NYOJ 61 传纸条（费用流 或 DP)

题目连接：http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=61
这道题目两种解法，费用流 或 DP。最近在搞图论，DP还没想到怎么做，不过费用流比DP效率高多了。。。

解析：首先问题是要从矩阵的G[1,1]到G[m,n]找出不相交的两条路径，使得他们的和最大。可以想到用最大费用最大流，注意是最大费用哦。从每个点[ i,j] 的左边邻点[ i,j-1 ] 和上边邻点[ i-1,j ]到点[ i,j ]建边，流量为1，费用为[ i,j ]的权值。但是这样是不是可能经过一个点多次？ 答案是可能，所以要进行拆点，每个点拆成两个点，这两个点边的流量为1，费用为0,。点[1,1]和[m,n]拆开后流量为2（因为要找两条边）,费用为0，接下来就是简单的费用流了。。。

好了，上代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;

#define CLR(arr,v) memset(arr,v,sizeof(arr))
const int INF = 1<<29 ;

template<int MaxV,int MaxE>
class MinCostMaxFlow{
public:
	void Clear(){
		pos = 0;
		CLR(h,-1); CLR(Flow,0);
	}
	void add(int u,int v,int f,int c){
		num[pos] = v;
		cap[pos] = f;
		cost[pos] = c;
		next[pos] = h[u];
		h[u] = pos++;

		num[pos] = u;
		cap[pos] = 0;
		cost[pos] = -c;
		next[pos] = h[v];
		h[v] = pos++;
	}
	int GetMinCostMaxFlow(int start,int end){
		maxflow = 0; mincost = 0;
		int cur;
		while(SPFA(start,end)){
			cur = end;
			minflow = INF;
			while(cur != start){
				if(minflow > cap[ pre_e[cur] ] - Flow[ pre_e[cur] ])
					minflow = cap[ pre_e[cur] ] - Flow[ pre_e[cur] ];
				cur = pre_v[cur];
			}
			cur = end;
			maxflow += minflow;
			while(cur != start){
				Flow[ pre_e[cur] ] += minflow;
				Flow[ pre_e[cur]^1] -= minflow;
				mincost += minflow * cost[ pre_e[cur] ];
				cur = pre_v[cur];
			}
		}
		return mincost;
	}
private:
	int h[MaxV],dis[MaxV],num[MaxE],cap[MaxE],cost[MaxE],Flow[MaxE],next[MaxE],pre_e[MaxV],pre_v[MaxV],Q[MaxV],InQ[MaxV],pos,minflow,mincost,maxflow;
	bool SPFA(int s,int t){
		CLR(dis,0);
		CLR(InQ,0);
		dis[s] = 1;
		int head = 0,total = 0,cur;
		Q[total++] = s;
		while(head != total){
			cur = Q[head++];
			if(head >= MaxV) head = 0;
			for(int i = h[cur];i != -1;i = next[i]){
				if(cap[i] > Flow[i] && dis[cur] + cost[i] > dis[ num[i] ]){
					dis[ num[i] ] = dis[cur] + cost[i];
					pre_v[ num[i] ] = cur;
					pre_e[ num[i] ] = i;
					if(!InQ[ num[i] ])
					{
						if(total >= MaxV) total = 0;
						InQ[ num[i] ] = true;
						Q[total++] = num[i];
					}
				}
			}
			InQ[cur] = false;
		}
		return dis[t] > 1;
	}
};

MinCostMaxFlow<5500,20050> g;

int main()
{
	//freopen("1.txt","r",stdin);
	//freopen("2.txt","w",stdout);
	int T;
	scanf("%d",&T);
	while(T--)
	{
		g.Clear();
		int m,n,u,U,v;
		scanf("%d%d",&m,&n);
		for(int i = 1;i <= m;++i)
		{
			for(int j = 1;j <= n;++j)
			{
				scanf("%d",&v);
				u = (i-1)*n + j;
				U = u + m*n;
				if(i == 1 && j == 1) 
					g.add(u,U,2,0);
				else if(i == m && j == n)
					g.add(u,U,2,0);
				else 
					g.add(u,U,1,0);
				if(i-1 != 0)
					g.add(U-n,u,1,v);
				if(j-1 != 0)
					g.add(U-1,u,1,v);
			}
		}
		printf("%d\n",g.GetMinCostMaxFlow(1,m*n*2));
	}
	return 0;
}