LA3276
2024-09-08 09:41:40
费用流
这种棋盘模型大概都是网络流吧
首先我们知道棋子之间不会影响到达目标的步数,那么就好做了,枚举终点,然后就是最小权匹配了,因为就是寻找总和最小,然后费用流就行了。
#include<bits/stdc++.h>
using namespace std;
const int N = , inf = 0x3f3f3f3f;
struct data {
int x, y;
} a[N];
struct edge {
int nxt, to, f, c;
} e[N * N];
int n, S, T, tot, cnt = , ans, kase;
int head[N], dis[N], q[N], Map[N][N], pree[N], prevv[N], inq[N];
void link(int u, int v, int f, int c)
{
e[++cnt].nxt = head[u];
head[u] = cnt;
e[cnt].to = v;
e[cnt].f = f;
e[cnt].c = c;
}
void insert(int u, int v, int f, int c)
{
link(u, v, f, c);
link(v, u, , -c);
}
bool spfa()
{
int l = , r = ;
memset(dis, 0x3f3f, sizeof(dis));
dis[] = ;
q[++r] = ;
while(l <= r)
{
int u = q[l++];
inq[u] = ;
for(int i = head[u]; i; i = e[i].nxt) if(e[i].f && dis[e[i].to] > dis[u] + e[i].c)
{
pree[e[i].to] = i;
prevv[e[i].to] = u;
dis[e[i].to] = dis[u] + e[i].c;
if(!inq[e[i].to])
{
inq[e[i].to] = ;
q[++r] = e[i].to;
}
}
}
return dis[T] != 0x3f3f3f3f;
}
int getflow()
{
int now = T, delta = inf;
while(now)
{
delta = min(delta, e[pree[now]].f);
now = prevv[now];
}
now = T;
while(now)
{
e[pree[now]].f -= delta;
e[pree[now] ^ ].f += delta;
now = prevv[now];
}
return delta * dis[T];
}
int maxcostflow()
{
int ret = ;
while(spfa()) ret += getflow();
return ret;
}
void build()
{
memset(head, , sizeof(head));
cnt = ;
for(int i = ; i <= n; ++i) insert(S, i, , );
for(int i = ; i <= n; ++i)
for(int x = ; x <= n; ++x)
for(int y = ; y <= n; ++y) if(Map[x][y])
insert(i, Map[x][y], , abs(a[i].x - x) + abs(a[i].y - y));
for(int i = ; i <= n; ++i)
for(int j = ; j <= n; ++j) if(Map[i][j])
insert(Map[i][j], T, , );
}
int main()
{
while(scanf("%d", &n))
{
if(!n) break;
T = * n + ;
ans = inf;
for(int i = ; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
for(int i = ; i <= n; ++i)
{
memset(Map, , sizeof(Map));
tot = n;
for(int j = ; j <= n; ++j) Map[i][j] = ++tot;
build();
ans = min(ans, maxcostflow());
memset(Map, , sizeof(Map));
tot = n;
for(int j = ; j <= n; ++j) Map[j][i] = ++tot;
build();
ans = min(ans, maxcostflow());
}
memset(Map, , sizeof(Map));
tot = n;
for(int i = ; i <= n; ++i) Map[i][i] = ++tot;
build();
ans = min(ans, maxcostflow());
memset(Map, , sizeof(Map));
tot = n;
for(int i = ; i <= n; ++i) Map[i][n - i + ] = ++tot;
build();
ans = min(ans, maxcostflow());
printf("Board %d: %d moves required.\n\n", ++kase, ans);
}
return ;
}
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