HDU 1045 Fire Net 二分图建图

HDU 1045

题意：

　　在一个n*n地图中，有许多可以挡住子弹的墙，问最多可以放几个炮台，使得炮台不会相互损害。炮台会向四面发射子弹。

思路：

　　把行列分开做，先处理行，把同一行中相互联通的点缩成一个点。再处理列，同样缩成一个点。然后把行列中，交点不是墙的点连一条边。对这个图跑网络流或者二分图匹配即可。

 

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
#include <unordered_map>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue
#define max3(a,b,c) max(max(a,b),c)



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/

            char mp[5][5];
            int idx[5][5];


            struct edge{
                int u,v,cap,flag;
                edge(){}
                edge(int u,int v,int cap,int flag):u(u),v(v),cap(cap),flag(flag){}
            }es[100009];

            int tot,s,t;
            vector<int>tab[2009];
            int dis[2009],cur[2009];
            void addedge(int u,int v,int cap){
             //   debug(u);
                tab[u].pb(tot);
                es[tot++] = edge(u,v,cap,1);
                tab[v].pb(tot);
                es[tot++] = edge(v,u,0,0);
            }

            bool bfs(){
                queue<int>q; q.push(s);
                memset(dis,inf,sizeof(dis));
                dis[s] = 0;
                while(!q.empty()){
                    int h = q.front(); q.pop();
                    for(int i=0; i<tab[h].size(); i++){
                        edge & e = es[tab[h][i]];
                        if(e.cap > 0 && dis[e.v] >= inf){
                            dis[e.v] = dis[h] + 1;
                            q.push(e.v);
                        }
                    }
                }
                return dis[t] < inf;
            }

            int dfs(int x,int maxflow){
                if(x == t || maxflow == 0) return maxflow;
                for(int i=cur[x] ; i<tab[x].size(); i++){
                    cur[x] = i;
                    edge & e = es[tab[x][i]];
                    if(dis[e.v] == dis[x] + 1 && e.cap > 0){
                        int flow = dfs(e.v, min(maxflow, e.cap));
                        if(flow){
                            e.cap -= flow;

                            es[tab[x][i] ^ 1].cap += flow;
                            return flow;
                        }
                    }
                }
                return 0;
            }

            int dinic(){
                int ans = 0;
                while(bfs()){

                    int flow;
                    memset(cur,0,sizeof(cur));
                    do{
                        flow = dfs(s,inf);
                        if(flow) ans += flow;
                    }while(flow);

                }
                return ans;
            }
int main(){
            int n;  scanf("%d", &n);
            while(~scanf("%d", &n) && n){
                memset(idx, -1, sizeof(idx));
                
                tot = 0;
                for(int i=0; i<n; i++) scanf("%s", mp[i]);
                s = 0, t = 1000;
                for(int i=s; i<=t; i++)tab[i].clear();
                int totx = 1;
                for(int i=0; i<n; i++){
                    for(int j=0; j<n; j++){
                        if(mp[i][j] == '.') idx[i][j] = totx;
                        else totx++;
                    }
                    totx++;
                }
                int p = totx;
                for(int i=1; i<totx; i++) addedge(s, i, 1);
                for(int i=0; i<n; i++){
                    for(int j=0; j<n; j++){
                        if(mp[j][i] == '.' ) {
                            // vis[idx[j][i]] = 1;
                            addedge(idx[j][i], totx, 1);
                        } 
                        else if(mp[j][i] == 'X') totx++;
                    }   
                    totx++;
                }
                for(int i=p; i<=totx; i++) addedge(i, t, 1);
                printf("%d
", dinic());
            }

            return 0;
}