首页 > 代码库 > POJ 3422Kaka's Matrix Travels（最小费用最大流）
POJ 3422Kaka's Matrix Travels（最小费用最大流）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9460 | Accepted: 3844 |
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 21 2 30 2 11 4 2
Sample Output
15
Source
【分析】下面是模板。
#include <iostream>#include <cstring>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <time.h>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#define inf 0x7fffffff#define mod 10000#define met(a,b) memset(a,b,sizeof a)typedef long long ll;using namespace std;const int N = 100;const int M = 100000;struct Edge { int to,next,cap,flow,cost;} edge[M];int head[N],tol;int pre[N],dis[N];bool vis[N];int T;//节点总个数，节点编号从0~N-1void init(int n) { T = 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;}//返回的是最大流，cost存的是最小费用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 a[N][N];int main() { int n,k; while(~scanf("%d%d",&n,&k) ) { for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) scanf("%d",&a[i][j]); init(2*n*n+2); for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) { addedge(n*i+j+1,n*n+n*i+j+1,1,-a[i][j]); addedge(n*i+j+1,n*n+n*i+j+1,inf,0); } for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) { if(i < n-1) addedge(n*n+n*i+j+1,n*(i+1)+j+1,inf,0); if(j < n-1) addedge(n*n+n*i+j+1,n*i+j+1+1,inf,0); } addedge(0,1,k,0); addedge(2*n*n,2*n*n+1,inf,0); int cost; minCostMaxflow(0,2*n*n+1,cost); printf("%d\n",-cost); } return 0;}
POJ 3422Kaka's Matrix Travels（最小费用最大流）