luogu P4013 数字梯形问题

三倍经验，三个条件，分别对应了常见的3种模型，第一种是限制每个点只能一次且无交点，我们可以把这个点拆成一个出点一个入点，capacity为1，这样就限制了只选择一次，第二种是可以有交点，但不能有交边，那我们就不需要拆点，限制每条capacity都为1就可以了，第三种直接连，没有限制(

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e6+;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, cost, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, fa[maxm], d[maxm], buf[][], num[][], ID;
bool inq[maxm];

void init() {
    memset(head, -, sizeof(head));
}

void add(int u, int v, int cap, int cost) {
    edges[cnt] = edge{u, v, cap, , cost, head[u]};
    head[u] = cnt++;
}

void addedge(int u, int v, int cap, int cost) {
    add(u, v, cap, cost), add(v, u, , -cost);
}

bool spfa(int s, int t, int &flow, LL &cost) {
    //for(int i = 0; i <= n+1; ++i) d[i] = INF; //init()
    memset(d, , sizeof(d));
    memset(inq, false, sizeof(inq));
    d[s] = , inq[s] = true;
    fa[s] = -, cur[s] = INF;
    queue<int> q;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for(int i = head[u]; i != -; i = edges[i].nex) {
            edge& now = edges[i];
            int v = now.v;
            if(now.cap > now.flow && d[v] > d[u] + now.cost) {
                d[v] = d[u] + now.cost;
                fa[v] = i;
                cur[v] = min(cur[u], now.cap - now.flow);
                if(!inq[v]) {q.push(v); inq[v] = true;}
            }
        }
    }
    if(d[t] == INF) return false;
    flow += cur[t];
    cost += 1LL*d[t]*cur[t];
    for(int u = t; u != s; u = edges[fa[u]].u) {
        edges[fa[u]].flow += cur[t];
        edges[fa[u]^].flow -= cur[t];
    }
    return true;
}

int MincostMaxflow(int s, int t, LL &cost) {
    cost = ;
    int flow = ;
    while(spfa(s, t, flow, cost));
    return flow;
}

void run_case() {
    int m, n; cin >> m >> n;
    int s = , t = 1e6;
    // first
    init();
    for(int i = m; i < n+m; ++i)
        for(int j = ; j <= i; ++j) {
            num[i][j] = ++ID;
            cin >> buf[i][j];
            addedge(ID<<, (ID<<)|, , -buf[i][j]);
        }
    for(int i = m; i < n+m-; ++i) {
        for(int j = ; j <= i; ++j) {
            addedge((num[i][j]<<)|, num[i+][j]<<, , );
            addedge((num[i][j]<<)|, num[i+][j+]<<, , );
        }
    }
    for(int i = ; i <= m; ++i) addedge(s, num[m][i]<<, , );
    for(int i = ; i < n+m; ++i) addedge((num[n+m-][i]<<)|, t, , );
    LL cost = ;
    MincostMaxflow(s, t, cost);
    cout << -cost << "\n";
    // second
    init();
    for(int i = m; i < n+m-; ++i)
        for(int j = ; j <= i; ++j) {
            addedge(num[i][j], num[i+][j], , -buf[i+][j]);
            addedge(num[i][j], num[i+][j+], , -buf[i+][j+]);
        }
    for(int i = ; i <= m; ++i) addedge(s, num[m][i], , -buf[m][i]);
    for(int i = ; i < n+m; ++i) addedge(num[n+m-][i], t, INF, );
    cost = , MincostMaxflow(s, t, cost);
    cout << -cost << "\n";
    // third
    init();
    for(int i = m; i < n+m-; ++i)
        for(int j = ; j <= i; ++j) {
            addedge(num[i][j], num[i+][j], INF, -buf[i+][j]);
            addedge(num[i][j], num[i+][j+], INF, -buf[i+][j+]);
        }
    for(int i = ; i <= m; ++i) addedge(s, num[m][i], , -buf[m][i]);
    for(int i = ; i < n+m; ++i) addedge(num[n+m-][i], t, INF, );
    cost = , MincostMaxflow(s, t, cost);
    cout << -cost << "\n";
}

int main() {
    ios::sync_with_stdio(false), cin.tie();
    run_case();
    cout.flush();
    return ;
}