$ color{#0066ff}{ 题目描述 }$
给出一个n*n的矩阵,每一格有一个非负整数Aij,(Aij <= 1000)现在从(1,1)出发,可以往右或者往下走,最后到达(n,n),每达到一格,把该格子的数取出来,该格子的数就变成0,这样一共走K次,现在要求K次所达到的方格的数的和最大
(color{#0066ff}{输入格式})
第一行两个数n,k(1<=n<=50, 0<=k<=10)
接下来n行,每行n个数,分别表示矩阵的每个格子的数
(color{#0066ff}{输出格式})
一个数,为最大和
(color{#0066ff}{输入样例})
3 1
1 2 3
0 2 1
1 4 2
(color{#0066ff}{输出样例})
11
(color{#0066ff}{数据范围与提示})
每个格子中的数不超过1000
(color{#0066ff}{题解})
拆点,矩阵每个元素拆成入点和出点
入点向出点连容量为1, 权值为点权的边,代表只能选一次
入点再向出点连容量为inf,权值为0的边,代表之后也能走这个格子,但是无法获得权值
最大费用最大流即可(这里用了ZKWqwq)
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int inf = 0x7fffffff;
const int maxn = 1e5 + 10;
struct node {
int to, can, dis;
node *nxt, *rev;
node(int to = 0, int can = 0, int dis = 0, node *nxt = NULL): to(to), can(can), dis(dis), nxt(nxt) {
rev = NULL;
}
};
node *head[maxn];
bool vis[maxn];
int dis[maxn], mp[100][100];
int n, k, s, t;
void add(int from, int to, int can, int dis) {
head[from] = new node(to, can, dis, head[from]);
}
void link(int from, int to, int can, int dis) {
add(from, to, can, dis);
add(to, from, 0, -dis);
head[from]->rev = head[to];
head[to]->rev = head[from];
}
bool spfa() {
for(int i = s; i <= t; i++) vis[i] = false, dis[i] = -inf;
std::deque<int> q;
q.push_back(t);
dis[t] = 0;
while(!q.empty()) {
int tp = q.front(); q.pop_front();
vis[tp] = false;
for(node *i = head[tp]; i; i = i->nxt) {
if(dis[i->to] < dis[tp] - i->dis && i->rev->can) {
dis[i->to] = dis[tp] - i->dis;
if(!vis[i->to]) {
if(!q.empty() && dis[i->to] > dis[q.front()]) q.push_front(i->to);
else q.push_back(i->to);
vis[i->to] = true;
}
}
}
}
return dis[s] != -inf;
}
int dfs(int x, int change) {
if(x == t || !change) return change;
int flow = 0, ls;
vis[x] = true;
for(node *i = head[x]; i; i = i->nxt) {
if(!vis[i->to] && dis[i->to] == dis[x] - i->dis && (ls = dfs(i->to, std::min(change, i->can)))) {
flow += ls;
change -= ls;
i->can -= ls;
i->rev->can += ls;
if(!change) break;
}
}
return flow;
}
int zkw() {
int cost = 0;
while(spfa()) {
vis[t] = true;
while(vis[t]) {
for(int i = s; i <= t; i++) vis[i] = false;
cost += dis[s] * dfs(s, inf);
}
}
return cost;
}
int id(int x, int y) { return (x - 1) * n + y; }
int main() {
n = in(), k = in();
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
mp[i][j] = in();
s = 0, t = 2 * n * n + 1;
link(s, id(1, 1), k, 0);
link(id(n, n) + n * n, t, k, 0);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++) {
if(i + 1 <= n) link(n * n + id(i, j), id(i + 1, j), inf, 0);
if(j + 1 <= n) link(n * n + id(i, j), id(i, j + 1), inf, 0);
link(id(i, j), n * n + id(i, j), 1, mp[i][j]);
link(id(i, j), n * n + id(i, j), inf, 0);
}
printf("%d", zkw());
return 0;
}