Solution -「NOI 2008」「洛谷 P3980」志愿者招募

\(\mathcal{Description}\)

  Link.

  一项持续 \(n\) 天的任务，第 \(i\) 天需要至少 \(a_i\) 人工作。还有 \(m\) 种雇佣方式，第 \(i\) 种每雇佣一人代价为 \(c_i\)，此人会从第 \(s_i\) 天工作到第 \(t_i\) 天（包括边界）。求满足条件的最小代价和。

  \(n\le10^3\)，\(m\le10^4\)。

\(\mathcal{Solution}\)

  非常巧妙地题意转化。

  初始时刻，有 \(+\infty\) 人待业，限制第 \(i\) 天至多 \(+\infty-a_i\) 人待业（那么至少就有 \(a_i\) 人上工）。雇佣一个人相当于用 \(c_i\) 的费用让一个人从 \(s_i\) 天起上工，到 \(t_i+1\) 天恢复待业状态。所以最小费用最大流建图为：

• \(i\rightarrow i+1\)，流量 \(+\infty-a_i\)，费用为 \(0\)；
• \(s_i\rightarrow t_i+1\)（建一个虚点 \(T=n+1\)），流量 \(+\infty\)，费用 \(c_i\)。

  跑最小费用最大流即可。复杂度 \(\mathcal O(\operatorname{Dinic}(n,n+m))\)。

\(\mathcal{Code}\)

/* Clearink */

#include <queue>
#include <cstdio>

typedef long long LL;
typedef std::pair<LL, LL> PLL;

const int MAXN = 1e3, MAXM = 1e4;
const LL INF = 1ll << 60;
int n, m;

inline LL imin ( const LL a, const LL b ) { return a < b ? a : b; }

// 这个 MFCG 其实是错的，SPFA d[i] 应初始化为 INF 而非 -1.
struct MaxFlowCostGraph {
	static const int MAXND = MAXN + 2, MAXEG = MAXN + MAXM + 1;
	int ecnt, head[MAXND + 5], S, T, bound, curh[MAXND + 5];
	LL d[MAXND + 5];
	bool inq[MAXND + 5];
	struct Edge { int to; LL flw; int cst, nxt; } graph[MAXEG * 2 + 5];

	MaxFlowCostGraph (): ecnt ( 1 ) {}

	inline void link ( const int s, const int t, const LL f, const int w ) {
		graph[++ecnt] = { t, f, w, head[s] };
		head[s] = ecnt;
	}

	inline Edge& operator [] ( const int k ) { return graph[k]; }

	inline void operator () ( const int s, const int t, const LL f, const int w ) {
		#ifdef RYBY
			printf ( "%d %d ", s, t );
			if ( f == INF ) printf ( "INF " );
			else printf ( "%lld ", f );
			printf ( "%d\n", w );
		#endif
		link ( s, t, f, w ), link ( t, s, 0, -w );
	}

	inline bool spfa () {
		static std::queue<int> que;
		for ( int i = 0; i <= bound; ++i ) d[i] = -1, inq[i] = false;
		d[S] = 0, inq[S] = true, que.push ( S );
		while ( !que.empty () ) {
			int u = que.front (); que.pop ();
			inq[u] = false;
			for ( int i = head[u], v; i; i = graph[i].nxt ) {
				if ( graph[i].flw
				&& ( !~d[v = graph[i].to] || d[v] > d[u] + graph[i].cst ) ) {
					d[v] = d[u] + graph[i].cst;
					if ( !inq[v] ) que.push ( v ), inq[v] = true;
				}
			}
		}
		return ~d[T];
	}

	inline PLL dfs ( const int u, const LL iflw ) {
		if ( u == T ) return { iflw, 0 };
		inq[u] = true; PLL ret ( 0, 0 );
		for ( int& i = curh[u], v; i; i = graph[i].nxt ) {
			if ( graph[i].flw && !inq[v = graph[i].to]
			&& d[v] == d[u] + graph[i].cst ) {
				PLL oflw ( dfs ( v, imin ( iflw - ret.first, graph[i].flw ) ) );
				graph[i].flw -= oflw.first, graph[i ^ 1].flw += oflw.first;
				ret.first += oflw.first;
				ret.second += graph[i].cst * oflw.first + oflw.second;
				if ( ret.first == iflw ) break;
			}
		}
		if ( !ret.first ) d[u] = -1;
		return inq[u] = false, ret;
	}

	inline PLL calc ( const int s, const int t ) {
		S = s, T = t;
		PLL ret ( 0, 0 );
		while ( spfa () ) {
			for ( int i = 0; i <= bound; ++i ) inq[i] = false, curh[i] = head[i];
			PLL tmp ( dfs ( S, INF ) );
			ret.first += tmp.first, ret.second += tmp.second;
		}
		return ret;
	}
} graph;

int main () {
	scanf ( "%d %d", &n, &m );
	int S = 0, T = graph.bound = n + 1;
	graph ( S, 1, INF, 0 );
	for ( int i = 1, a; i <= n; ++i ) {
		scanf ( "%d", &a );
		graph ( i, i + 1, INF - a, 0 );
	}
	for ( int i = 1, s, t, c; i <= m; ++i ) {
		scanf ( "%d %d %d", &s, &t, &c );
		graph ( s, t + 1, INF, c );
	}
	printf ( "%lld\n", graph.calc ( S, T ).second );
	return 0;
}