【HDU - 2686 and HDU4322 】 Matrix 【拆点最大费用流】

Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.

Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end.

Input

The input contains multiple test cases.

Each case first line given the integer n (2 < n < 30)

Than n lines,each line include n positive integers.(<100)

Output

For each test case output the maximal values yifenfei can get.

Sample Input

2

10 3

5 10

3

10 3 3

2 5 3

6 7 10

5

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

5 6 7 8 9

Sample Output

28

46

80

虽然提上说的是正反跑两次，其实我们可以认为从起点到终点走有两条流就行了。

再看提上有说明每个点只可以走一次，我就想到了拆点网络流，又因为是带点权的，所以就是最大费用流。

我建的图

………………左点………………….右点

S—[2,0]—-1 —–[2,cost]— n*n+1

……………….2——[1,cost]—n*n+2

……………….3——[1,cost]—n*n+3

……………..n*n—–[2,cost]—n*n+n*n —[2,0]——T

左点为入点，右点为出点。 这种拆点很常见，理解了就不难了。

HDU 2686 和4322 都一样啊 ，就改了下范围就过了。

代码

#include<bits/stdc++.h>
using namespace std ;
typedef long long LL ;
#define debug() puts("======")

const int MAXN = 5*(30+10)*(30+10)+100;
const int MAXM =  1e5+100 ;
const int mod  = 1e9+7 ;
const int inf  = 0x3f3f3f3f;

struct Edge {
int  to,cap,flow,cost,next;
}edge[MAXM];
int head[MAXN],top;
void init(){
memset(head,-1,sizeof(head));
top=0;
}
void addedge(int a,int  b
4000
,int w,int  c){
edge[top].to=b;edge[top].cap=w;edge[top].flow=0;edge[top].cost=c;edge[top].next=head[a];
head[a]=top++;
edge[top].to=a;edge[top].cap=0;edge[top].flow=0;edge[top].cost=-c;edge[top].next=head[b];
head[b]=top++;
}

int n;
int S,T;
int a,b;
void getmap(){  //  建图
S=0;   T=2*n*n+1;
int temp=n*n;
addedge(S,1,2,0);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
int val;scanf("%d",&val);
if(i==1&&j==1){
a=val;
addedge((i-1)*n+j,(i-1)*n+j+temp,2,val);
int nx=i+1;int ny=j;
if(nx>=1&&nx<=n&&ny>=1&&ny<=n)
addedge((i-1)*n+j+temp,(nx-1)*n+ny,1,0);
nx=i;  ny=j+1;
if(nx>=1&&nx<=n&&ny>=1&&ny<=n)
addedge((i-1)*n+j+temp,(nx-1)*n+ny,1,0);
continue;
}
if(i==n&&j==n) {
b=val;
addedge((i-1)*n+j,(i-1)*n+j+temp,2,val);
addedge((i-1)*n+j+temp,T,2,0);
continue;
}

addedge((i-1)*n+j,(i-1)*n+j+temp,1,val);
int nx=i+1;int ny=j;
if(nx>=1&&nx<=n&&ny>=1&&ny<=n)
addedge((i-1)*n+j+temp,(nx-1)*n+ny,1,0);
nx=i;  ny=j+1;
if(nx>=1&&nx<=n&&ny>=1&&ny<=n)
addedge((i-1)*n+j+temp,(nx-1)*n+ny,1,0);
}
}
}

int  pre[MAXN],dis[MAXN];
bool vis[MAXN];
bool  spfa(int s,int t){
queue<int>Q;
memset(dis,-inf,sizeof(dis));
memset(vis,false,sizeof(vis));
memset(pre,-1,sizeof(pre));
dis[s]=0;vis[s]=true;Q.push(s);
while(!Q.empty()){
int now=Q.front();Q.pop();vis[now]=0;
for(int i=head[now];i!=-1;i=edge[i].next){
Edge  e=edge[i];
if(e.cap>e.flow&&dis[e.to]<dis[now]+e.cost){
dis[e.to]=dis[now]+e.cost;
pre[e.to]=i;
if(!vis[e.to]){
vis[e.to]=true;
Q.push(e.to);
}
}
}
}
return pre[t]!=-1;
}

void MCMF(int s, int t, int &flow, int &cost)
{
cost = flow = 0;
while(spfa(s, t)) {
int Min = inf;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]){
Edge E = edge[i];
Min = min(Min, E.cap-E.flow);
}
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
}

int main(){
while(scanf("%d",&n)!=EOF){
init();
getmap();
int flow,ans;
MCMF(S,T,flow,ans);
printf("%d\n",ans-a-b);//每个边的花费=每条边上的费用*流过这个边的流量，所以我建的图，始末点的费用多了一倍，要减掉。
}
return  0;
}