Uva 10888 Warehouse（KM）

題目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1829

思路：

1.首先求出每個箱子到達每個目的地的最短距離w。

2.將箱子看做x點，將目的地看做y點，x向y連邊，權值爲w，求二分圖最優匹配即可。

3.是否存在移動過程中路徑上有其他箱子阻擋。例如，從A到C的路徑上存在B箱子，則此時可以看做將B箱子移動到C後，A移動到B箱子。即最優匹配一定對應着一組可行解。

#include<map>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define F first
#define S second
#define pii pair<int,int>
#define debug
using namespace std;
const int N = 20;
const int maxn = 45;
const int INF = 0x3f3f3f3f;

int nx,ny;//兩邊的點數
int g[N][N];//二分圖描述
int linker[N],lx[N],ly[N];
int slack[N];
bool visx[N],visy[N];

bool DFS(int x)
{
    visx[x] = true;
    for(int y = 0; y < ny; y++)
    {
        if(visy[y])continue;
        int tmp = lx[x] + ly[y] - g[x][y];
        if(tmp == 0)
        {
            visy[y] = true;
            if(linker[y] == -1 || DFS(linker[y]))
            {
                linker[y] = x;
                return true;
            }
        }
        else if(slack[y] > tmp)
            slack[y] = tmp;
    }
    return false;
}
int KM()
{
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(int i = 0; i < nx; i++)
    {
        lx[i] = -INF;
        for(int j = 0; j < ny; j++)
            if(g[i][j] > lx[i])
                lx[i] = g[i][j];
    }
    for(int x = 0; x < nx; x++)
    {
        for(int i = 0; i < ny; i++)
            slack[i] = INF;
        while(true)
        {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(DFS(x))break;
            int d = INF;
            for(int i = 0; i < ny; i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 0; i < nx; i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 0; i < ny; i++)
            {
                if(visy[i])ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 0; i < ny; i++)
        if(linker[i] != -1)
            res += g[linker[i]][i];
    return res;
}

struct Node
{
    pii p;
    int step;
    Node() {}
    Node(pii p,int step):p(p),step(step) {}
};

int n,m;
int cntB,cntE;
int vis[maxn*maxn];
char ch[maxn][maxn];
int dx[]= {-1,1,0,0};
int dy[]= {0,0,-1,1};
queue<Node> q;
map<pii,int> box,ed;

void init()
{
    cntB=cntE=0;
    box.clear(),ed.clear();
}

inline int ID(int x,int y)
{
    return x*m+y;
}

inline int isIn(int x,int y)
{
    return x>=1&&x<=n&&y>=1&&y<=m;
}

void build(pii p,int id)
{
    memset(vis,0,sizeof(vis));
    while(!q.empty()) q.pop();

    q.push(Node(p,0));
    vis[ID(p.F,p.S)]=1;

    while(!q.empty())
    {
        Node now=q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            int x=now.p.F+dx[i];
            int y=now.p.S+dy[i];
            if(isIn(x,y)&&!vis[ID(x,y)]&&ch[x][y]!='#')
            {
                if(ch[x][y]=='X')
                {
                    g[id][ed[make_pair(x,y)]]=-(now.step+1);
                }
                vis[ID(x,y)]=1;
                q.push(Node(make_pair(x,y),now.step+1));
            }
        }
    }
}

void solve()
{
    for(int i=0;i<box.size();i++)
    {
        for(int j=0;j<box.size();j++)
        {
            g[i][j]=-INF;
        }
    }

    int cnt=0;
    map<pii,int>::iterator it=box.begin();
    for(; it!=box.end(); it++)
    {
        build(it->F,cnt++);
    }
    nx=ny=cnt;

    printf("%d\n",-KM());
}

int main()
{
#ifdef debu
    freopen("in.txt","r",stdin);
#endif // debug
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            getchar();
            for(int j=1; j<=m; j++)
            {
                scanf("%c",&ch[i][j]);
                if(ch[i][j]=='B') box[make_pair(i,j)]=cntB++;
                else if(ch[i][j]=='X') ed[make_pair(i,j)]=cntE++;
            }
        }
        solve();
    }
    return 0;
}