poj_3422_Kaka's Matrix Travels

Description

On an N × N chessboard with a non-negative number in each grid, Kaka starts his matrix travels with SUM = 0. For each travel, Kaka moves one rook from the left-upper grid to the right-bottom one, taking care that the rook moves only to the right or down. Kaka adds the number to SUM in each grid the rook visited, and replaces it with zero. It is not difficult to know the maximum SUM Kaka can obtain for his first travel. Now Kaka is wondering what is the maximum SUM he can obtain after his Kth travel. Note the SUM is accumulative during the K travels.

Input

The first line contains two integers N and K (1 ≤ N ≤ 50, 0 ≤ K ≤ 10) described above. The following N lines represents the matrix. You can assume the numbers in the matrix are no more than 1000.

Output

The maximum SUM Kaka can obtain after his Kth travel.

题解

建图：拆点，把一个点拆成两个点，每两个点连一条边。 设置一个源点和汇点，上模版。

代码：

#include <iostream>#include <queue>#include <string.h>using namespace std;#define inf  1 << 30#define M 5005int n, m, nm, ans;int map[51][51],head[M],qu[M],dis[M];bool v[M];struct node {    int x,y,v,w,next;} e[100005];void add(int x, int y, int w, int f){    e[nm].x=x; e[nm].y=y; e[nm].v=w; e[nm].w=f;    e[nm].next=head[x];    head[x]=nm;    nm++;    e[nm].x=y; e[nm].y=x; e[nm].v=-w; e[nm].w=0;    e[nm].next=head[y];    head[y]=nm;    nm++;}bool spfa() {    for (int i=0;i<=n*n*2+1;i++){        qu[i] = -1; dis[i] = inf;          v[i] = false;      }      queue<int>q;    dis[n*n*2]=0; v[n*n*2]=true;    q.push(n*n*2);    int i;    while (!q.empty()){        int t=q.front();        q.pop();        i=head[t]; v[t]=false;        while (i!=-1){            if (e[i].w>0&&dis[e[i].y]>dis[t]+e[i].v){                dis[e[i].y]=dis[t]+e[i].v;                qu[e[i].y]=i;                if (!v[e[i].y]){                    v[e[i].y]=true;                    q.push(e[i].y);                }            }            i=e[i].next;        }    }    if (qu[n*n*2+1]==-1) return false;    return true;}void getflow() {    while (spfa()) {        int max1=inf;        int p=qu[n*n*2+1];        while (p!=-1) {            max1=min(max1,e[p].w);            p=qu[e[p].x];        }        p=qu[n*n*2+1];        while (p!=-1){            e[p].w-=max1;            e[p^1].w+=max1;            ans+=max1*e[p].v;            p=qu[e[p].x];        }    }}int main() {    while (cin>>n>>m) {        int i,j;        nm=0;        for (i=1;i<=n;i++)            for (j=1;j<=n;j++)                cin>>map[i][j];        memset(head,-1,sizeof(head));        for (i=1;i<=n;i++)            for (j=1;j<=n;j++){                int x=(i-1)*n+j-1;                add(x*2,x*2+1,-map[i][j],1);                add(x*2,x*2+1,0,m-1);            }        for (i=1;i<=n;i++)            for (j=1;j<n;j++){                int x=(i-1)*n+j-1;                add(x*2+1,(x+1)*2,0,m);            }        for (i=1;i<n;i++)              for (j=1;j<=n;j++){                int x=(i-1)*n+j-1;                  add(x*2+1,(x+n)*2,0,m);              }        add(n*n*2,0,0,m);        add(n*n*2-1,n*n*2+1,0,m);        ans=0;        getflow();        cout<<-ans<<endl;    }    return 0;}