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P4013 数字梯形问题

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P4013 数字梯形问题

题面

题目链接

解法

三次询问分别拆点构建网络流即可。

费用流不能在残余网络上加边重新跑，否则无法保证最大/最小费用。

数组应当开$20 \times 40$而不是$20 \times 20$。

AC代码

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/**
 * @file:           P4013.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4013
*/
// #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);
}
template <const int MAXV, const int MAXE>
struct MCMF {
    const int INF = 0x3f3f3f3f3f3f3f3f;
    struct Edge {
        int v, flow, cost;
    } edge[MAXE * 2];
    int tot = 1, head[MAXV], nxt[MAXE];
    int flow, cost, cur[MAXV], dis[MAXV];
    bool vis[MAXV];
    void addedge(int u, int v, int flow, int cost) {
        edge[++tot] = {v, flow, cost};
        nxt[tot] = head[u], head[u] = tot;
        edge[++tot] = {u, 0, -cost};
        nxt[tot] = head[v], head[v] = tot;
    }
    bool spfa(int s, int t) {
        fill(vis, vis + MAXV, 0);
        fill(dis, dis + MAXV, -INF);
        queue<int> que;
        que.push(s);
        dis[s] = 0;
        vis[s] = 1;
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            vis[u] = 0;
            for (int i = head[u]; i; i = nxt[i]) {
                int v = edge[i].v;
                if (edge[i].flow && dis[v] < dis[u] + edge[i].cost) {
                    dis[v] = dis[u] + edge[i].cost;
                    if (!vis[v]) {
                        que.push(v);
                        vis[v] = 1;
                    }
                }
            }
        }
        return dis[t] != -INF;
    }
    int augment(int u, int t, int mx) {
        if (u == t || mx == 0)
            return mx;
        vis[u] = 1;
        int ret = 0;
        for (int &i = cur[u]; i; i = nxt[i]) {
            int v = edge[i].v;
            if (vis[v] || dis[v] != dis[u] + edge[i].cost)
                continue;
            int tmp = augment(v, t, min(mx, edge[i].flow));
            cost += tmp * edge[i].cost;
            mx -= tmp, ret += tmp;
            edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
            if (mx == 0)
                break;
        }
        vis[u] = 0;
        return ret;
    }
    pair<int, int> mcmf(int s, int t) {
        while (spfa(s, t)) {
            copy(head, head + MAXV, cur);
            flow += augment(s, t, INF);
        }
        return make_pair(flow, cost);
    }
};
const int MAXN = 45;
const int MAXV = 4e3 + 10;
const int MAXE = 4e4 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
int m, n, s, t, num, a[MAXN][MAXN], pt[MAXN][MAXN][2];
MCMF<MAXV, MAXE> nt[3];
signed main() {
    read(m), read(n);
    s = ++num, t = ++num;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j < m + i; ++j) {
            read(a[i][j]);
            pt[i][j][0] = ++num;
            pt[i][j][1] = ++num;
        }
    }
    for (int w = 0; w < 3; ++w) {
        for (int i = 1; i <= m; ++i)
            nt[w].addedge(s, pt[1][i][0], 1, 0);
        for (int i = 1; i < m + n; ++i)
            nt[w].addedge(pt[n][i][1], t, INF, 0);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j < m + i; ++j)
                nt[w].addedge(pt[i][j][0], pt[i][j][1], 1, a[i][j]);
        for (int i = 1; i < n; ++i) {
            for (int j = 1; j < m + i; ++j) {
                nt[w].addedge(pt[i][j][1], pt[i + 1][j][0], 1, 0);
                nt[w].addedge(pt[i][j][1], pt[i + 1][j + 1][0], 1, 0);
            }
        }
    }
    for (int w = 1; w < 3; ++w) {
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j < m + i; ++j)
                nt[w].addedge(pt[i][j][0], pt[i][j][1], INF, a[i][j]);
    }
    for (int w = 2; w < 3; ++w) {
        for (int i = 1; i < n; ++i) {
            for (int j = 1; j < m + i; ++j) {
                nt[w].addedge(pt[i][j][1], pt[i + 1][j][0], INF, 0);
                nt[w].addedge(pt[i][j][1], pt[i + 1][j + 1][0], INF, 0);
            }
        }
    }
    for (int w = 0; w < 3; ++w)
        write(nt[w].mcmf(s, t).second), putchar('\n');
    return 0;
}