Luogu P4014 「 网络流 24 题 」分配问题

解题思路

还是建立超级源点和超级汇点，又因为题目给出规定一个人只能修一个工件，所以建图的时候还要讲容量都设为$1$。

人的编号是$1\rightarrow n$，工件的编号是$n+1\rightarrow 2\times n$。人和超级源点连边，工件和超级汇点连边，跑一个最小费用最大流和最大费用最大流。

附上代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int maxnode = , maxedge = , INF = ;
inline int read() {
    int x = , f = ; char c = getchar();
    while (c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while (c >= '' && c <= '') {x = x* + c-''; c = getchar();}
    return x * f;
}
int n, m, dis[maxnode], Depth[maxnode], head[maxnode], cnt = , s, t, Ans;
struct edge {
    int nxt, u, v, w;
}ed[maxedge];
inline void addedge(int x, int y, int cap) {
    ed[++cnt].nxt = head[x];
    ed[cnt].v = y, ed[cnt].w = cap, ed[cnt].u = x;
    head[x] = cnt;
}
inline bool BFS() {
    queue<int> Q;
    memset(Depth, , sizeof(Depth));
    Depth[s] = , Q.push(s);
    int u;
    while(!Q.empty()) {
        u = Q.front();
        Q.pop();
        for(int i=head[u]; i; i=ed[i].nxt) {
            if(ed[i].w >  && Depth[ed[i].v] == ) {
                Depth[ed[i].v] = Depth[u] + ;
                Q.push(ed[i].v);
                if(ed[i].v == t) return true;
            }
        }
    }
    return false;
}
inline int Dinic(int u, int cap) {
    if(u == t) return cap;
    int delta;
    for(int i=head[u]; i; i=ed[i].nxt) {
        if(Depth[ed[i].v] == Depth[u] +  && ed[i].w > ) {
            delta = Dinic(ed[i].v, min(cap, ed[i].w));
            if(delta > ) {
                ed[i].w -= delta;
                ed[i^].w += delta;
                return delta;
            }
        }
    }
    return ;
}
int main() {
    n = read(), m = read();
    s = , t = n+m+;
    for(int i=; i<=n; i++) addedge(s, i, ), addedge(i, s, );
    for(int i=n+; i<=m; i++) addedge(i, t, ), addedge(t, i, );
    static int x, y;
    while () {
        x = read(), y = read();
        addedge(x, y, ), addedge(y, x, );
        if(x == - && y == -) break;
    }
    while (BFS()) Ans += Dinic(s, INF);
    if(Ans) printf("%d\n", Ans);
    else printf("No Solution!\n");
    if(Ans) {
        for(int i=; i<=cnt; i+=) {
            if(ed[i].v != s && ed[i].v != t && ed[i^].v != s && ed[i^].v != t)
                if(ed[i^].w != )printf("%d %d\n", ed[i].u, ed[i].v);
        }
    }
    return ;
}