线性规划费用流解法(Bzoj1061: [Noi2008]志愿者招募)

题面

传送门

Sol

线性规划费用流解法用与求解未知数为非负数的问题

这道题可以列出一堆形如
\(x[i]+x[j]+x[k]+...>=a[p]\)
的不等式
我们强行给每个式子减去一个东西，使他变成这样
\(x[i]+x[j]+x[k]+...-y[p]==a[p]\)

然后相邻两个式子差分一下
把每个式子看成一个点
那么这样后，在这个题中所有的未知数只会出现在一个方程中
等式左边符号是正的向符号为负的方程连边，费用为代价，如果是补的未知数\(y\)，那么费用为零
右边的数是正的连\(s\)，否则连\(t\)
费用流出解

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(1005);
const int inf(1e9);

int n, m, first[maxn], cnt, ans, s, t;
int dis[maxn], pre1[maxn], pre2[maxn], vis[maxn];
queue <int> q;

struct Edge{
	int to, next, f, w;
} edge[maxn * 25];

IL void Add(RG int u, RG int v, RG int f, RG int w){
	edge[cnt] = (Edge){v, first[u], f, w}, first[u] = cnt++;
	edge[cnt] = (Edge){u, first[v], 0, -w}, first[v] = cnt++;
}

IL int Aug(){
	for(RG int i = s; i <= t; ++i) dis[i] = inf;
	q.push(s), dis[s] = 0, vis[s] = 1;
	while(!q.empty()){
		RG int u = q.front(); q.pop();
		for(RG int e = first[u]; e != -1; e = edge[e].next){
			RG int v = edge[e].to;
			if(edge[e].f && dis[v] > dis[u] + edge[e].w){
				dis[v] = dis[u] + edge[e].w;
				pre1[v] = e, pre2[v] = u;
				if(!vis[v]) q.push(v), vis[v] = 1;
			}
		}
		vis[u] = 0;
	}
	if(dis[t] == inf) return 0;
	RG int ret = inf;
	for(RG int p = t; p; p = pre2[p]) ret = min(ret, edge[pre1[p]].f);
	ans += ret * dis[t];
	for(RG int p = t; p; p = pre2[p])
		edge[pre1[p]].f -= ret, edge[pre1[p] ^ 1].f += ret;
	return 1;
}

int main(){
	n = Input(), m = Input();
	s = 0, t = n + 2;
	for(RG int i = s; i <= t; ++i) first[i] = -1;
	RG int last = 0;
	for(RG int i = 1; i <= n; ++i){
		RG int v = Input();
		if(v - last > 0) Add(s, i, v - last, 0);
		else if(v - last < 0) Add(i, t, last - v, 0);
		last = v, Add(i + 1, i, inf, 0);
	}
	Add(n + 1, t, last, 0);
	for(RG int i = 1; i <= m; ++i){
		RG int l = Input(), r = Input(), c = Input();
		Add(l, r + 1, inf, c);
	}
	while(Aug());
	printf("%d\n", ans);
	return 0;
}