bzoj1061: [Noi2008]志愿者招募

orz

题解戳这里 https://www.byvoid.com/blog/noi-2008-employee/

暂时只能理解下

由于不等式不能作差，加上Yi变成等式作差

作差后每个X最多出现一次，这样就可以考虑两个之间的影响了（通过连边，这里流出去这么多那里就要流进来这么对，对应一正一负）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>

using namespace std;

void setIO(const string& s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
}
template<typename Q> Q read(Q& x) {
    static char c, f;
    for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
    for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if(f) x = -x;
    return x;
}
template<typename Q> Q read() {
    static Q x; read(x); return x;
}
// MCMF

const int N = 1000 + 10, M = 25000 + 10, INF = ~0u >> 1;

struct Edge {
    int to, adv, cost;
    Edge *next;
    Edge(int to = 0, int adv = 0, int cost = 0, Edge *next = 0) : to(to), adv(adv), cost(cost), next(next) {}
}pool[M * 2], *pis = pool, *fir[N];

void AddEdge(int u, int v, int w, int c) {
    fir[u] = new(pis++) Edge(v, w, c, fir[u]);
    fir[v] = new(pis++) Edge(u, 0, -c, fir[v]);
}

#define inv(p) (pool + ( ((p) - pool)^1 ))

namespace MCMF {
    int n, s, t;
    int d[N], q[N], ql, qr, a[N];
    Edge *pre[N];
    bool inq[N];
    
    bool insert(int u, int dis, int aug) {
        if(d[u] > dis) {
            a[u] = aug;
            d[u] = dis;
            if(!inq[u]) {
                q[qr++] = u;
                if(qr == N) qr = 0;
                inq[u] = 1;
            }
            return 1;
        }return 0;
    }
    
    bool SPFA(int& flow, int& cost) {
        for(int u = 1; u <= n; u++) d[u] = INF, inq[u] = 0;
        insert(s, 0, INF);
        while(ql ^ qr) {
            int u = q[ql++]; if(ql == N) ql = 0; inq[u] = 0;
            for(Edge *p = fir[u]; p; p = p->next) if(p->adv) {
                if(insert(p->to, d[u] + p->cost, min(a[u], p->adv))) {
                    pre[p->to] = p;
                }
            }
        }
        if(d[t] == INF) return 0;
        flow += a[t];
        cost += d[t] * a[t];
        for(int u = t; u != s; u = inv(pre[u])->to) {
            pre[u]->adv -= a[t];
            inv(pre[u])->adv += a[t];
        }
        return 1;
    }
    
    int main(int _n ,int _s, int _t) {
        n = _n, s = _s, t = _t;
        int flow = 0, cost = 0;
        while(SPFA(flow, cost));
        return cost;
    }
}

const int maxn = 1000 + 10, maxm = 10000 + 10;
int demand[maxn];

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    
    int n = read<int>(), m = read<int>();
    int s = n + 2, t = s + 1;
    for(int i = 1; i <= n; i++) {
        read(demand[i]);
        AddEdge(i + 1, i, INF, 0);
    }
    for(int i = 1; i <= n + 1; i++) {
        int dlt = demand[i] - demand[i - 1];
        if(dlt >= 0) AddEdge(s, i, dlt, 0);
        else AddEdge(i, t, -dlt, 0);
    }
    int l, r, c;
    for(int i = 1; i <= m; i++) {
        read(l), read(r), read(c);
        AddEdge(l, r + 1, INF, c);
    }
    
    printf("%d
", MCMF::main(t, s, t));
    
    return 0;
}

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