HDU 2676 Matrix

拆点跑费用流

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define maxn 2010
#define maxm 10100
#define INF 1<<30
using namespace std;
struct MCMF{
    int src,sink,e,n;
    int first[maxn];
    int cap[maxm],cost[maxm],v[maxm],next[maxm];
    bool flag;
    void init(){
        e = 0;
        memset(first,-1,sizeof(first));
    }

    void add_edge(int a,int b,int cc,int ww){
        //printf("add:%d to %d,cap = %d,cost = %d
",a,b,cc,ww);
        cap[e] = cc;cost[e] = ww;v[e] = b;
        next[e] = first[a];first[a] = e++;
        cap[e] = 0;cost[e] = -ww;v[e] = a;
        next[e] = first[b];first[b] = e++;
    }

    int d[maxn],pre[maxn],pos[maxn];
    bool vis[maxn];

    bool spfa(int s,int t){
        memset(pre,-1,sizeof(pre));
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        for(int i = 0;i <= n;i++)   d[i] = INF;
        Q.push(s);pre[s] = s;d[s] = 0;vis[s] = 1;
        while(!Q.empty()){
            int u = Q.front();Q.pop();
            vis[u] = 0;
            for(int i = first[u];i != -1;i = next[i]){
                if(cap[i] > 0 && d[u] + cost[i] < d[v[i]]){
                    d[v[i]] = d[u] + cost[i];
                    pre[v[i]] = u;pos[v[i]] = i;
                    if(!vis[v[i]])  vis[v[i]] = 1,Q.push(v[i]);
                }
            }
        }
        return pre[t] != -1;
    }

    int Mincost;
    int Maxflow;

    int MinCostFlow(int s,int t,int nn){
        Mincost = 0,Maxflow = 0,n = nn;
        while(spfa(s,t)){
            int min_f = INF;
            for(int i = t;i != s;i = pre[i])
                if(cap[pos[i]] < min_f) min_f = cap[pos[i]];
            Mincost += d[t] * min_f;
            Maxflow += min_f;
            for(int i = t;i != s;i = pre[i]){
                cap[pos[i]] -= min_f;
                cap[pos[i]^1] += min_f;
            }
        }
        return Mincost;
    }
};
MCMF g;

int N;
inline int id(int i,int j){
    return (i-1)*N + j;
}

int main(){
    while(scanf("%d",&N) != EOF){
        g.init();
        int k = N*N,ans = 0,a;
        int S = 0,T = 2*k+1;
        for(int i = 1;i <= N;i++){
            for(int j = 1;j <= N;j++){
                scanf("%d",&a);
                if(i+j == 2 || i+j == 2*N){
                    g.add_edge(id(i,j),id(i,j)+k,2,-a);
                    ans += a;
                }else{
                    g.add_edge(id(i,j),id(i,j)+k,1,-a);
                }
                if(i < N)   g.add_edge(id(i,j)+k,id(i+1,j),1,0);
                if(j < N)   g.add_edge(id(i,j)+k,id(i,j+1),1,0);
            }
        }
        g.add_edge(S,id(1,1),2,0);
        g.add_edge(id(N,N)+k,T,2,0);
        ans = -g.MinCostFlow(S,T,T) - ans;
        printf("%d
",ans);
    }
    return 0;
}