loj6068. 「2017 山东一轮集训 Day4」棋盘

链接

https://loj.ac/problem/6068

思路

上来没头绪，后来套算法，套了个网络流
经典二分图
左边横，右边列
先重新算一下行和列，就是他们x相通的的算一个
然后就去掉了障碍的作用
然后每一行贡献是递增的(0,1,2,3,4………)
直接暴力连上每条可能有的流量为1的边就行了
下面的图可能没啥用就是个普通二分图

错误

有的数组开小了
有的memset(1e6)
T成40

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 57, M = 3e5 + 7, inf = 0x3f3f3f3f;
int read() {
    int x = 0, f = 1;
    char s = getchar();
    for (; s > '9' || s < '0'; s = getchar())if (s == '-')f = -1;
    for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}
int n, m, S, T, dsr[N * N];
char s[N][N];
struct node {
    int u, v, nxt, cost, flow;
} e[M];
int head[M], tot = 1;
void add_edge(int u, int v, int flow, int cost) {
    e[++tot].v = v, e[tot].u = u;
    e[tot].cost = cost, e[tot].flow = flow, e[tot].nxt = head[u], head[u] = tot;
}
void add(int u, int v, int flow, int cost) {
    add_edge(u, v, flow, cost);
    add_edge(v, u, 0, -cost);
}
int dis[M], vis[M], frm[M];
int spfa() {
    for (int i = 1; i <= T; ++i) dis[i] = inf, vis[i] = 0;
    queue<int> q;
    q.push(S);
    dis[S] = 0;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].v;
            if (dis[u] + e[i].cost < dis[v] && e[i].flow > 0) {
                dis[v] = dis[u] + e[i].cost;
                frm[v] = i;
                if (!vis[v]) {
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
    return dis[T] != inf;
}
int ans, maxflow;
void work() {
    spfa();
    int mn = inf;
    for (int i = T; i != S; i = e[frm[i]].u) mn = min(e[frm[i]].flow, mn);
    for (int i = T; i != S; i = e[frm[i]].u) {
        e[frm[i]].flow -= mn;
        e[frm[i] ^ 1].flow += mn;
        ans += e[frm[i]].cost * mn;
    }
    maxflow += mn;
}
int belong_x[N][N], belong_y[N][N], hav_x[N * N], hav_y[N * N], num_x, num_y;
int main() {
    n = read();
    for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    int js = n * n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (s[i][j] == '#') {
                js--;
                continue;
            }
            belong_x[i][j] = (s[i][j - 1] == '.') ? belong_x[i][j - 1] : ++num_x;
            belong_y[i][j] = (s[i - 1][j] == '.') ? belong_y[i - 1][j] : ++num_y;
            hav_x[belong_x[i][j]]++;
            hav_y[belong_y[i][j]]++;
        }
    }
    S = num_x + num_y + 1, T = num_x + num_y + 2;
    for (int i = 1; i <= num_x; ++i)
        for (int j = 0; j < hav_x[i]; ++j) add(S, i, 1, j);
    for (int i = 1; i <= num_y; ++i)
        for (int j = 0; j < hav_y[i]; ++j) add(i + num_x, T, 1, j);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (s[i][j] == '.')
                add(belong_x[i][j], num_x + belong_y[i][j], 1, 0);
    for (int i = 1; i <= js; ++i) work(), dsr[i] = ans;
    m = read();
    while (m--) {
        int x = read();
        if (x > js)
            puts("0");
        else
            printf("%d\n", dsr[x]);
    }

    return 0;
}