题目

给出一个NN的矩阵B和一个1N的矩阵C。求出一个1*N的01矩阵A.使得

D=（AB-C）A^T最大。其中A^T为A的转置。输出D

输入格式

第一行输入一个整数N，接下来N行输入B矩阵，第i行第J个数字代表Bij.
接下来一行输入N个整数，代表矩阵C。矩阵B和矩阵C中每个数字都是不超过1000的非负整数。

输出格式

输出最大的D

输入样例

3

1 2 1

3 1 0

1 2 3

2 3 7

输出样例

2

提示

1<=N<=500

题解

我们将式子化简，就是：
\(\sum_{i = 1}^{n} \sum_{j = 1}^{n} Bij * Ai * Aj - \sum_{i}^{n} Ci * Ai\)
相当于，有n个物品，选择每个物品都有代价，任意两个物品同时选择时都有价值，求最大价值

用最小割解决
对于任意两个点i和j，建一个新点u，两点向u连边INF，u向T连边，容量为两个物品选择的权值
S向所有点连边，容量为该物品价值

我们假设一开始拥有所有价值
这样一来，要割去，每个点要么要花费其代价S->u，要么花费其与其它物品的共同代价u->T

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 300005,maxm = 5000005,INF = 1000000000;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
int h[maxn],ne = 2,n,m;
struct EDGE{int to,nxt,f;}ed[maxm];
inline void build(int u,int v,int w){ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;ed[ne] = (EDGE){u,h[v],0}; h[v] = ne++;
}
int d[maxn],vis[maxn],S,T,cur[maxn];
bool bfs(){for (int i = S; i <= T; i++) vis[i] = 0,d[i] = INF;queue<int> q;q.push(S); vis[S] = true; d[S] = 0;int u;while (!q.empty()){u = q.front(); q.pop();Redge(u) if (ed[k].f && !vis[to = ed[k].to]){d[to] = d[u] + 1; vis[to] = true;q.push(to);}}return vis[T];
}
int dfs(int u,int minf){if (u == T || !minf) return minf;int f,flow = 0,to;if (cur[u] == -1) cur[u] = h[u];for (int& k = cur[u]; k; k = ed[k].nxt)if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ed[k].f -= f; ed[k ^ 1].f += f;flow += f; minf -= f;if (!minf) break;}return flow;
}
int maxflow(){int flow = 0;while (bfs()){memset(cur,-1,sizeof(cur));flow += dfs(S,INF);}return flow;
}
int main(){n = read(); S = 0; T = n + n * n + 1;int x,ans = 0,id = n;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){ans += (x = read()); id++;build(i,id,INF);build(j,id,INF);build(id,T,x);}}for (int i = 1; i <= n; i++) build(S,i,read());printf("%d\n",ans - maxflow());return 0;
}