洛谷P3980 [NOI2008]志愿者招募

题解

最小费用最大流

每一天是一条边\((inf-a[i], 0)\)

然后对于一类志愿者,

区间两端连一条\((inf, c[i])\)

\(S\)向第一个点连\((inf, 0)\)

最后一个点向\(T\)连\((inf, 0)\)

然后跑最小费用最大流

这为什么是对的?

我们的目的变成用加的那些边,把最大流量填成\(inf\)

求最小费用

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 2010, inf = 2147483647;
int a[N];

struct node {
    int to, nxt, w, v;
}g[2000000];
int last[N], gl = 1;
void add(int x, int y, int w, int v) {
    g[++gl] = (node) {y, last[x], w, v};
    last[x] = gl;
    g[++gl] = (node) {x, last[y], 0, -v};
    last[y] = gl;
}

int s, t;

int dis[N], from[N], pre[N];
bool vis[N];
queue<int> q;

bool spfa() {
    q.push(s);
    memset(dis, 127, sizeof(dis));
    dis[s] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = last[u]; i; i = g[i].nxt) {
            int v = g[i].to;
            if (dis[v] > dis[u] + g[i].v && g[i].w) {
                dis[v] = dis[u] + g[i].v;
                from[v] = i; pre[v] = u;
                if (!vis[v]) {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
        vis[u] = 0;
    }
    return dis[0] != dis[t];
}


int McMf() {
    int ans = 0;
    while (spfa()) {
        int di = inf;
        for (int i = t; i != s; i = pre[i]) di = min(di, g[from[i]].w);
        ans  += di * dis[t];
        for (int i = t; i != s; i = pre[i])
            g[from[i]].w -= di, g[from[i]^1].w += di;
    }   
    return ans;
}

int main() {
    int n, m;
    read(n), read(m);
    for (int i = 1; i <= n; i++) read(a[i]);
    s = n + 2, t = s + 1;
    for (int i = 1; i <= n; i++)
        add(i, i+1, inf - a[i], 0);
    add(s, 1, inf, 0), add(n + 1, t, inf, 0);
    for (int i = 1; i <= m; i++) {
        int S, T, C;
        read(S), read(T), read(C);
        add(S, T+1, inf, C);
    }
    printf("%d\n", McMf());
    return 0;
}