• hdu3376 Matrix Again


    最大费用最大流
    咋写?取个相反数就可以了……

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <queue>
    using namespace std;
    int n, a[605][605], ss, tt, hea[750005], cnt, minCost, dis[750005], pre[750005];
    queue<int> d;
    bool vis[750005];
    struct Edge{
    	int too, nxt, val, cst;
    }edge[2200005];
    const int oo=0x3f3f3f3f;
    const int dx[]={0, 1, 0};
    const int dy[]={0, 0, 1};
    int f(int i, int j){
    	return (i-1)*n+j;
    }
    void init(){
    	minCost = cnt = 0;
    	memset(hea, -1, sizeof(hea));
    }
    void add_edge(int fro, int too, int val, int cst){
    	edge[cnt].nxt = hea[fro];
    	edge[cnt].too = too;
    	edge[cnt].val = val;
    	edge[cnt].cst = cst;
    	hea[fro] = cnt++;
    }
    void addEdge(int fro, int too, int val, int cst){
    	add_edge(fro, too, val, cst);
    	add_edge(too, fro, 0, -cst);
    }
    bool spfa(){
    	memset(dis, 0x3f, sizeof(dis));
    	memset(pre, -1, sizeof(pre));
    	d.push(ss);
    	vis[ss] = true;
    	dis[ss] = 0;
    	while(!d.empty()){
    		int x=d.front();
    		d.pop();
    		vis[x] = false;
    		for(int i=hea[x]; i!=-1; i=edge[i].nxt){
    			int t=edge[i].too;
    			if(dis[t]>dis[x]+edge[i].cst && edge[i].val>0){
    				dis[t] = dis[x] + edge[i].cst;
    				pre[t] = i;
    				if(!vis[t]){
    					vis[t] = true;
    					d.push(t);
    				}
    			}
    		}
    	}
    	return dis[tt]!=oo;
    }
    void mcmf(){
    	while(spfa()){
    		int tmp=0x3f3f3f3f;
    		for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too])
    			tmp = min(tmp, edge[i].val);
    		for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too]){
    			edge[i].val -= tmp;
    			edge[i^1].val += tmp;
    			minCost += tmp * edge[i].cst;
    		}
    	}
    }
    int main(){
    	while(scanf("%d", &n)!=EOF){
    		init();
    		for(int i=1; i<=n; i++)
    			for(int j=1; j<=n; j++)
    				scanf("%d", &a[i][j]);
    		ss = 0; tt = 2 * n * n + 1;
    		addEdge(ss, 1, 2, 0);
    		addEdge(1, 1+n*n, 1, 0);
    		addEdge(f(n,n), f(n,n)+n*n, 1, 0);
    		addEdge(f(n,n)+n*n, tt, 2, 0);
    		for(int i=1; i<=n; i++)
    			for(int j=1; j<=n; j++){
    				addEdge(f(i,j), f(i,j)+n*n, 1, -a[i][j]);
    				for(int k=1; k<=2; k++){
    					int kx=i+dx[k];
    					int ky=j+dy[k];
    					if(kx>n || ky>n)	continue;
    					addEdge(f(i,j)+n*n, f(kx,ky), oo, 0);
    				}
    			}
    		mcmf();
    		printf("%d
    ", -minCost);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8337881.html
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