标签：poj3422
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8006 | Accepted: 3204 |
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 <queue>
#define inf 0x3f3f3f3f
#define maxN 55
#define maxn maxN * maxN * 2
#define maxm maxn * 4
using std::queue;
int head[maxn], n, k, id;
struct Node {
int u, v, c, f, next;
} E[maxm];
int dist[maxn], map[maxN][maxN];
int pre[maxn], source, sink;
bool vis[maxn];
void addEdge(int u, int v, int c, int f) {
E[id].u = u; E[id].v = v; E[id].f = f;
E[id].c = c; E[id].next = head[u];
head[u] = id++;
E[id].u = v; E[id].v = u; E[id].f = -f;
E[id].c = 0; E[id].next = head[v];
head[v] = id++;
}
void getMap() {
memset(head, -1, sizeof(head));
int i, j, f, pos, down, right; id = 0;
for(i = 0; i < n; ++i)
for(j = 0; j < n; ++j) {
scanf("%d", &map[i][j]);
pos = i * n + j; right = pos + 1;
down = pos + n;
addEdge(pos, pos + n*n, 1, map[i][j]); // 拆点
addEdge(pos, pos + n*n, inf, 0);
if(i != n - 1) {
addEdge(pos + n*n, down, inf, 0);
}
if(j != n - 1) {
addEdge(pos + n*n, right, inf, 0);
}
}
source = 2 * n * n; sink = source + 1;
map[n][0] = map[n][1] = 0;
addEdge(source, 0, k, 0);
addEdge(source - 1, sink, k, 0);
}
bool SPFA(int start, int end) {
memset(pre, -1, sizeof(pre));
memset(vis, 0, sizeof(vis));
memset(dist, -1, sizeof(dist));
queue<int> Q; Q.push(start);
int u, v, i; vis[start] = 1; dist[start] = 0;
while(!Q.empty()) {
u = Q.front(); Q.pop(); vis[u] = 0;
for(i = head[u]; i != -1; i = E[i].next) {
v = E[i].v;
if(E[i].c && dist[v] < dist[u] + E[i].f) {
dist[v] = dist[u] + E[i].f;
pre[v] = i;
if(!vis[v]) {
vis[v] = 1; Q.push(v);
}
}
}
}
return dist[end] != -1;
}
void solve() {
int sum = 0, i, u, v, minCut;
while(SPFA(source, sink)) {
minCut = inf;
for(i = pre[sink]; i != -1; i = pre[E[i].u]) {
if(minCut > E[i].c) minCut = E[i].c;
}
sum += minCut * dist[sink];
for(i = pre[sink]; i != -1; i = pre[E[i].u]) {
E[i].c -= minCut;
E[i^1].c += minCut;
}
}
printf("%d\n", sum);
}
int main() {
// freopen("stdin.txt", "r", stdin);
while(scanf("%d%d", &n, &k) == 2) {
getMap();
solve();
}
return 0;
}POJ3422 Kaka's Matrix Travels 【最大费用最大流】
标签：poj3422
原文地址：http://blog.csdn.net/chang_mu/article/details/39801397
