POJ训练计划3422_Kaka's Matrix Travels(网络流/费用流)

解题报告

题目传送门

题意：

从ｎ×ｎ的矩阵的左上角走到右下角，每次仅仅能向右和向下走，走到一个格子上加上格子的数，能够走ｋ次。问最大的和是多少。

思路：

建图：每一个格子掰成两个点，分别叫“出点”，“入点”，
入点到出点间连一个容量1。费用为格子数的边。以及一个容量∞，费用0的边。
同一时候。一个格子的“出点”向它右、下的格子的“入点”连边。容量∞，费用0。
源点向（0，0）的入点连一个容量K的边。(N-1,N-1)的出点向汇点连一个容量inf的边。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
struct node {
int v,cost,cap,next;
} edge[101000];
int head[10000],dis[10000],pre[10000],vis[10000],f[10000],mmap[100][100],cnt,s,t,n,m,k,flow,cost;
void add(int u,int v,int cost,int cap) {
edge[cnt].v=v;
edge[cnt].cost=cost;
edge[cnt].cap=cap;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].cost=-cost;
edge[cnt].cap=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
int _spfa() {
for(int i=s; i<=t; i++) {
dis[i]=-1;
pre[i]=f[i]=vis[i]=0;
}
dis[s]=0;
f[s]=inf;
pre[s]=-1;
vis[s]=1;
queue<int>Q;
Q.push(s);
while(!Q.empty()) {
int u=Q.front();
Q.pop();
vis[u]=0;
for(int i=head[u]; i!=-1; i=edge[i].next) {
int v=edge[i].v;
if(edge[i].cap&&dis[v]<dis[u]+edge[i].cost) {
dis[v]=dis[u]+edge[i].cost;
f[v]=min(f[u],edge[i].cap);
pre[v]=i;
if(!vis[v]) {
vis[v]=1;
Q.push(v);
}
}
}
}
if(dis[t]==-1)return 0;
cost+=dis[t];
flow+=f[t];
for(int i=pre[t]; i!=-1; i=pre[edge[i^1].v]) {
edge[i].cap-=f[t];
edge[i^1].cap+=f[t];
}
return 1;
}
void mcmf() {
cost=flow=0;
while(_spfa());
printf("%d\n",cost);
}
int dx[]= {0,1};
int dy[]= {1,0};
int main() {
int i,j;
scanf("%d%d",&n,&k);
memset(head,-1,sizeof(head));
cnt=0;
m=n*n;
s=0;
t=2*m+1;
for(i=0; i<n; i++) {
for(j=0; j<n; j++) {
scanf("%d",&mmap[i][j]);
}
}
add(s,1,0,k);
for(i=0; i<n; i++) {
for(j=0; j<n; j++) {
add(i*n+j+1,m+i*n+j+1,mmap[i][j],1);
add(i*n+j+1,m+i*n+j+1,0,inf);
for(int l=0; l<2; l++) {
int x=i+dx[l];
int y=j+dy[l];
if(x>=0&&x<n&&y>=0&&y<n) {
add(m+i*n+j+1,x*n+y+1,0,inf);
}
}
}
}
add(2*m,t,0,inf);
mcmf();
return 0;
}

[/code]

Kaka's Matrix Travels

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7807		Accepted: 3140

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.

Sample Input

3 2
1 2 3
0 2 1
1 4 2

Sample Output

15