HDOJ 4862 Jump

K路径覆盖问题，最小费用最大流。。。。

最小K路径覆盖的模型，用费用流或者KM算法解决，构造二部图，X部有N*M个节点，源点向X部每个节点连一条边，流量1，费用0，Y部有N*M个节点，每个节点向汇点连一条边，流量1，费用0，如果X部的节点x可以在一步之内到达Y部的节点y，那么就连边x->y，费用为从x格子到y格子的花费能量减去得到的能量，流量1，再在X部增加一个新的节点，表示可以从任意节点出发K次，源点向其连边，费用0，流量K，这个点向Y部每个点连边，费用0，流量1，最这个图跑最小费用最大流，如果满流就是存在解，反之不存在，最小费用的相反数就是可以获得的最大能量

Jump

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 546    Accepted Submission(s): 238

Problem Description

There are n*m grids, each grid contains a number, ranging from 0-9. Your initial energy is zero. You can play up to K times the game, every time you can choose any one of the grid as a starting point (but not traveled before) then you can choose a grid on the right or below the current grid to jump, but it has not traveled before. Every time you can jump as many times as you want, as long as you do not violate rules. If you are from (x1, y1) to (x2, y2), then you consume |x1-x2|+|y1-y2|-1 energies. Energy can be negative. 
However, in a jump, if you start position and end position has same numbers S, then you can increase the energy value by S. 
Give me the maximum energy you can get. Notice that you have to go each grid exactly once and you don’t have to play exactly K times.

 

Input

The first line is an integer T, stands for the number of the text cases.
Then T cases followed and each case begin with three numbers N, M and K. Means there are N rows and M columns, you have K times to play.
Then N lines follow, each line is a string which is made up by M numbers.
The grids only contain numbers from 0 to 9.
(T<=100, N<=10，M<=10，K<=100)

 

Output

Each case, The first you should output “Case x : ”,(x starting at 1),then output The maximum number of energy value you can get. If you can’t reach every grid in no more than K times, just output -1.

 

Sample Input

   
   
   
   

    
    
    
    5
1 5 1
91929
1 5 2
91929
1 5 3
91929
3 3 3
333
333
333
3 3 2
333
333
333
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1 : 0
Case 2 : 15
Case 3 : 16
Case 4 : 18
Case 5 : -1
   
   
   
   

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn=110000;
const int INF=0x3f3f3f3f;

int n,m,k;
int a[20][20];

struct Edge
{
	int to,next,cap,flow,cost;
}edge[maxn];

int Adj[maxn],Size,N;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void addedge(int u,int v,int cap,int cost)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].cost=cost;
	edge[Size].cap=cap;
	edge[Size].flow=0;
	Adj[u]=Size++;
}

void Add_Edge(int u,int v,int cap,int cost)
{
	addedge(u,v,cap,cost);
	addedge(v,u,0,-cost);
}

int dist[1000],vis[1000],pre[1000];

bool spfa(int s,int t)
{
	queue<int> q;
	for(int i=0;i<N;i++)
	{
		dist[i]=INF;vis[i]=false; pre[i]=-1;
	}
	dist[s]=0; vis[s]=true; q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=false;
		for(int i=Adj[u];~i;i=edge[i].next)
		{
			int v=edge[i].to;
			if(edge[i].cap>edge[i].flow&&
				dist[v]>dist[u]+edge[i].cost)
			{
				dist[v]=dist[u]+edge[i].cost;
				pre[v]=i;
				if(!vis[v])
				{
					vis[v]=true;
					q.push(v);
				}
			}
		}
	}
	if(pre[t]==-1) return false;
	return true;
}

int MinCostMaxFlow(int s,int t,int& cost)
{
	int flow=0;
	cost=0;
	while(spfa(s,t))
	{
		int Min=INF;
		for(int i=pre[t];~i;i=pre[edge[i^1].to])
		{
			if(Min>edge[i].cap-edge[i].flow)
				Min=edge[i].cap-edge[i].flow;
		}
		for(int i=pre[t];~i;i=pre[edge[i^1].to])
		{
			edge[i].flow+=Min;
			edge[i^1].flow-=Min;
			cost+=edge[i].cost*Min;
		}
		flow+=Min;
	}
	return flow;
}
char in[10010];
int main()
{
	int T_T,cas=1;
	scanf("%d",&T_T);
while(T_T--)
{
	init();
	scanf("%d%d%d",&n,&m,&k);
	memset(a,0,sizeof(a));
	for(int i=0;i<n;i++)	
	{
		scanf("%s",in);
		for(int j=0;j<m;j++)
		{
 			a[i][j]=in[j]-'0';
		}
	}
	///source:2*n*m sink:2*n*m+1 mid:2*n*m+2;
	int source=2*n*m,sink=2*n*m+1,mid=2*n*m+2;	
	N=mid+1;
	Add_Edge(source,mid,k,0);
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			int from=i*m+j;
			Add_Edge(source,from,1,0);
			Add_Edge(from+n*m,sink,1,0);
			Add_Edge(mid,from+n*m,1,0);
			for(int ii=i+1;ii<n;ii++)
			{
				int to=ii*m+j+n*m;
				int cost=0;
				if(a[i][j]==a[ii][j])
					cost=a[i][j];
				cost-=ii-i-1;
				Add_Edge(from,to,1,-cost);
			}
			for(int jj=j+1;jj<m;jj++)
			{
				int to=i*m+jj+n*m;
				int cost=0;
				if(a[i][j]==a[i][jj])
					cost=a[i][j];
				cost-=jj-j-1;
				Add_Edge(from,to,1,-cost);
			}
		}
	}
	int C,F;
	F=MinCostMaxFlow(source,sink,C);
	C=-C;
	if(F!=n*m) C=-1;
	printf("Case %d : %d\n",cas++,C);
}
	return 0;
}