洛谷P2762 太空飞行计划问题(最大权闭合图)

题意

有$m$个实验，$n$中器材，每个实验需要使用一些器材

每个实验有收入，每个器材有花费

最大化收入 - 花费

Sol

最大权闭合图的经典应用

从$S$向每个实验连流量为该实验收入的边

从每个器材箱$T$连流量为花费的边

每个实验向其需要其器材连边权为$INF$的边

答案为：总收入 - 最小割

 

考虑如何统计方案

在最小割中，割去实验表示不选该实验。

那么我们从源点出发，不经过边权为$0$的边，走到的就是需要选的。

这正好是Dinic最后一次增光的deep数组

 

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
char c;
inline int read() {
    if(c == '\r') return 0;
    c = getchar(); int x = 0, f = 1; 
    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 M, N, S, T;
struct Edge {
    int u, v, f, nxt;
}E[MAXN];
int head[MAXN], cur[MAXN], num;
inline void add_edge(int x, int y, int f) {
    E[num] = (Edge){x, y, f, head[x]};
    head[x] = num++;
}
inline void AddEdge(int x, int y, int z) {
    add_edge(x, y, z);
    add_edge(y, x, 0);
}
int sum = 0, deep[MAXN];
bool BFS() {
    queue<int> q; q.push(S);
    memset(deep, 0, sizeof(deep)); deep[S] = 1;
    while(!q.empty()) {
        int p = q.front(); q.pop();
        for(int i = head[p]; i != -1; i = E[i].nxt) {
            int to = E[i].v;
            if(!deep[to] && E[i].f) {
                deep[to] = deep[p] + 1;
                q.push(to);
            }
        }
    }
    return deep[T] > 0;
}
int DFS(int x, int flow) {
    if(x == T) return flow;
    int ansflow = 0;
    for(int &i = cur[x]; i != -1; i = E[i].nxt) {
        int to = E[i].v;
        if(deep[to] == deep[x] + 1 && E[i].f) {
            int nowflow = DFS(to, min(flow, E[i].f));
            E[i].f -= nowflow; E[i ^ 1].f += nowflow;
            ansflow += nowflow; flow -= nowflow;
            if(flow <= 0) break;
        }
    }
    return ansflow;
}
int Dinic() {
    int ans = 0;
    while(BFS()) {
        memcpy(cur, head, sizeof(head));
        ans += DFS(S, INF);
    }
    return ans;
}
int main() {
    memset(head, -1, sizeof(head));
    scanf("%d %d\n", &M, &N); S = 0; T = N + M + 1;
    int ans = 0;
    for(int i = 1; i <= M; i++) {
        c = '+';
        int val = read(), x; ans += val;
        AddEdge(S, i, val);
        while(x = read()) 
            AddEdge(i, x + M, INF);
    }
    for(int i = 1; i <= N; i++) {
        int x; scanf("%d", &x);
        AddEdge(i + M, T, x);
    }
    int cut = Dinic();
    for(int i = head[S]; i != -1; i = E[i].nxt)
        if(E[i].f)
            printf("%d ", E[i].v); puts("");
    for(int x = M + 1; x <= N + M; x++)
        for(int i = head[x]; i != -1; i = E[i].nxt)
            if(E[i].f)
                {printf("%d ", x - M); break;}
    puts("");
    printf("%d", ans - cut);
    return 0;
}