Matrix

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

28
46
80

思路：这里用到费用流求解，首先添加一个超级源点s=0和超级汇点t=n*n+1，然后对每个点拆点， i 向 i` 连边，容量为1，花费为该点的权值mp[i][j]，然后s与 1` 连边，容量为2，花费为0，n*n向t连边，容量为2，花费为0，最后矩阵中的点之间连边，容量为1，花费为0。最后答案为cost+mp[1][1]+mp[n][n]。注意数组的大小。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

const int MAXN = 2000;  //注意数组的大小
const int MAXM = 100000;

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

int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N,n,m;

void init(int n)
{
    N=n;
    tol=0;
    memset(head,-1,sizeof(head));
}

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

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

int mp[MAXN][MAXN];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t;
    while (~sf(n))
    {
        init(2*n*n+2);
        FRE(i,1,n)
            FRE(j,1,n)
                sf(mp[i][j]);
        addedge(0,1+n*n,2,0);
        addedge(n*n,2*n*n+1,2,0);
        FRE(i,1,n)
            FRE(j,1,n)
            {
                addedge((i-1)*n+j,(i-1)*n+j+n*n,1,mp[i][j]);
                if (j+1<=n)
                    addedge((i-1)*n+n*n+j,(i-1)*n+j+1,1,0);
                if (i+1<=n)
                    addedge((i-1)*n+j+n*n,i*n+j,1,0);
            }
        int cost;
        int ans=minCostMaxflow(0,N-1,cost);
        printf("%d\n",cost+mp[1][1]+mp[n][n]);
    }
    return 0;
}

Matrix (hdu 2686 最大费用最大流)

原文：http://blog.csdn.net/u014422052/article/details/46763039