lightoj 1243 - Guardian Knights 最小费用流

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

const int maxe = 100000;
const int maxn = 205;
const int INF  = 0x3f3f3f3f;

struct Edge{
    int u,v,flow,cap,cost;
    int next;
    Edge(int u=0,int v=0,int flow=0,int cap=0,int cost=0,int next=0):
        u(u), v(v), flow(flow), cap(cap), cost(cost), next(next){}
};

struct MCMF{
    Edge edges[maxe];
    int head[maxn],cnt;
    int d[maxn];
    bool inq[maxn];
    int pa[maxn];
    int res[maxn];

    void init(){
        memset(head,-1,sizeof(head));
        cnt = 0;
    }

    void addedge(int u,int v,int cap,int cost){
        edges[cnt] = Edge(u,v,0,cap,cost,head[u]);
        head[u] = cnt++;
        edges[cnt] = Edge(v,u,0,0,-cost,head[v]);
        head[v] = cnt++;
    }

    bool SPFA(int s,int t,int& flow,int& cost){
        memset(d,0x3f,sizeof(d));
        memset(inq,0,sizeof(inq));

        queue<int> Q;
        Q.push(s);
        inq[s] = true;  d[s] = 0;
        res[s] = INF;  res[t] = 0;

        while(!Q.empty()){
            int u = Q.front();   Q.pop();
            inq[u] = false;

            for(int i=head[u];i!=-1;i=edges[i].next){
                Edge&  e = edges[i];
                if(e.cap>e.flow && d[e.v] > d[u] + e.cost){
                    d[e.v] = d[u] + e.cost;
                    pa[e.v] = i;
                    res[e.v] = min(res[u],e.cap-e.flow);
                    if(!inq[e.v]){
                        Q.push(e.v);
                        inq[e.v] = true;
                    }
                }
            }
        }
        if(res[t] == 0)  return false;
        flow += res[t];
        cost += res[t]*d[t];
        for(int i=t;i!=s;i=edges[pa[i]].u){
            edges[pa[i]].flow   += res[t];
            edges[pa[i]^1].flow -= res[t];
        }
        return true;
    }

    int MinCost(int s,int t){
        int flow = 0,cost = 0;
        while(SPFA(s,t,flow,cost)){}

        return cost;
    }
}solver;

struct Node{
    int x,y;
    int dist;
    Node(int x=0,int y=0,int dist = 0): x(x), y(y) ,dist(dist) {}
};
Node Knight[35];
int n,k,m;
int dist[35][105];

int Mymap[35][35];  // == 0代表rock，-1代表space ,1-m代表mill.
int next_x[4] = {-1,0,1,0};
int next_y[4] = {0,-1,0,1};

void bfs(){
    bool vis[35][35];
    memset(dist,0x3f,sizeof(dist));
    queue<Node> Q;
    for(int i=1;i<=k;i++){
        while(!Q.empty())  Q.pop();
        Q.push(Node(Knight[i].x,Knight[i].y,0));

        memset(vis,0,sizeof(vis));
        vis[Knight[i].x][Knight[i].y] = true;

        while(!Q.empty()){
            Node out = Q.front(); Q.pop();
            for(int j=0;j<4;j++){
                int x = out.x + next_x[j];
                int y = out.y + next_y[j];
                if(vis[x][y]) continue;
                vis[x][y] = true;
                if(Mymap[x][y] == -1){
                    Q.push(Node(x,y,out.dist+1));
                }
                else if(Mymap[x][y]>0){
                    dist[i][Mymap[x][y]] = out.dist+1;
                    Q.push(Node(x,y,out.dist+1));
                }
            }
        }
    }
}

int main()
{
    //freopen("E:\\acm\\input.txt","r",stdin);
    int T;
    cin>>T;
    for(int cas=1;cas<=T;cas++){
        cin>>n>>k>>m;
        char ch[35];
        memset(Mymap,0,sizeof(Mymap));
        int cntm = 1;
        for(int i=1;i<=n;i++){
            scanf("%s",ch+1);
            for(int j=1;j<=n;j++){
               if(ch[j]>='A' && ch[j]<='Z'){
                  int num = ch[j] - 'A' + 1;
                  Knight[num] = Node(i,j,0);
                  Mymap[i][j] = -1;
               }
               else if(ch[j] == 'm'){
                  Mymap[i][j] = cntm++;
               }
               else if(ch[j] == '.'){
                  Mymap[i][j] = -1;
               }
            }
        }
        bfs();
        solver.init();
        int s = 0,  t = k+m+1;
        for(int i=1;i<=k;i++){
            int a;
            scanf("%d",&a);
            solver.addedge(s,i,a,0);
        }
        for(int i=1;i<=k;i++)
            for(int j=1;j<=m;j++){
                solver.addedge(i,k+j,1,dist[i][j]);
        }

        for(int i=1;i<=m;i++){
            solver.addedge(k+i,t,1,0);
        }

        printf("Case %d: %d\n",cas,solver.MinCost(s,t));
    }
}