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[COGS 740]分配问题

2016年9月03日 22:30 | Comments(0) | Category:题解 | Tags:

很明显是二分图最大权匹配。可以用匈牙利算法求解,但我比较弱……所以就写了个费用流。

注意最优解和最差解都要给出!

代码:

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=205;
namespace MCMF{
	struct Edge{
		int u,v,cap,flow,cost;
	};
	vector<Edge> edges;
	vector<int> G[maxn];
	int m;
	inline void AddEdge(int u,int v,int cap,int cost){
		edges.push_back((Edge){u,v,cap,0,cost});
		edges.push_back((Edge){v,u,0,0,-cost});
		m=edges.size();
		G[u].push_back(m-2);
		G[v].push_back(m-1);
	}
	inline void ClearGraph(){
		register int i;
		edges.clear();
		for(i=0;i<maxn;i++)
			G[i].clear();
	}
	int a[maxn];
	bool inQueue[maxn];
	int d[maxn],p[maxn];
	bool BellmanFord(int s,int t,long long& cost){
		register int i,u;
		queue<int> Q;
		memset(d,0x7f,sizeof(d));
		memset(inQueue,0,sizeof(inQueue));
		d[s]=0;
		inQueue[s]=true;
		p[s]=0;
		a[s]=0x7f7f7f7f;
		Q.push(s);
		while(!Q.empty()){
			u=Q.front();
			Q.pop();
			inQueue[u]=false;
			for(i=0;i<G[u].size();i++){
				Edge& e=edges[G[u][i]];
				if(e.cap>e.flow && d[e.v]>d[u]+e.cost){
					d[e.v]=d[u]+e.cost;
					p[e.v]=G[u][i];
					a[e.v]=min(a[u],e.cap-e.flow);
					if(!inQueue[e.v]){
						Q.push(e.v);
						inQueue[e.v]=true;
					}
				}
			}
		}
		if(d[t]==0x7f7f7f7f)
			return false;
		cost+=((long long)d[t])*((long long)a[t]);
		for(u=t;u!=s;u=edges[p[u]].u){
			edges[p[u]].flow+=a[t];
			edges[p[u]^1].flow-=a[t];
		}
		return true;
	}
	long long MincostMaxflow(int s,int t){
		register long long ans=0;
		while(BellmanFord(s,t,ans));
		return ans;
	}
};
int A[105][105];
int main(){
	int n;
	register int i,j;
	freopen("job.in","r",stdin);
	freopen("job.out","w+",stdout);
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++){
			scanf("%d",&A[i][j]);
			MCMF::AddEdge(i,j+n,1,A[i][j]);
		}
	for(i=1;i<=n;i++)
		MCMF::AddEdge(0,i,1,0);
	for(i=n+1;i<=2*n;i++)
		MCMF::AddEdge(i,2*n+1,1,0);
	printf("%lld\n",MCMF::MincostMaxflow(0,2*n+1));
	MCMF::ClearGraph();
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++){
			MCMF::AddEdge(i,j+n,1,-A[i][j]);
		}
	for(i=1;i<=n;i++)
		MCMF::AddEdge(0,i,1,0);
	for(i=n+1;i<=2*n;i++)
		MCMF::AddEdge(i,2*n+1,1,0);
	printf("%lld\n",-(MCMF::MincostMaxflow(0,2*n+1)));
	return 0;
}