POJ 3422 Kaka's Matrix Travels 费用流
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7465 | Accepted: 3004 |
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
Source
#include <stdio.h> #include <string.h> #include <algorithm> #define min(a, b) ((a) < (b) ? (a) : (b)) #define REP(i, n) for(int i = 1; i <= n; ++i) #define MS0(X) memset(X, 0, sizeof X) #define MS1(X) memset(X, -1, sizeof X) using namespace std; const int maxE = 3000000; const int maxN = 5005; const int maxM = 55; const int oo = 0x3f3f3f3f; struct Edge{ int v, c, w, n; }; Edge edge[maxE]; int adj[maxN], l; int d[maxN], cur[maxN], Minflow; int inq[maxN], Q[maxE], head, tail; int cost, flow, s, t; int n, m, nn, a[maxM][maxM]; void addedge(int u, int v, int c, int w){ edge[l].v = v; edge[l].c = c; edge[l].w = w; edge[l].n = adj[u]; adj[u] = l++; edge[l].v = u; edge[l].c = 0; edge[l].w = -w; edge[l].n = adj[v]; adj[v] = l++; } int SPFA(){ memset(d, oo, sizeof d); memset(inq, 0, sizeof inq); head = tail = 0; d[s] = 0; Minflow = oo; cur[s] = -1; Q[tail++] = s; while(head != tail){ int u = Q[head++]; inq[u] = 0; for(int i = adj[u]; ~i; i = edge[i].n){ int v = edge[i].v; if(edge[i].c && d[v] > d[u] + edge[i].w){ d[v] = d[u] + edge[i].w; cur[v] = i; Minflow = min(edge[i].c, Minflow); if(!inq[v]){ inq[v] = 1; Q[tail++] = v; } } } } if(d[t] == oo) return 0; flow += Minflow; cost += Minflow * d[t]; for(int i = cur[t]; ~i; i = cur[edge[i ^ 1].v]){ edge[i].c -= Minflow; edge[i ^ 1].c += Minflow; } return 1; } int MCMF(){ flow = cost = 0; while(SPFA()); return cost; } void work(){ REP(i, n) REP(j, n) scanf("%d", &a[i][j]); MS1(adj); l = 0; nn = n * n; s = 0; t = (nn << 1) + 1; addedge(s, 1, m, 0); addedge(nn << 1, t, m, 0); REP(i, n) REP(j, n){ int ij = (i - 1) * n + j; addedge(ij, ij + nn, 1, -a[i][j]); addedge(ij, ij + nn, oo, 0); if(i < n) addedge(ij + nn, i * n + j, oo, 0); if(j < n) addedge(ij + nn, (i - 1) * n + j + 1, oo, 0); } printf("%d\n", -MCMF()); } int main(){ while(~scanf("%d%d", &n, &m)) work(); return 0; }POJ 3422 优质内容筛选与推荐>>
1、微软邀请黑客访问总部 决定每年召开黑客大会
2、php伪静态--隐藏地址实际路径方法
3、编程3总结
4、面向对象-面向对象和面向过程的区别
5、显示的接口实现