poj3422 最小费用最大流

上一题的模板应用,好久没做图论,结果建图一塌糊涂。。。

题意,给出一个矩阵,求从左上角到右下角走k次使路上所得的权值最大,其中经过一个权值后,该点权值清零。

思路:建图,每一个格子为一个点,同时衍生另外一个点,用来存清零后的路线,则每个点有两条路,一条是容量为1,权值为给出的值,一条为容量为k,权值为0;

因为求得是最大,所以建图的时候用1000减去原来的数,求出的最小费再用k*2(n-1)*1000减去,所得的值就为最大值了。

算是理解一下别人的模板吧,建图用时间比较长,最后建完图后,发现TLE了,结果去把范围给开大后,AC了,难道范围小了不应该RE么。。。囧

 

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
#define typef int
#define typec int
#define maxn 6000
#define maxm 30005
#define N maxn + 2
#define E maxm * 4 + 4
const typef inff = 0x3f3f3f3f;
const typec infc = 0x3f3f3f3f;
int map[55][55];
struct network
{
    int nv, ne, pnt[E], nxt[E];
    int vis[N], que[N], head[N], pv[N], pe[N];
    typef flow, cap[E];
    typec cost, dis[E], d[N];
    void addedge(int u, int v, typef c, typec w)
    {
        pnt[ne] = v;
        cap[ne] = c;
        dis[ne] = +w;
        nxt[ne] = head[u];
        head[u] = (ne++);
        pnt[ne] = u;
        cap[ne] = 0;
        dis[ne] = -w;
        nxt[ne] = head[v];
        head[v] = (ne++);
    }
    int mincost(int src, int sink)
    {
        int i, k, f, r;
        typef mxf;
        for (flow = 0, cost = 0;;)
        {
            memset(pv, -1, sizeof(pv));
            memset(vis, 0, sizeof(vis));
            for (i = 0; i < nv; ++i)
                d[i] = infc;
            d[src] = 0;
            pv[src] = src;
            vis[src] = 1;
            for (f = 0, r = 1, que[0] = src; r != f;)
            {
                i = que[f++];
                vis[i] = 0;
                if (N == f)
                    f = 0;
                for (k = head[i]; k != -1; k = nxt[k])
                    if (cap[k] && dis[k] + d[i] < d[pnt[k]])
                    {
                        d[pnt[k]] = dis[k] + d[i];
                        if (0 == vis[pnt[k]])
                        {
                            vis[pnt[k]] = 1;
                            que[r++] = pnt[k];
                            if (N == r)
                                r = 0;
                        }
                        pv[pnt[k]] = i;
                        pe[pnt[k]] = k;
                    }
            }
            if (-1 == pv[sink])
                break;
            for (k = sink, mxf = inff; k != src; k = pv[k])
                if (cap[pe[k]] < mxf)
                    mxf = cap[pe[k]];
            flow += mxf;
            cost += d[sink] * mxf;
            for (k = sink; k != src; k = pv[k])
            {
                cap[pe[k]] -= mxf;
                cap[pe[k] ^ 1] += mxf;
            }
        }
        return cost-1000+map[1][1];
    }
    void build(int v,int n,int k)
    {
        nv = v;
        ne = 0;
        memset(head, -1, sizeof(head));
        int x, y;
        typec w;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            {
                scanf("%d",&map[i][j]);
                map[i][j]=1000-map[i][j];
            }
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(i-1>=1&&j-1<=0)
            {
                addedge(n*(i-2)+j, n*(i-1)+j, 1, map[i][j]);
                addedge(n*(i-2)+j,n*(i-1)+j+n*n,k,1000);
                addedge(n*(i-1)+j+n*n, n*(i-1)+j, k, 0);
            }
            else if(j-1>=1&&i-1<=0)
            {
                addedge(n*(i-1)+j-1,n*(i-1)+j, 1, map[i][j]);
                addedge(n*(i-1)+j-1,n*(i-1)+j+n*n,k,1000);
                addedge(n*(i-1)+j+n*n, n*(i-1)+j, k, 0);
            }    //2*n+n*n
            else if(j-1>=1&&i-1>=1)
            {
                addedge(n*(i-1)+j-1,n*(i-1)+j+n*n,k,0);
                addedge(n*(i-2)+j,  n*(i-1)+j+n*n,k,0);
                addedge(n*(i-2)+j,  n*(i-1)+j, k, 1000);
                addedge(n*(i-1)+j-1,n*(i-1)+j, k, 1000);
                addedge(n*(i-1)+j+n*n,n*(i-1)+j, 1, map[i][j]);

            }

        }
        addedge(0, 1, k, 1000);     //答案加上map【1】【1】;
        addedge(n*n, 2*n*n+1, k, 0);
    }
} g;

int n, m,k;

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d%d", &n, &k);
    g.build(2*n*n + 2,n,k);
    if(k==0)
        printf("0\n");
    else
    printf("%d\n",k*(2*n-1)*1000-g.mincost(0, 2*n*n + 1));
    return 0;
}


 

posted @ 2013-05-31 22:29  amourjun  阅读(107)  评论(0编辑  收藏  举报