loj#6073. 「2017 山东一轮集训 Day5」距离(费用流)

题意

sol

我们可以把图行列拆开，同时对于行/列拆成很多个联通块，然后考虑每个点所在的行联通块/列联通块的贡献。

可以这样建边

从s向每个行联通块连联通块大小条边，每条边的容量为1，费用为\(i\)(i表示这是第几条边)。

从每个点所在的行联通块向列联通块连边，容量为1，费用为0

从每个列联通块向t连联通块大小条边，每条边的容量为1，费用为\(i\)(i表示这是第几条边)。

这样跑最小费用最大流，每增光一次的费用就是答案。预处理后o(1)回答即可

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define ull unsigned long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 5001, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
template <typename a> a inv(a x) {return fp(x, mod - 2);}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

int n, s, t , tt;
char s[51][51];
int id[51][51][2], c1 = 1, c2 = 1, ans[maxn * maxn], tot1[20 * maxn], tot2[20 * maxn ], num1, num2;
struct edge {
    int u, v, w, f, nxt;
}e[2 * maxn * maxn];
int head[maxn  * 20 + 1], num;
void add_edge(int x, int y, int w, int f) {
    e[num] = (edge){x, y, w, f, head[x]};
    head[x] = num++;
}
void addedge(int x, int y, int w, int f) {
    //printf("%d %d %d %d\n", x, y, w, f);
    add_edge(x, y, w, f);
    add_edge(y, x, -w, 0);
}
int dis[maxn * 10], vis[maxn * 10], pre[maxn * 10];
int spfa() {
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    queue<int> q; q.push(s); dis[s] = 0;
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = 0;
        for(int i = head[p]; ~i; i = e[i].nxt) {
            int to = e[i].v, w = e[i].w;
            if(dis[to] > dis[p] + w && e[i].f) {
                dis[to] = dis[p] + w; pre[to] = i;
                if(!vis[to]) vis[to] = 1, q.push(to);
            }
        }
    }
    return dis[tt];
}
int mcmf() {
    int val = spfa(), dec = inf;
    for(int k = tt; k != s; k = e[pre[k]].u) chmin(dec, e[pre[k]].f);
    for(int k = tt; k != s; k = e[pre[k]].u) e[pre[k]].f -= dec, e[pre[k] ^ 1].f += dec;
    return dec * val;
}
signed main() {
    //freopen("a.in", "r", stdin);
    memset(head, -1, sizeof(head));
    n = read(); s = 0; t = n * n * 10, tt = t + 1; c2 = n * n * 3 + 1;
    for(int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            if(s[i][j] == '#') tot1[c1] = num1, num1 = 0, c1++; 
            else id[i][j][0] = c1, num1++;
            if(s[j][i] == '#') tot2[c2] = num2, num2 = 0, c2++;
            else id[j][i][1] = c2, num2++;
        }
        if(num1) tot1[c1++] = num1, num1 = 0;
        if(num2) tot2[c2++] = num2, num2 = 0;
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(id[i][j][0] && id[i][j][1])
                addedge(id[i][j][0], id[i][j][1], 0, 1);
    for(int i = 1; i <= c1; i++) 
        for(int j = 0; j < tot1[i]; j++)
            addedge(s, i, j, 1);
    for(int i = n * n * 3 + 1; i <= c2; i++)
        for(int j = 0; j < tot2[i]; j++)
            addedge(i, t, j, 1);
    for(int i = 1; i <= 2 * n * n; i++)
        addedge(t, tt, 0, 1);
    for(int i = 1; i <= n * n; i++)
        ans[i] = ans[i - 1] + mcmf();
    int q = read();
    while(q--) cout << ans[read()] << '\n';
    return 0;
}