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 = 208, maxedge = 40010, INF = 2147483647; inline int read() { int x = 0, f = 1; char c = getchar(); while (c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while (c >= '0' && c <= '9') {x = x*10 + c-'0'; c = getchar();} return x * f; } int n, m, dis[maxnode], Depth[maxnode], head[maxnode], cnt = 1, 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, 0, sizeof(Depth)); Depth[s] = 1, 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 > 0 && Depth[ed[i].v] == 0) { Depth[ed[i].v] = Depth[u] + 1; 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] + 1 && ed[i].w > 0) { delta = Dinic(ed[i].v, min(cap, ed[i].w)); if(delta > 0) { ed[i].w -= delta; ed[i^1].w += delta; return delta; } } } return 0; } int main() { n = read(), m = read(); s = 0, t = n+m+1; for(int i=1; i<=n; i++) addedge(s, i, 1), addedge(i, s, 0); for(int i=n+1; i<=m; i++) addedge(i, t, 1), addedge(t, i, 0); static int x, y; while (1) { x = read(), y = read(); addedge(x, y, 1), addedge(y, x, 0); if(x == -1 && y == -1) break; } while (BFS()) Ans += Dinic(s, INF); if(Ans) printf("%d\n", Ans); else printf("No Solution!\n"); if(Ans) { for(int i=2; i<=cnt; i+=2) { if(ed[i].v != s && ed[i].v != t && ed[i^1].v != s && ed[i^1].v != t) if(ed[i^1].w != 0)printf("%d %d\n", ed[i].u, ed[i].v); } } return 0; }