被某神犇吐槽:
你自己寫的題解太少了!
跪。。蒟蒻怎麼能想出來辣麼神的題目OTZ
但還是要認認(len)真真寫題解了=-=。。
話說這道題的題面好不着調=-=
神犇告訴我:
1 2 3
4 5 6
7 8 9
的轉置是
1 4 7
2 5 8
3 6 9
= =然並卵
呢。蘭後就這樣啦
求一個D=【A[1*N的矩陣] * B[N*N的矩陣] - C[1*N的矩陣]】 * [A的轉置N*1的矩陣]最大化
注意到D是一個數值
然後我們發現
D = x1(x1 * a1,1 + x2 * a1,2 + x3 * a1,3... - c1) + x2(x1 * a2,1 + x2 * a2,2 + x3 * a2,3 .. -c2) + x3...
是不是很神奇!!(有什麼神奇的)
然後變一變
注意到x1,x2,x3...的取值爲0或1
D = x1 * x1 * a1,1 +
x1 * x2 * a1,2 +
x1 * x3 * a1,3... + x2 * x1 * a2,1 +
x2 * x2 * a2,2 +
x2 *x3 * a2,3 .. + x3...
- x1 * c1 - x2 * c2
然後就轉化成了!!
選了某個數xi要花費ci的代價,然後選了xi和xj會有ai, j的貢獻!!
最小割即可
#include
#define maxn 510
using namespace std;
const int inf = 0x7fffffff;
queueQ;
int S, T, a[maxn][maxn], n;
struct Edge{
int to, next, w;
}edge[5000010];
int h[100010], cnt = 1;
void add(int u, int v, int w){
cnt ++;
edge[cnt].to = v;
edge[cnt].next = h[u];
edge[cnt].w = w;
h[u] = cnt;
swap(u, v), w = 0;
cnt ++;
edge[cnt].to = v;
edge[cnt].next = h[u];
edge[cnt].w = w;
h[u] = cnt;
}
int d[maxn];
bool BFS(){
memset(d, -1, sizeof d);
Q.push(S);d[S] = 0;
while(!Q.empty()){
int u = Q.front();Q.pop();
for(int i = h[u]; i; i = edge[i].next){
if(edge[i].w == 0)continue;
int v = edge[i].to;
if(d[v] == -1)d[v] = d[u] + 1, Q.push(v);
}
}
return d[T] != -1;
}
int DFS(int x, int a){
if(x == T || a == 0)return a;
int used = 0, f;
for(int i = h[x]; i; i = edge[i].next){
int v = edge[i].to;
if(d[v] == d[x] + 1){
f = DFS(v, min(a-used, edge[i].w));
edge[i].w -= f;
edge[i^1].w += f;
used += f;
if(used == a)return used;
}
}
if(!used)d[x] = -1;
return used;
}
int Dinic(){
int ret = 0;
while(BFS())
ret += DFS(S, inf);
return ret;
}
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= n; j ++)
scanf("%d", &a[i][j]);
S = 0, T = n+1;
int total = 0;
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j ++){
total += a[i][j];
add(S, i, 2 * a[i][j]);
add(i, j, a[i][j]);
add(j, i, a[i][j]);
}
}
int x = 0;
for(int i = 1; i <= n; i ++){
scanf("%d", &x);
add(i, T, 2 * x);
}
printf("%d\n", total - Dinic() / 2);
return 0;
}