bzoj 3996: [TJOI2015]线性代数 [最小割]

3996: [TJOI2015]线性代数

题意：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

\(D=（A * B-C）* A^T\)最大。其中A^T为A的转置。输出D。每个数非负。

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分析一下这个乘法的性质或者化简一下容易发现，\(C_i\)代价生效需要\(A_i=1\)，\(B_{ij}\)贡献生效需要\(A_i =A_j=1\)

最小割

我成功的把dinic里的括号打错了...gg

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=300005, M=2e6+5, INF = 1e9;
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, b[505][505], c[505], sum;
struct edge{int v, ne, c, f;} e[M];
int cnt=1, h[N], s, t;
inline void ins(int u, int v, int c) { 
	e[++cnt] = (edge){v, h[u], c, 0}; h[u] = cnt;
	e[++cnt] = (edge){u, h[v], 0, 0}; h[v] = cnt;
}

namespace mf {
	int q[N], head, tail, vis[N], d[N];
	bool bfs() {
		memset(vis, 0, sizeof(vis));
		head = tail = 1;
		q[tail++] = s; d[s] = 0; vis[s] = 1;
		while(head != tail) {
			int u = q[head++]; 
			for(int i=h[u];i;i=e[i].ne) 
				if(!vis[e[i].v] && e[i].c > e[i].f) {
					int v = e[i].v;
					vis[v] = 1; d[v] = d[u]+1;
					q[tail++] = v;
					if(v == t) return true;
				}
		}
		return false;
	}
	int cur[N];
	int dfs(int u, int a) {
		if(u == t || a == 0) return a;
		int flow = 0, f;
		for(int &i=cur[u];i;i=e[i].ne) 
			if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c - e[i].f)) ) >0 ) {
				flow += f;
				e[i].f += f;
				e[i^1].f -= f;
				a -= f;
				if(a == 0) break;
			}
		if(a) d[u] = -1;
		return flow;
	}
	int dinic() {
		int flow = 0;
		while(bfs()) { 
			for(int i=s; i<=t; i++) cur[i] = h[i];
			flow += dfs(s, INF); 
		}
		return flow;
	}
}

void build() {
	s = 0; t = n + n*n + 1;
	for(int i=1; i<=n; i++) ins(s, i, c[i]), ins(i, t, b[i][i]);
	for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) if(i != j) {
		int id = i*n+j;
		ins(i, id, INF); ins(j, id, INF); ins(id, t, b[i][j]);
	}
}

int main() {
	freopen("in", "r", stdin);
	n = read();
	for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) b[i][j] = read(), sum += b[i][j];
	for(int i=1; i<=n; i++) c[i] = read();
	build();
	int ans = mf::dinic();
	printf("%d\n", sum - ans);
}