14更多学校的第二个问题
网络流量 分别以行,列作为结点建图
i行表示的结点到j列表示的结点的流量便是(i, j)的值
跑遍最大流 若满流了便是有解 推断是否unique 就是在残余网络中dfs。走能够添加流量的边,找到环即不唯一
dfs的时候一定要回溯!!
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#include <cstdio> #include <ctime> #include <cstdlib> #include <cstring> #include <queue> #include <string> #include <set> #include <stack> #include <map> #include <cmath> #include <vector> #include <iostream> #include <algorithm> #include <bitset> #include <fstream> using namespace std; //LOOP #define FF(i, a, b) for(int i = (a); i < (b); ++i) #define FE(i, a, b) for(int i = (a); i <= (b); ++i) #define FED(i, b, a) for(int i = (b); i>= (a); --i) #define REP(i, N) for(int i = 0; i < (N); ++i) #define CLR(A,value) memset(A,value,sizeof(A)) #define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++) //OTHER #define SZ(V) (int)V.size() #define PB push_back #define MP make_pair #define all(x) (x).begin(),(x).end() //INPUT #define RI(n) scanf("%d", &n) #define RII(n, m) scanf("%d%d", &n, &m) #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k) #define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p) #define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q) #define RS(s) scanf("%s", s) //OUTPUT #define WI(n) printf("%d\n", n) #define WS(n) printf("%s\n", n) //debug //#define online_judge #ifndef online_judge #define dt(a) << (#a) << "=" << a << " " #define debugI(a) cout dt(a) << endl #define debugII(a, b) cout dt(a) dt(b) << endl #define debugIII(a, b, c) cout dt(a) dt(b) dt(c) << endl #define debugIV(a, b, c, d) cout dt(a) dt(b) dt(c) dt(d) << endl #define debugV(a, b, c, d, e) cout dt(a) dt(b) dt(c) dt(d) dt(e) << endl #else #define debugI(v) #define debugII(a, b) #define debugIII(a, b, c) #define debugIV(a, b, c, d) #endif #define sqr(x) (x) * (x) typedef long long LL; typedef unsigned long long ULL; typedef vector <int> VI; const double eps = 1e-9; const int MOD = 1000000007; const double PI = acos(-1.0); const int INF = 0x3f3f3f3f; const int maxn = 900; bool use[maxn]; struct Edge{ int from, to, cap, flow; }; int MAX; struct Dinic{ int n, m ,s, t; vector<Edge> edges; VI G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn] ; void init(int nn) { this->n = nn; REP(i, n) G[i].clear(); edges.clear(); } void addEdge(int from, int to, int cap) { edges.PB((Edge){from, to, cap, 0}); edges.PB((Edge){to, from, 0, 0}); m = edges.size(); G[from].PB(m - 2); G[to].PB(m - 1); } bool bfs() { CLR(vis, 0); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = 1; while (!Q.empty()) { int x = Q.front(); Q.pop(); REP(i, G[x].size()) { Edge& e = edges[G[x][i]]; if (!vis[e.to] && e.cap > e.flow) { vis[e.to] = 1; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int dfs(int x, int a) { if (x == t || a == 0) return a; int flow = 0, f; for (int& i = cur[x]; i < G[x].size(); i++) { Edge& e = edges[G[x][i]]; if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[G[x][i] ^ 1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } int maxflow(int s, int t) { this-> s = s, this-> t = t; int flow = 0; while (bfs()) { CLR(cur, 0); flow += dfs(s, INF); } return flow; } bool visit(int u, int fa) { if (u == 0 || u == MAX) return false; use[u] = 1; REP(i, G[u].size()) { Edge& e = edges[G[u][i]]; // debugII(e.to, use[e.to]); if (e.to != fa && e.cap > e.flow) if (use[e.to] || visit(e.to, u)) return true; } use[u] = 0; return false; } }di; int main() { int n, m, k; while (~RIII(n, m, k)) { int x, sum1 = 0, sum2 = 0; MAX = n + m + 1; di.init(n + m + 2); FE(i, 1, n) { RI(x); sum1 += x; di.addEdge(0, i, x); } FE(i, n + 1, n + m) { RI(x); sum2 += x; di.addEdge(i, n + m + 1, x); } FE(i, 1, n) FE(j, n + 1, n + m) di.addEdge(i, j, k); if (sum2 != sum1) { puts("Impossible"); continue; } int ans = di.maxflow(0, n + m + 1); if (ans < sum1) { puts("Impossible"); continue; } FE(i, 1, n) { CLR(use, 0); if (di.visit(i, -1)) { puts("Not Unique"); goto end; } } puts("Unique"); for (int i = 2 * n + 2 * m; i < di.edges.size(); i += 2) { printf("%d", di.edges[i].flow); if (i + 2 >= di.edges.size()) printf("\n"); else printf(" "); } end:; } return 0; }
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时间: 2024-10-25 04:36:49