线性规划费用流解法(Bzoj1061: [Noi2008]志愿者招募)

Posted 2020-11-09 Cyhlnj

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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;
}