[NOI2008] 志愿者招募【最大流最小费用 经典问题】

洛谷连接

COGS题目连接

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  很好的一道题，我们要去处理区间问题，这里用到了一种很特殊，但是很好的算法。

  因为天数是连续的，所以把第i天与第i+1天连接起来，费用为0、流量INF - a[i]。

  然后再放一个第N+1天连接到T为INF的流，费用为0；

  之后，我们要放对应的{ s[i], t[i], c[i] }是不是就可以往里面填充了，是不是可以直接s[i] -> t[i] + 1建立费用为c[i]的边即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(x, y) make_pair(x, y)
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e3 + 7, S = 0;
int N, M, T;
ll h[maxN], dist[maxN];
struct Eddge
{
    int nex, u, v;
    ll flow, cost;
    Eddge(int a=-1, int b=0, int c=0, ll d=0, ll f=0):nex(a), u(b), v(c), flow(d), cost(f) {}
};
vector<Eddge> G[maxN];
inline void _add(int u, int v, ll flow, ll cost)
{
    G[u].push_back(Eddge((int)G[v].size(), u, v, flow, cost));
    G[v].push_back(Eddge((int)G[u].size() - 1, v, u, 0, -cost));
}
struct node
{
    int id; ll val;
    node(int a=0, ll b=0):id(a), val(b) {}
    friend bool operator < (node e1, node e2) { return e1.val > e2.val; }
};
priority_queue<node> Q;
int preP[maxN], preE[maxN];
inline ll MaxFlow_MinCost(ll Flow)
{
    ll ans = 0;
    for(int i=0; i<=T; i++) h[i] = 0;
    while(Flow)
    {
        for(int i=0; i<=T; i++) dist[i] = INF;
        dist[S] = 0;
        while(!Q.empty()) Q.pop();
        Q.push(node(S, 0));
        while(!Q.empty())
        {
            node now = Q.top(); Q.pop();
            int u = now.id;
            if(dist[u] < now.val) continue;
            int len = (int)G[u].size();
            for(int i=0, v; i<len; i++)
            {
                v = G[u][i].v; ll f = G[u][i].flow, c = G[u][i].cost;
                if(f && dist[v] > dist[u] + c - h[v] + h[u])
                {
                    dist[v] = dist[u] + c - h[v] + h[u];
                    preP[v] = u; preE[v] = i;
                    Q.push(node(v, dist[v]));
                }
            }
        }
        if(dist[T] == INF) break;
        for(int i=0; i<=T; i++) h[i] += dist[i];
        ll Capa = Flow;
        for(int u=T; u != S; u = preP[u]) Capa = min(Capa, G[preP[u]][preE[u]].flow);
        Flow -= Capa;
        ans += Capa * h[T];
        for(int u = T; u != S; u = preP[u])
        {
            Eddge &E = G[preP[u]][preE[u]];
            E.flow -= Capa;
            G[E.v][E.nex].flow += Capa;
        }
    }
    return ans;
}
inline void init()
{
    T = N + 2;
    for(int i=0; i<=T; i++) G[i].clear();
}
int main()
{
//    freopen("employee.in", "r", stdin);
//    freopen("employee.out", "w", stdout);
    scanf("%d%d", &N, &M);
    init();
    _add(S, 1, INF, 0);
    for(int i=1, val; i<=N; i++)
    {
        scanf("%d", &val);
        _add(i, i + 1, INF - val, 0);   //免费部分
    }
    for(int i=1, si, ti, ci; i<=M; i++)
    {
        scanf("%d%d%d", &si, &ti, &ci);
        _add(si, ti + 1, INF, ci);
    }
    _add(N + 1, T, INF, 0);
    printf("%lld\n", MaxFlow_MinCost(INF));
    return 0;
}