BZOJ3996: [TJOI2015]线性代数

3996: [TJOI2015]线性代数

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 1850 Solved: 1097
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Description

给出一个NN的矩阵B和一个1N的矩阵C。求出一个1N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

Input

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。

Output

输出最大的D

Sample Input

3

1 2 1

3 1 0

1 2 3

2 3 7

Sample Output

2

HINT

1<=N<=500

Source

题解

分析一下就会发现

[Ans = sum_{i=1}^{n}sum_{j=1}^{n}(A_i imes B{_i}{_j} imes A_j - C_i imes A_i) ]

即：
选(i)并且选(j)，才能获得(B{_i}{_j})的价值
选(i)会损失(C_i)的价值
选(j)会损失(C_j)的价值

如果做过(BZOJ2127)或者(BZOJ2132)，就会觉得这题真水。
对于i,j,这样建模：

一遍过

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a < b ? a : b;}
inline int abs(int x){return x < 0 ? -x : x;}
inline void swap(int &x, int &y){int tmp = x;x = y;y = tmp;}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;
struct Edge
{
	int u,v,w,nxt;
	Edge(int _u, int _v, int _w, int _nxt){u = _u;v = _v;w = _w;nxt = _nxt;}
	Edge(){}
}edge[3000010];
int head[1000010], cnt = 1, S, T, cur[1000010], q[1000010], he, ta, h[1000010], ans;
inline void insert(int a, int b, int c)
{
	edge[++ cnt] = Edge(a, b, c, head[a]), head[a] = cnt;
	edge[++ cnt] = Edge(b, a, 0, head[b]), head[b] = cnt;
}
bool bfs()
{
	memset(h, -1, sizeof(h)), h[S] = 0, he = ta = 0, q[ta ++] = S;
	while(he < ta)
	{
		register int now = q[he ++];
		for(register int pos = head[now];pos;pos = edge[pos].nxt)
		{
			register int v = edge[pos].v;
			if(edge[pos].w && h[v] == -1)
				h[v] = h[now] + 1, q[ta ++] = v;
		}
	}
	return h[T] != -1;
}
int dfs(int x, int f)
{
	if(x == T || f == 0) return f;
	int used = 0, w;
	for(register int& pos = cur[x];pos;pos = edge[pos].nxt)
	{
		register int v = edge[pos].v;
		if(h[v] == h[x] + 1 && edge[pos].w > 0)
		{
			w = dfs(v, min(edge[pos].w, f - used));
			if(w > 0)
				edge[pos].w -= w, edge[pos ^ 1].w += w, used += w;
			if(used == f) return f;
		}
	}
	if(!used) h[x] = -1;
	return used;
}
void dinic()
{
	while(bfs()) 
	{
		memcpy(cur, head, sizeof(head));
		ans += dfs(S, INF);
	}
}
int n, tot, sum, tmp;
int main()
{
	read(n);
	tot = n;
	S = ++ tot, T = ++ tot;
	for(int i = 1;i <= n;++ i)
		for(int j = 1;j <= n;++ j)
		{
			read(tmp);
			++ tot;
			insert(i, tot, INF);
			insert(j, tot, INF);
			insert(tot, T, tmp);
			sum += tmp;
		}
	for(int i = 1;i <= n;++ i) 
	{ 
		read(tmp);
		insert(S, i, tmp);
	} 
	dinic();
	printf("%d", sum - ans);
	return 0;
}