HDU_3376_Matrix Again(最小费用流)

Matrix Again

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 3509    Accepted Submission(s): 1032

Problem Description

Starvae 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 starvae should to do is that choose a detour which from 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 starvae 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 starvae can not pass the same area of the Matrix except the start and end..
Do you know why call this problem as “Matrix Again”? AS it is like the problem 2686 of HDU.

 

Input

The input contains multiple test cases.
Each case first line given the integer n (2<=n<=600)
Then n lines, each line include n positive integers. (<100)

 

Output

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

 

Sample Input

210 35 10310 3 32 5 36 7 1051 2 3 4 52 3 4 5 63 4 5 6 74 5 6 7 85 6 7 8 9

 

Sample Output

284680

 

题意：从矩阵起点(1,1) -> (n,n) -> (1,1) 经过点的权值和最大是多少。每个点只能走一次，从(1,1) -> (n,n)只能往右或往下走，从(n,n) -> (1,1)只能往左或往上走。

分析：跟HDU_2686题意一模一样，只不过点数增加了很多，所有用vector会超时，转为结构体数组就行了，然而C++ TLE了，G++ AC。不知为何？

建图：首先要想到拆点。加入超级源点s(0)，s连向1，容量为2，费用为0；加入超级汇点t(N*N*2+1)，点N*N*2连向t，容量为2，费用为0；对于一个点u(I,j)，如果u为(1,1)或(N,N)，那么连边 u -> u+N*N，容量为2，费用为-m[i][j]，否则连边 u -> u+N*N，容量为1，费用为-m[I][j]；然后对于其右边点以及下边点v，连边 u+N*N -> v，容量为1，费用为0。然后跑一遍最小费用 最大流即可。注意得到的mincost里包含了两个点(1,1),(N,N)，所以结果为 -mincost - m[1][1] - m[N][N]。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3376

代码清单：

#include<map>#include<set>#include<cmath>#include<queue>#include<stack>#include<ctime>#include<cctype>#include<string>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;#define end() return 0typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;const int maxn = 1000000 + 5;const int INF = 0x7f7f7f7f;struct Edge{    int from,to,cap,flow,cost,next;    Edge(){}    Edge(int u,int v,int c,int f,int w,int n):from(u),to(v),cap(c),flow(f),cost(w),next(n){}};struct MCMF{    int n,m,cnt;    Edge edge[4*maxn];    int head[maxn];    int inq[maxn]; //是否在队列    int d[maxn]; //Bellman-Ford    int p[maxn]; //上一条弧    int a[maxn]; //可改进量    void init(int n){        this -> n = n;        memset(head,-1,sizeof(head));        cnt=0;    }    void addEdge(int from,int to,int cap,int cost){        edge[cnt]=Edge(from,to,cap,0,cost,head[from]); head[from]=cnt++;        edge[cnt]=Edge(to,from,0,0,-cost,head[to]); head[to]=cnt++;    }    bool BellmanFord(int s,int t,int& flow,int& cost){        memset(d,INF,sizeof(d));        memset(inq,0,sizeof(inq));        //memset(p,-1,sizeof(p));        d[s]=0; inq[s]=1; p[s]=0; a[s]=INF;        queue<int>q;        q.push(s);        while(!q.empty()){            int u=q.front();q.pop();            inq[u]=0;            for(int i=head[u];i!=-1;i=edge[i].next){                Edge& e=edge[i];                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){                    d[e.to]=d[u]+e.cost;                    p[e.to]=i;                    a[e.to]=min(a[u],e.cap-e.flow);                    if(!inq[e.to]){                        q.push(e.to);                        inq[e.to]=1;                    }                }            }        }        if(d[t]==INF) return false;        flow+=a[t];        cost+=d[t]*a[t];        //if(flow==2) return false;        for(int u=t;u!=s;u=edge[p[u]].from){            edge[p[u]].flow+=a[t];            edge[p[u]^1].flow-=a[t];        }        return true;    }    //需要保证初始网络中没有负权圈    int MincostMaxflow(int s,int t){        int flow=0,cost=0;        while(BellmanFord(s,t,flow,cost));        return cost;    }};int N,tail;int m[605][605];MCMF mcmf;void input(){    for(int i=1;i<=N;i++){        for(int j=1;j<=N;j++){            scanf("%d",&m[i][j]);        }    }}void createGraph(){    tail=N*N*2+1;    mcmf.init(tail+1);    mcmf.addEdge(0,1,2,0);    mcmf.addEdge(N*N*2,tail,2,0);    for(int i=1;i<=N;i++){        for(int j=1;j<=N;j++){            int u=(i-1)*N+j;            int v1=(i-1)*N+j+1; //(i,j+1)            int v2=i*N+j; //(i+1,j)            if(i==1&&j==1){                mcmf.addEdge(u,u+N*N,2,-m[i][j]);            }            else if(i==N&&j==N){                mcmf.addEdge(u,u+N*N,2,-m[i][j]);            }            else{                mcmf.addEdge(u,u+N*N,1,-m[i][j]);            }            if(j+1<=N)                mcmf.addEdge(u+N*N,v1,1,0);            if(i+1<=N)                mcmf.addEdge(u+N*N,v2,1,0);        }    }}void solve(){    createGraph();    printf("%d\n",-m[1][1]-m[N][N]-mcmf.MincostMaxflow(0,tail));}int main(){    while(scanf("%d",&N)!=EOF){        input();        solve();    }end();}