LA3276

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 = 110, 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 = 1, 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, 0, -c);
}
bool spfa()
{
    int l = 1, r = 0;
    memset(dis, 0x3f3f, sizeof(dis));
    dis[0] = 0;
    q[++r] = 0;
    while(l <= r)
    {
        int u = q[l++];
        inq[u] = 0;
        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] = 1;
                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] ^ 1].f += delta;
        now = prevv[now];
    }
    return delta * dis[T];
}
int maxcostflow()
{
    int ret = 0;
    while(spfa()) ret += getflow();
    return ret;
}
void build()
{
    memset(head, 0, sizeof(head));
    cnt = 1;
    for(int i = 1; i <= n; ++i) insert(S, i, 1, 0);
    for(int i = 1; i <= n; ++i)
        for(int x = 1; x <= n; ++x)
            for(int y = 1; y <= n; ++y) if(Map[x][y])
                insert(i, Map[x][y], 1, abs(a[i].x - x) + abs(a[i].y - y));
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) if(Map[i][j])
            insert(Map[i][j], T, 1, 0);
}
int main()
{
    while(scanf("%d", &n))
    {
        if(!n) break;
        T = 2 * n + 1;
        ans = inf;
        for(int i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
        for(int i = 1; i <= n; ++i)
        {
            memset(Map, 0, sizeof(Map));
            tot = n;
            for(int j = 1; j <= n; ++j) Map[i][j] = ++tot;
            build();
            ans = min(ans, maxcostflow());
            memset(Map, 0, sizeof(Map));
            tot = n;
            for(int j = 1; j <= n; ++j) Map[j][i] = ++tot;
            build();
            ans = min(ans, maxcostflow());        
        }
        memset(Map, 0, sizeof(Map));
        tot = n;
        for(int i = 1; i <= n; ++i) Map[i][i] = ++tot;
        build();
        ans = min(ans, maxcostflow());
        memset(Map, 0, sizeof(Map));
        tot = n;
        for(int i = 1; i <= n; ++i) Map[i][n - i + 1] = ++tot;
        build();
        ans = min(ans, maxcostflow());
        printf("Board %d: %d moves required.

", ++kase, ans); 
    }
    return 0;
}