【TJOI2015】线性代数

题面

题解

要求的是

\[\sum_{i=1}^n\sum_{j=1}^na_ia_jb_{i,j} - \sum_{i=1}^na_ic_i
\]

可以看出这是一个最大权闭合子图问题

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int N(510), maxn(3000010), INF(0x3f3f3f3f);
struct edge { int next, to, cap; } e[maxn << 1];
int head[maxn], e_num = -1, n, q[maxn], tail, lev[maxn], cur[maxn];
int S, T, id_b[N][N], id_c[N], idcnt, ans;

inline void add_edge(int from, int to, int cap)
{
	e[++e_num] = (edge) {head[from], to, cap}; head[from] = e_num;
	e[++e_num] = (edge) {head[to], from, cap}; head[to]   = e_num;
}

int bfs()
{
	clear(lev, 0); q[tail = lev[S] = 1] = S;
	for(RG int i = 1; i <= tail; i++)
	{
		int x = q[i];
		for(RG int j = head[x]; ~j; j = e[j].next)
		{
			int to = e[j].to; if(lev[to] || (!e[j].cap)) continue;
			q[++tail] = to, lev[to] = lev[x] + 1;
		}
	}
	return lev[T];
}

int dfs(int x, int f)
{
	if(x == T || (!f)) return f;
	int ans = 0, cap;
	for(RG int &i = cur[x]; ~i; i = e[i].next)
	{
		int to = e[i].to;
		if(e[i].cap && lev[to] == lev[x] + 1)
		{
			cap = dfs(to, std::min(f - ans, e[i].cap));
			e[i].cap -= cap, e[i ^ 1].cap += cap, ans += cap;
			if(ans == f) break;
		}
	}
	return ans;
}

inline int Dinic()
{
	int ans = 0;
	while(bfs())
	{
		for(RG int i = S; i <= T; i++) cur[i] = head[i];
		ans += dfs(S, INF);
	}
	return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	clear(head, -1); n = read(); S = ++idcnt;
	for(RG int i = 1; i <= n; i++)
		for(RG int j = 1; j <= n; j++)
			id_b[i][j] = ++idcnt;
	for(RG int i = 1; i <= n; i++) id_c[i] = ++idcnt;
	T = ++idcnt;
	for(RG int i = 1, x; i <= n; i++)
		for(RG int j = 1; j <= n; j++)
			ans += (x = read()), add_edge(S, id_b[i][j], x),
			add_edge(id_b[i][j], id_c[i], INF),
			add_edge(id_b[i][j], id_c[j], INF);
	for(RG int i = 1, x; i <= n; i++)
		x = read(), add_edge(id_c[i], T, x);
	printf("%d\n", ans - Dinic());
	return 0;
}