果枚举总芯片数（即枚举最大流），需要跑$O(n^2)$次费用流，会超时。可以枚举每行/列的最大芯片总数$k$（即$y_i$向$x_i$连容量为$k$费用为-1的有向边）求出最大费用$c$（即最大芯片总数）：若$c\times A<k\times B$则不满足任意行/列的芯片不能超过总部件数的A/B，否则更新答案。

  注意：上下界网络拆边调整后跑最大流，附加源点$s$的出边都满流时才可行；循环流消除负权边，也是加源点$s$的出边都满流时才可行。因此要检验最大流是否为源点$s$流出的总容量。

  利用补集的思想，可以得到更优的建图方法：已经放了芯片的格子不连边；可以放或者不放的格子$(i,j)$连容量为1费用为1的有向边$x_i\to y_j$（跑完费用流后此边若有流量则表示该格子不放芯片）；$x_i$向$y_i$连容量为$k$费用为0的有向边来满足第i行的总芯片个数等于第i列的约束；源点$s$向$x_i$连容量为第$i$行所有已经放了芯片的格子总数与可以放或者不放的格子总数之和费用为0的有向边；$y_i$向汇点$t$连容量为第$i$列所有已经放了芯片的格子总数与可以放或者不放的格子总数之和费用为0的有向边。跑最小费用最大流，得到不能放的芯片数最小值（即最大芯片数的补）。

AC 代码

上下界循环费用流版本

#include <iostream>
#include <cstring>
using namespace std;#define M 3520
#define N 82
struct edge {int u, v, cap, flow, cost;} e[M];
int g[N][N], q[M*N], a[N], d[N], p[N], cnt[N], hu[N], c, n, h, b, kase = 0; bool vis[N]; void add_edge(int u, int v, int cap, int cc) {e[c] = {u, v, cap, 0, cc}; g[u][cnt[u]++] = c++; e[c] = {v, u, 0, 0, -cc}; g[v][cnt[v]++] = c++;
}int solve() {int s = 0, t = 2*n+1, f = 0, cc = 0;memset(cnt, c = 0, sizeof(cnt)); memset(a, 0, sizeof(a)); memset(d, 0, sizeof(d));for (int i=1; i<=n; ++i) for (int j=1; j<=n; ++j) {char x; cin >> x;if (x == 'C') ++a[j+n], ++a[i], ++f, ++d[i], --d[j];else if (x == '.') add_edge(i, j+n, 1, 0);}for (int i=1; i<=n; ++i) {hu[i] = c; add_edge(i, i+n, n, 1); add_edge(s, i, n, 0); add_edge(i+n, t, n, 0);if (a[i]) add_edge(i, t, a[i], 0);if (a[i+n]) add_edge(s, i+n, a[i+n], 0);if (d[i] || f*h < a[i]*b) cc = -1;}for (int k=n; k>0; --k) {for (int i=0; i<c; ++i) e[i].flow = 0;for (int i=1; i<=n; ++i) e[hu[i]].cap = e[hu[i]+2].cap = e[hu[i]+4].cap = k;int cost = n*k, flow = cost+f;while (true) {memset(d, 1, sizeof(d)); memset(vis, 0, sizeof(vis)); d[s] = 0; q[0] = s; a[s] = M;int head = 0, tail = 1;while (head < tail) {int u = q[head++]; vis[u] = false;for (int i=0; i<cnt[u]; ++i) {const edge& ee = e[g[u][i]];if (ee.cap > ee.flow && d[ee.v] > d[u]+ee.cost) {d[ee.v] = d[u]+ee.cost;p[ee.v] = g[u][i];a[ee.v] = min(a[u], ee.cap-ee.flow);if (!vis[ee.v]) vis[q[tail++] = ee.v] = true;}}}if (d[t] > M) break;flow -= a[t]; cost -= d[t] * a[t];for (int u=t; u!=s; u=e[p[u]].u) e[p[u]].flow += a[t], e[p[u]^1].flow -= a[t];}if (flow || cost*h < k*b) continue;cc = max(cc, cost-f);}return cc;
}int main() {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);while (cin >> n >> h >> b && n) {int cc = solve();cout << "Case " << ++kase << ": ";cc < 0 ? cout << "impossible" << endl : cout << cc << endl;}return 0;
}

优化建图版本

#include <iostream>
#include <cstring>
using namespace std;#define M 3440
#define N 82
struct edge {int u, v, cap, flow, cost;} e[M];
int g[N][N], q[M*N], a[N], d[N], p[N], cnt[N], hu[N], c, n, h, b, kase = 0; bool vis[N]; void add_edge(int u, int v, int cap, int cc) {e[c] = {u, v, cap, 0, cc}; g[u][cnt[u]++] = c++; e[c] = {v, u, 0, 0, -cc}; g[v][cnt[v]++] = c++;
}int solve() {int s = 0, t = 2*n+1, f = 0, c0 = 0, cc = -1;memset(cnt, c = 0, sizeof(cnt)); memset(a, 0, sizeof(a));for (int i=1; i<=n; ++i) for (int j=1; j<=n; ++j) {char x; cin >> x;if (x == 'C') ++a[j+n], ++a[i], ++f, ++c0;else if (x == '.') ++a[j+n], ++a[i], ++f, add_edge(i, j+n, 1, 1);}for (int i=1; i<=n; ++i) {hu[i] = c; add_edge(i, i+n, n, 0);if (a[i]) add_edge(s, i, a[i], 0);if (a[i+n]) add_edge(i+n, t, a[i+n], 0);}for (int k=0; k<=n; ++k) {for (int i=0; i<c; ++i) e[i].flow = 0;for (int i=1; i<=n; ++i) e[hu[i]].cap = k;int flow = f, cost = f;while (true) {memset(d, 1, sizeof(d)); memset(vis, 0, sizeof(vis)); d[s] = 0; q[0] = s; a[s] = M;int head = 0, tail = 1;while (head < tail) {int u = q[head++]; vis[u] = false;for (int i=0; i<cnt[u]; ++i) {const edge& ee = e[g[u][i]];if (ee.cap > ee.flow && d[ee.v] > d[u]+ee.cost) {d[ee.v] = d[u]+ee.cost;p[ee.v] = g[u][i];a[ee.v] = min(a[u], ee.cap-ee.flow);if (!vis[ee.v]) vis[q[tail++] = ee.v] = true;}}}if (d[t] > M) break;flow -= a[t]; cost -= d[t] * a[t];for (int u=t; u!=s; u=e[p[u]].u) e[p[u]].flow += a[t], e[p[u]^1].flow -= a[t];}if (flow || cost*h < k*b) continue;cc = max(cc, cost - c0);}return cc;
}int main() {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);while (cin >> n >> h >> b && n) {int cc = solve();cout << "Case " << ++kase << ": ";cc < 0 ? cout << "impossible" << endl : cout << cc << endl;}return 0;
}