BZOJ 3996: [TJOI2015]线性代数(最大权闭合子图)

3996: [TJOI2015]线性代数

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 724  Solved: 485

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Description

给出一个N*N的矩阵B和一个1*N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

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

解题思路：

观察矩阵发现，若要取得B[i][j]的权值，则必须取A[i] 和A[j], 而取A[i]的花费为C[i]，这样题目就转化为了最大权闭合子图，详细见胡伯涛论文《最小割模型在信息学竞赛中的应用》。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
using namespace std;
const int MAXN = 300000;
const int MAXM = 2000000;
const int INF = 0x3f3f3f3f;
int read()
{
int rs = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f *= -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {rs = rs * 10 + ch - '0'; ch = getchar();}
return rs;
}
struct Edge
{
int to, next, cap, flow;
}edge[MAXM];
int tot, head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w, int rw = 0)
{
edge[tot].to = v; edge[tot].cap = w; edge[tot].next = head[u];
edge[tot].flow = 0; head[u] = tot++;
edge[tot].to = u; edge[tot].cap = rw; edge[tot].next = head[v];
edge[tot].flow = 0; head[v] = tot++;
}
long long sap(int start, int end, int N)
{
memset(gap, 0, sizeof(gap));
memset(dep, 0, sizeof(dep));
memcpy(cur, head, sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
long long ans = 0;
while(dep[start] < N)
{
if(u == end)
{
long long Min = INF;
for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for(int i=cur[u];i!=-1;i=edge[i].next)
{
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag)
{
u = v;
continue;
}
int Min = N;
for(int i=head[u];i!=-1;i=edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if(u!=start) u = edge[pre[u]^1].to;
}
return ans;
}
int n;
int B[510][510], C[510];
int main()
{
n = read();
long long tot = 0;
int x, id = n ;init();
int T = n * n + n + 1,S = 0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
x = read();
tot += x;++id;
addedge(i, id, INF);
addedge(j, id, INF);
addedge(id, T, x);
}
}
for(int i=1;i<=n;i++)
{
x = read();
addedge(S, i, x);
}
long long ans = sap(S, T, n * n + n + 2);
// cout << tot << ' ' << ans << endl;
printf("%lld\n", tot - ans);
return 0;
}