yaoxi-std 的博客

$\text{开}\mathop{\text{卷}}\limits^{ju\check{a}n}\text{有益}$

0%

P2774 方格取数问题

发表于 更新于 分类于

P2774 方格取数问题

题面

题目链接

解法

使用割模型来做。由于相邻的数不能同时取，于是很容易想到建立二分图，二分图两侧的点之间连$+\infty$的边，并且分别向源点和汇点连边，流量为对应权值。这样最小割一定只能割掉两侧与源点和汇点的连边，被割掉的边一定不取。

AC代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
/**
 * @file:           P2774.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P2774
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 1e4 + 10;
const int MAXM = 5e4 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
    struct Edge {
        int v, flow;
    } edge[MAXM];
    int tot = 1, flow = 0;
    int head[MAXN], nxt[MAXM], lev[MAXN], cur[MAXN];
    void addedge(int u, int v, int flow) {
        edge[++tot] = {v, flow};
        nxt[tot] = head[u], head[u] = tot;
        edge[++tot] = {u, 0};
        nxt[tot] = head[v], head[v] = tot;
    }
    bool bfs(int s, int t) {
        fill(lev, lev + MAXN, -1);
        queue<int> que;
        que.push(s);
        lev[s] = 0;
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            for (int i = head[u]; i; i = nxt[i]) {
                int v = edge[i].v;
                if (edge[i].flow && lev[v] == -1) {
                    lev[v] = lev[u] + 1;
                    que.push(v);
                }
            }
        }
        return lev[t] != -1;
    }
    int augment(int u, int t, int mx) {
        if (u == t || mx == 0)
            return mx;
        int ret = 0;
        for (int &i = cur[u]; i; i = nxt[i]) {
            int v = edge[i].v;
            if (lev[v] != lev[u] + 1)
                continue;
            int tmp = augment(v, t, min(mx, edge[i].flow));
            mx -= tmp, ret += tmp;
            edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
            if (mx == 0)
                break;
        }
        return ret;
    }
    int maxflow(int s, int t) {
        while (bfs(s, t)) {
            copy(head, head + MAXN, cur);
            flow += augment(s, t, INF);
        }
        return flow;
    }
};
int n, m, s, t, num, a[MAXN][MAXN], pt[MAXN][MAXN];
Dinic network;
signed main() {
    read(n), read(m);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            read(a[i][j]);
    s = ++num, t = ++num;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            pt[i][j] = ++num;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if ((i ^ j) & 1)
                network.addedge(s, pt[i][j], a[i][j]);
            else
                network.addedge(pt[i][j], t, a[i][j]);
            if (((i ^ j) & 1) == 0)
                continue;
            if (i > 1)
                network.addedge(pt[i][j], pt[i - 1][j], INF);
            if (j > 1)
                network.addedge(pt[i][j], pt[i][j - 1], INF);
            if (i < n)
                network.addedge(pt[i][j], pt[i + 1][j], INF);
            if (j < m)
                network.addedge(pt[i][j], pt[i][j + 1], INF);
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            ans += a[i][j];
    ans -= network.maxflow(s, t);
    write(ans), putchar('\n');
    return 0;
}