日常训练 20170603 棋盘

给定一个 n×n 的棋盘，棋盘上每个位置要么为空要么为障碍。定义棋盘上两个位置 (x,y),(u,v) 能互相攻击当前仅当满足以下两个条件：
1.x=u 或 y=v
2.对于 (x,y) 与 (u,v) 之间的所有位置，均不是障碍。
现在有 q 个询问，每个询问给定 ki，要求从棋盘中选出 ki 个空位置来放棋子，问最少互相能攻击到的棋子对数是多少？
n≤50 , q≤10000 , k≤棋盘中空位置数量
本来是道费用流大裸题，结果只有我没想出来。考虑棋盘模型，一排表示行，一排表示列，然后棋子相互攻击的贡献是递增的，所以 S 到表示行的点和 表示列的点到 T 之间连递增费用的边，如果一个位置不是障碍，就把对应行列连起来，障碍隔开的行和列只要建不同的点即可。

#include<bits/stdc++.h>const int N = 55;const int P = N * N * 2;const int INF = 1e9;template <typename T> void read(T &x) {    x = 0; char c = getchar();    for (; !isdigit(c); c = getchar());    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';}int n, a[N][N], b[N][N], first[P], s, q[P + 10], dis[P], from[P], ans[N * N], cnt, line, S, T, Q, x;bool inq[P];char ch[N][N];struct edge{    int x, y, f, v, next;}mp[N*N*N*N*2];void ins(int x, int y, int f, int v) {    //printf("ins x=%d y=%d f=%d v=%d\n", x, y, f, v);    mp[++s] = (edge) {x, y, f, v, first[x]}; first[x] = s;    mp[++s] = (edge) {y, x, 0, -v, first[y]}; first[y] = s;}bool SPFA() {    for (int i=S; i <= T; i++)        inq[i] = 0, dis[i] = INF;    int head = 1, tail = 2;    inq[q[head] = S] = 1;    dis[q[head]] = 0;    while (head != tail) {        int x = q[head];        //printf("x=%d\n", x);        for (int t=first[x]; t; t=mp[t].next) {            //printf("walk x=%d y=%d f=%d v=%d\n", mp[t].x, mp[t].y, mp[t].f, mp[t].v);            if (mp[t].f && dis[x] + mp[t].v < dis[mp[t].y]) {                dis[mp[t].y] = dis[x] + mp[t].v;                from[mp[t].y] = t;                if (!inq[mp[t].y]) {                    inq[q[tail++] = mp[t].y] = 1;                    if (tail > P) tail = 1;                }            } }        inq[q[head++]] = 0;        if (head > P) head = 1;     }    return dis[T] < INF;}void mcf(int i) {    int fl = INF;    for (int x=T; x != S; x=mp[from[x]].x)        fl = std::min(fl, mp[from[x]].f);    for (int x=T; x != S; x=mp[from[x]].x)        mp[from[x]].f -= fl,        mp[from[x]^1].f += fl;    ans[i] = ans[i - 1] + dis[T];    //printf("i=%d Dinic=%d\n", i, dis[T]);}int main() {    //freopen("A.in", "r", stdin);    //freopen("A.out", "w", stdout);    read(n);    for (int i=1; i <= n; i++)        scanf("%s", ch[i] + 1);    for (int i=0; i <= n + 1; i++)        ch[0][i] = ch[n + 1][i] = ch[i][0] = ch[i][n + 1] = '#';    for (int i=1; i <= n; i++)        for (int j=1; j <= n; j++) {            if (ch[i][j] == '#') continue;            if (ch[i][j - 1] == '.')                a[i][j] = a[i][j - 1];            else                a[i][j] = ++cnt;        }    line = cnt;    for (int j=1; j <= n; j++)        for (int i=1; i <= n; i++) {            if (ch[i][j] == '#') continue;            if (ch[i - 1][j] == '.')                b[i][j] = b[i - 1][j];            else                b[i][j] = ++cnt;        }    S = 0; T = cnt + 1; s = 1;    for (int j=1; j <= n; j++)        for (int i=1; i <= n; i++)            if (ch[i][j] == '.')                ins(a[i][j], b[i][j], 1, 0);    for (int i=1; i <= line; i++)        for (int j=0; j < n; j++)            ins(S, i, 1, j);    for (int i=line + 1; i <= cnt; i++)        for (int j=0; j < n; j++)            ins(i, T, 1, j);    for (int i=1; SPFA(); i++) mcf(i);    for (read(Q); Q--;)        read(x),        printf("%d\n", ans[x]);    return 0;}