[TJOI2015]线性代数 网络流

题面

题面

题解

先化一波式子：

\[D = (A \cdot B - C)A^T
\]

\[= \sum_{i = 1}^{n}H_{1i}\cdot A^T_{i1}
\]

\[H_{1i} = (\sum_{j = 1}^{n}A_{1j} \cdot B_{ji}) - C_{1i}
\]

\[D = \sum_{i = 1}^{n}{((\sum_{j = 1}^{n}A_{1j} \cdot B_{ji}) - C_{1i}) A^T_{i1}}
\]

\[D = \sum_{i = 1}^{n}{((\sum_{j = 1}^{n}A_{1j} \cdot B_{ji}) - C_{1i}) A_{1i}}
\]

\[= \sum_{i = 1}^{n} \sum_{j = 1}^{n}A_j \cdot B_{ji} \cdot A_i - \sum_{i = 1}^{n}C_iA_i
\]

\[= \sum_{i = 1}^{n} \sum_{j = 1}^{n}A_i \cdot B_{ij} \cdot A_j - \sum_{i = 1}^{n}C_iA_i
\]

因此选\(i\)和\(j\)则得到\(B_{ij}\)的贡献，选\(i\)则花费\(C_i\)的代价。

因此我们有如下关系:选\((i, j)\)则必选\(i, j\).

因此建图方式如下：

• 对于每个二元组\((i, j)\),我们连\(s --- > (i, j) : B_{ij}\)
• 对于每个二元组\((i, j)\),我们连\((i, j) ---> i : inf , (i, j) ---> j : inf\)
• 对于每个点\(i\),我们连\(i ---> t : C_i\)

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 550
#define ac 1000000
#define maxn 3000000
#define inf 1000000000
//#define D printf("in line %d\n", __LINE__);

int n, s, t, x, addflow, ans, all;
int B[AC][AC], C[AC];
int Head[ac], date[maxn], Next[maxn], haveflow[maxn], tot = 1;
int have[ac], good[ac], c[ac], last[ac];
int q[ac], head, tail;

inline int read()
{
    int x = 0;char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

inline void upmin(int &a, int b) {if(b < a) a = b;}
inline void upmax(int &a, int b) {if(b > a) a = b;}

inline void add(int f, int w, int S)
{
    date[++ tot] = w, Next[tot] = Head[f], Head[f] = tot, haveflow[tot] = S;
    date[++ tot] = f, Next[tot] = Head[w], Head[w] = tot, haveflow[tot] = 0;
    //printf("%d --- > %d : %d\n", f, w, S);
}

void bfs()
{
    q[++ tail] = t, have[1] = c[t] = 1;
    while(head < tail)
    {
        int x = q[++ head];
        for(R i = Head[x]; i; i = Next[i])
        {
            int now = date[i];
            if(!c[now] && haveflow[i ^ 1])
            {
                ++ have[c[now] = c[x] + 1];
                q[++ tail] = now;
            }
        }
    }
    memcpy(good, Head, sizeof(good));
}

void aru()
{
    while(x != s)
    {
        haveflow[last[x]] -= addflow;
        haveflow[last[x] ^ 1] += addflow;
        x = date[last[x] ^ 1];
    }
    ans -= addflow, addflow = inf;
}

void ISAP()
{
    bool done = false;
    addflow = inf, x = s;
    while(c[t] != all + 10)
    {
        if(x == t) aru();
        done = false;
        for(R i = good[x]; i; i = Next[i])
        {
            int now = date[i];
            good[x] = i;
            if(c[now] == c[x] - 1 && haveflow[i])
            {
                upmin(addflow, haveflow[i]);
                last[now] = i, x = now, done = true;
                break;
            }
        }

        if(!done)
        {
            int go = all + 9;
            for(R i = Head[x]; i; i = Next[i])
                if(c[date[i]] && haveflow[i]) upmin(go, c[date[i]]);
            good[x] = Head[x];
            if(!(-- have[c[x]])) break;
            have[c[x] = go + 1] ++;
            if(x != s) x = date[last[x] ^ 1];
        }
    }
    printf("%d\n", ans);
}

void pre()
{
    n = read(), all = n * n + n, s = all + 1, t = s + 1;
    for(R i = 1; i <= n; i ++)
        for(R j = 1; j <= n; j ++) B[i][j] = read(), ans += B[i][j];
    for(R i = 1; i <= n; i ++) C[i] = read();
}

inline int id(int i, int j){return (i - 1) * n + j;}

void build()
{
    for(R i = 1; i <= n; i ++)
        for(R j = 1; j <= n; j ++)
        {
            int ID = id(i, j);
            add(s, ID, B[i][j]);
            add(ID, n * n + i, inf), add(ID, n * n + j, inf);
        }
    for(R i = 1; i <= n; i ++) add(n * n + i, t, C[i]);
}

int main()
{
//	freopen("in.in", "r", stdin);
    pre();
    build();
    bfs();
    ISAP();
//	fclose(stdin);
    return 0;
}