网络费用流-最小k路径覆盖

多校联赛第一场（hdu4862）

Jump

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 644    Accepted Submission(s): 275

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

题目大意：给出一个n行m列的矩阵，每个坐标上都有一个权值，可以最多走k次（注意不是k步），每一次行走可以任意选择起点（必须是没有走过的点），然后可以向右或者向下走任意的步数，每一步可以跨越任意的格数，前提是满足题目的限制条件，每走一步耗费的能量是两坐标的曼哈顿距离-1，加入走的当前点的权值和前一步的权值一样，可以获得该权值的能量，问在k次行走内，走完每一格且仅走一次，最多可以获得多少能量：这道题目主要是建图，k次的建图非常巧妙。

官方题解：

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

可以这样考虑，首先建立一个二分图，在x部内的可以到达y部的点建边<u,v>；首先求得该二部图的最小路径覆盖=节点总数-最大流；（最小路径覆盖：用最少的路径覆盖住所有的节点，且每个节点只能在一条路径上），加入k值小于最小的路径数目，那么肯定是在k次之内是不可能完成任务的。

所以这样建图之后求得是原先的最大流+k，若最大流+k<n*m则无解，输出-1，否则求出的最小费用的相反数就是最大能量。

程序;

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#define M 333
#define eps 1e-10
#define inf 1000000000
#define mod 2333333
using namespace std;
struct node
{
    int u,v,w,cost,next;
}edge[M*M*3];
int t,head[M],dis[M],use[M],pre[M],Max_flow,Min_cost,top[M],q[M],work[M];
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w,int cost)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].w=w;
    edge[t].cost=cost;
    edge[t].next=head[u];
    head[u]=t++;

    edge[t].u=v;
    edge[t].v=u;
    edge[t].w=0;
    edge[t].cost=-cost;
    edge[t].next=head[v];
    head[v]=t++;
}
int min_flow(int S,int T)
{
    int ans=0;
    Max_flow=0;
    while(1)
    {
        int i;
        queue<int>q;
        for(i=0;i<=T+1;i++)
            dis[i]=inf;
        dis[S]=0;
        memset(use,0,sizeof(use));
        memset(pre,-1,sizeof(pre));
        q.push(S);
        while(!q.empty())
        {
            int u=q.front();
            use[u]=0;
            q.pop();
            for(i=head[u];i!=-1;i=edge[i].next)
            {
                int v=edge[i].v;
                if(edge[i].w&&dis[v]>dis[u]+edge[i].cost)
                {
                    dis[v]=dis[u]+edge[i].cost;
                    pre[v]=i;
                    if(!use[v])
                    {
                        use[v]=1;
                        q.push(v);
                    }
                }
            }
        }
        if(dis[T]==inf)
            break;
        int Min=inf+1;
        for(i=pre[T];i!=-1;i=pre[edge[i].u])
        {
            Min=min(Min,edge[i].w);
        }
        for(i=pre[T];i!=-1;i=pre[edge[i].u])
        {
            edge[i].w-=Min;
            edge[i^1].w+=Min;
        }
        Max_flow+=Min;
        ans+=Min*dis[T];
    }
    return ans;
}
int num[22][22],mp[22][22];
char Mp[22][22];
int main()
{
    int T,kk=1;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,k,i,j,r;
        scanf("%d%d%d",&n,&m,&k);
        int cnt=0;
        init();
        for(i=1;i<=n;i++)
        {
            scanf("%s",Mp[i]);
            for(j=1;j<=m;j++)
            {
                mp[i][j]=Mp[i][j-1]-'0';
                num[i][j]=++cnt;
            }
        }
        for(i=1;i<=m*n;i++)
        {
            add(0,i,1,0);
            add(i+m*n,m*n*2+1,1,0);
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                int Cost;
                for(r=j+1;r<=m;r++)
                {
                    if(mp[i][j]==mp[i][r])
                        Cost=r-j-1-mp[i][j];
                    else
                        Cost=r-j-1;
                    add(num[i][j],num[i][r]+m*n,1,Cost);
                }
                for(r=i+1;r<=n;r++)
                {
                    if(mp[i][j]==mp[r][j])
                        Cost=r-i-1-mp[i][j];
                    else
                        Cost=r-i-1;
                    add(num[i][j],num[r][j]+m*n,1,Cost);
                }
            }
        }
        for(i=n*m+1;i<=m*n*2;i++)
            add(m*n*2+2,i,1,0);
        add(0,m*n*2+2,k,0);
        Min_cost=min_flow(0,m*n*2+1);
        printf("Case %d : ",kk++);
        if(Max_flow!=n*m)
        {
            printf("-1
");
            continue;
        }
        printf("%d
",-Min_cost);
    }
}