P3980 [NOI2008]志愿者招募 费用流 （人有多大胆地有多大产

https://www.luogu.org/problemnew/show/P3980

感觉费用流比网络流的图更难想到，要更大胆。
首先由于日期是连续的，所以图中的点是横向排列的。

这道题有点绕道走的意思，由于一类志愿者是可以服务于一段时间，那我们给第$i$天连出去多条边，第一条边是流向$i+1$点的，容量为$inf-a[i]$，费用为0，
若有$i$到 其他点$t$，费用为$w$， 则连一条$(i， t+1， inf，w)$的边

跑一遍费用流，算出结果

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const ll mod = 2147483648;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 1e3 + 9;
int a[maxn];
struct E {
    int v, val, cost;
    int nxt;
} edge[maxn * maxn];
int head[maxn], gtot;
void addedge(int u, int v, int val, int cost) {
    edge[gtot].v = v;
    edge[gtot].val = val;
    edge[gtot].cost = cost;
    edge[gtot].nxt = head[u];
    head[u] = gtot++;

    edge[gtot].v = u;
    edge[gtot].val = 0;
    edge[gtot].cost = -cost;
    edge[gtot].nxt = head[v];
    head[v] = gtot++;
}

int dis[maxn], pre[maxn], vis[maxn], path[maxn];
bool spfa(int s, int t) {
    memset(dis, inf, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    memset(pre, -1, sizeof(pre));

    dis[s] = 0;
    vis[s] = 1;
    queue<int> que;
    que.push(s);

    while (!que.empty()) {
        int u = que.front();
        que.pop();
        vis[u] = 0;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v, val = edge[i].val, cost = edge[i].cost;
            if (val > 0 && dis[v] > dis[u] + cost) {
                dis[v] = dis[u] + cost;
                pre[v] = u;
                path[v] = i;
                if (vis[v] == 0) {
                    vis[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

int mcmf(int s, int t) {
    int flow = 0, cost = 0;
    while (spfa(s, t)) {
        int f = inf;
        for (int i = t; i != s; i = pre[i]) {
            f = min(f, edge[path[i]].val);
        }
        flow += f;
        cost += f * dis[t];
        for (int i = t; i != s; i = pre[i]) {
            edge[path[i]].val -= f;
            edge[path[i] ^ 1].val += f;
        }
    }
    return cost;
}

int main() {
    memset(head, -1, sizeof(head));
    int n, m;
    scanf("%d%d", &n, &m);
    rep(i, 1, n) scanf("%d", &a[i]);
    int s = 0, t = n + 2;
    addedge(s, 1, inf, 0);
    addedge(n + 1, t, inf, 0);
    rep(i, 1, n) addedge(i, i + 1, inf - a[i], 0);

    while (m--) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v + 1, inf, w);
    }
    printf("%d\n", mcmf(s, t));

    return 0;
}