洛谷P2045 方格取数加强版(费用流)

题意

题目链接

sol

这题能想到费用流就不难做了

从s向(1, 1)连费用为0，流量为k的边

从(n, n)向t连费用为0，流量为k的边

对于每个点我们可以拆点限流，同时为了保证每个点只被经过一次，需要拆点。

对于拆出来的每个点，在其中连两条边，一条为费用为点权，流量为1，另一条费用为0，流量为inf

相邻两个点之间连费用为0，流量为inf的边。

跑最大费用最大流即可

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? eof : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int maxn = 51, max = 1e5 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9, pi = acos(-1);
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = (x * 10 + c - '0') % mod, c = getchar();
    return x * f;
}
int n, k, s = 0, t = 1e5 - 1, a[maxn][maxn], dis[max], vis[max], pre[max], id[maxn][maxn][2], cnt, maxcost;
struct edge {
    int u, v, w, f, nxt;
}e[max];
int head[max], num;
inline void add_edge(int x, int y, int w, int f) {
    e[num] = (edge) {x, y, w, f, head[x]};
    head[x] = num++;
}
inline void ae(int x, int y, int w, int f) {
    add_edge(x, y, w, f);
    add_edge(y, x, -w, 0);
}
bool spfa() {
    queue<int> q; q.push(s);
    memset(dis, -0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[s] = 0;
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = 0;
        for(int i = head[p]; ~i; i = e[i].nxt) {
            int to = e[i].v;
            if(e[i].f && dis[to] < dis[p] + e[i].w) {
                dis[to] = dis[p] + e[i].w; pre[to] = i;
                if(!vis[to]) vis[to] = 1, q.push(to);
            }
        }
    }
    return dis[t] > -inf;
}
void f() {
    int canflow = inf;
    for(int i = t; i != s; i = e[pre[i]].u) chmin(canflow, e[pre[i]].f);
    for(int i = t; i != s; i = e[pre[i]].u) e[pre[i]].f -= canflow, e[pre[i] ^ 1].f += canflow;
    maxcost += canflow * dis[t];
}
void mcmf() {
    while(spfa()) f();
}
signed main() {
//  freopen("a.in", "r", stdin);
    memset(head, -1, sizeof(head));
    n = read(); k = read();
    for(int i = 1; i <= n; i++) 
        for(int j = 1; j <= n; j++) 
            a[i][j] = read(), id[i][j][0] = ++cnt, id[i][j][1] = ++cnt;
    ae(s, id[1][1][0], 0, k);
    ae(id[n][n][1], t, 0, k);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            ae(id[i][j][0], id[i][j][1], a[i][j], 1);
            ae(id[i][j][0], id[i][j][1], 0, inf);
            if(i + 1 <= n) ae(id[i][j][1], id[i + 1][j][0], 0, inf);
            if(j + 1 <= n) ae(id[i][j][1], id[i][j + 1][0], 0, inf);
        }
    }
    mcmf();
    printf("%d", maxcost);
    return 0;
}
/*
3 2
1 2 3
0 2 1
1 4 2
*/