P4014 分配问题

\(\color{#0066ff}{题目描述}\)

有 \(n\) 件工作要分配给 \(n\) 个人做。第 \(i\) 个人做第 \(j\) 件工作产生的效益为 \(c_{ij}\) 。试设计一个将 \(n\) 件工作分配给 \(n\) 个人做的分配方案，使产生的总效益最大。

\(\color{#0066ff}{输入格式}\)

文件的第 \(1\) 行有 \(1\) 个正整数 \(n\)，表示有 \(n\) 件工作要分配给 \(n\) 个人做。

接下来的 \(n\) 行中，每行有 \(n\) 个整数 \(c_{ij}\) ??，表示第 \(i\) 个人做第 \(j\) 件工作产生的效益为 \(c_{ij}\) 。

\(\color{#0066ff}{输出格式}\)

两行分别输出最小总效益和最大总效益。

\(\color{#0066ff}{输入样例}\)

5
2 2 2 1 2
2 3 1 2 4
2 0 1 1 1
2 3 4 3 3
3 2 1 2 1

\(\color{#0066ff}{输出样例}\)

5
14

\(\color{#0066ff}{数据范围与提示}\)

none

\(\color{#0066ff}{题解}\)

裸的费用流。。。

好像可以二分图最大带权匹配。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
#define _ 0
#define LL long long
inline LL in() {
    LL x = 0, f = 1; char ch;
    while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
    while(isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
    return x * f;
}
int n, m, s, t;
const int maxn = 105050;
const int inf = 0x7fffffff;
struct node {
    int to, dis, can;
    node *nxt, *rev;
    node(int to = 0, int dis = 0, int can = 0, node *nxt = NULL):to(to), dis(dis), can(can), nxt(nxt) {}
    void *operator new (size_t) {
        static node *S = NULL, *T = NULL;
        return (S == T)&&(T = (S = new node[1024]) + 1024) , S++;
    }
};
typedef node* nod;
nod head[maxn], road[maxn];
int dis[maxn], change[maxn];
bool vis[maxn];
int c[120][120];
std::queue<int> q;
inline void add(int from, int to, int dis, int can) {
    nod o = new node(to, dis, can, head[from]);
    head[from] = o;
}
inline void link(int from, int to, int dis, int can) {
    add(from, to, dis, can);
    add(to, from, -dis, 0);
    head[from]->rev = head[to];
    head[to]->rev = head[from];
}
inline bool spfa1() {
    for(int i = s; i <= t; i++) dis[i] = inf, change[i] = inf;
    dis[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int tp = q.front(); q.pop();
        vis[tp] = false;
        for(nod i = head[tp]; i; i = i->nxt) {
            if(dis[i->to] > dis[tp] + i->dis && i->can > 0) {
                dis[i->to] = dis[tp] + i->dis;
                road[i->to] = i;
                change[i->to] = std::min(change[tp], i->can);
                if(!vis[i->to]) vis[i->to] = true, q.push(i->to);
            }
        }
    }
    return change[t] != inf;
}
inline bool spfa2() {
    for(int i = s; i <= t; i++) dis[i] = -inf, change[i] = inf;
    dis[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int tp = q.front(); q.pop();
        vis[tp] = false;
        for(nod i = head[tp]; i; i = i->nxt) {
            if(dis[i->to] < dis[tp] + i->dis && i->can > 0) {
                dis[i->to] = dis[tp] + i->dis;
                road[i->to] = i;
                change[i->to] = std::min(change[tp], i->can);
                if(!vis[i->to]) vis[i->to] = true, q.push(i->to);
            }
        }
    }
    return change[t] != inf;
}
inline void mcmf1()
{
    int cost = 0;
    while(spfa1()) {
        cost += change[t] * dis[t];
        for(int i = t; i != s; i = road[i]->rev->to) {
            road[i]->can -= change[t];
            road[i]->rev->can += change[t];
        }
    }
    printf("%d\n", cost);
}
inline void mcmf2()
{
    int cost = 0;
    while(spfa2()) {
        cost += change[t] * dis[t];
        for(int i = t; i != s; i = road[i]->rev->to) {
            road[i]->can -= change[t];
            road[i]->rev->can += change[t];
        }
    }
    printf("%d", cost);
}
int main() {
    n = in();
    s = 0, t = (n << 1) + 1;
    for(int i = 1; i <= n; i++) {
        link(s, i, 0 ,1);
        link(i + n, t, 0, 1);
        for(int j = 1; j <= n; j++)
            link(i, j + n, c[i][j] = in(), 1);
    }
    mcmf1();
    for(int i = s; i <= t; i++) head[i] = NULL;
    for(int i = 1; i <= n; i++) {
        link(s, i, 0 ,1);
        link(i + n, t, 0, 1);
        for(int j = 1; j <= n; j++)
            link(i, j + n, c[i][j], 1);
    }
    mcmf2();
    return 0;
}

原文地址：https://www.cnblogs.com/olinr/p/10123650.html