HDU 3488--Tour（KM or 费用流）
因为每个点只能经过一次 所以考虑拆点
这题有坑，有重边。。
KM算法
把一个点拆成入点和出点 入点在X部，出点在Y步。
如果u，v之间有路径，就在X部的u点连接Y部的v点
求完美匹配。
当完美匹配的时候，每个点都有一个入度和一个出度，可知成环。
因为完美匹配求得是最大匹配
记得把每条边权值取相反数
#include <iostream> #include <cstring> #include <cstdio> using namespace std; const int MAXN = 205; const int INF = 0x3f3f3f3f; int G[MAXN][MAXN]; int vx[MAXN], vy[MAXN]; bool visx[MAXN], visy[MAXN]; int match[MAXN]; int slack[MAXN]; int N, M; bool dfs(int x) { visx[x] = true; for (int y = 0; y < N; ++y) { if (visy[y]) continue; int gap = vx[x] + vy[y] - G[x][y]; if (gap == 0) { visy[y] = true; if (match[y] == -1 || dfs( match[y] )) { match[y] = x; return true; } } else { slack[y] = min(slack[y], gap); } } return false; } int KM() { memset(match, -1, sizeof match); memset(vy, 0, sizeof vy); for (int i = 0; i < N; ++i) { vx[i] = G[i][0]; for (int j = 1; j < N; ++j) { vx[i] = max(vx[i], G[i][j]); } } for (int i = 0; i < N; ++i) { fill(slack, slack + N, INF); while (1) { memset(visx, false, sizeof visx); memset(visy, false, sizeof visy); if (dfs(i)) break; int d = INF; for (int j = 0; j < N; ++j) if (!visy[j]) d = min(d, slack[j]); for (int j = 0; j < N; ++j) { if (visx[j]) vx[j] -= d; if (visy[j]) vy[j] += d; else slack[j] -= d; } } } int res = 0; for (int i = 0; i < N; ++i) res += G[ match[i] ][i]; return res; } int Scan() { int res = 0, flag = 0; char ch; if((ch = getchar()) == '-') flag = 1; else if(ch >= '0' && ch <= '9') res = ch - '0'; while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0'); return flag ? -res : res; } int main() { int T = Scan(); while (T--) { N = Scan(), M = Scan(); for (int i = 0; i <= N; ++i) { for (int j = 0; j <= N; ++j) { G[i][j] = -INF; } } int u, v, w; while (M--) { u = Scan()-1, v = Scan()-1, w = Scan(); G[u][v] = max(G[u][v], -w); } printf("%d\n", -KM()); } return 0; }
费用流
就是把上面的完美匹配用网络流来求。。
和完美匹配一样
如果u，v之间有路径，就u的入点连v的出点，然后所有入点连起点，所有出点连汇点，求最大流最小花费即可。
费用流比起KM慢了几倍。。
#include <algorithm> #include <iostream> #include <cstring> #include <string> #include <vector> #include <bitset> #include <cstdio> #include <queue> #include <stack> #include <cmath> #include <list> #include <map> #include <set> #define pk(x) printf("%d\n", x) using namespace std; #define PI acos(-1.0) #define EPS 1E-6 #define clr(x,c) memset(x,c,sizeof(x)) //#pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; #define CLR(x, v, n) memset(x, v, sizeof(x[0])*n) const int N = 410; const int M = 1000000; const int INF = 0x3f3f3f3f; struct Edge { int to, next, cap, flow, cost; void init(int _to, int _cap, int _cost, int _next) { to = _to; cap = _cap; cost = _cost; next = _next; flow = 0; } } edge[M]; int head[N], cntE; int pre[N], dis[N]; bool vis[N]; int src, sink, tot; void dn(int &x, int y) { if(x>y) x=y; } void init() { cntE = 0; memset(head, -1, sizeof head); } void addedge(int u, int v, int cap, int cost) { edge[cntE].init(v, cap, cost, head[u]); head[u] = cntE++; edge[cntE].init(u, 0, -cost, head[v]); head[v] = cntE++; } bool spfa() { queue<int> q; fill(dis, dis+tot, INF); CLR(vis, false, tot); CLR(pre, -1, tot); dis[src] = 0; vis[src] = true; q.push(src); while (q.size()) { int u = q.front(); q.pop(); vis[u] = false; for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].to; if (edge[i].cap > edge[i].flow && dis[u]+edge[i].cost < dis[v]) { dis[v] = dis[u]+edge[i].cost; pre[v] = i; if (!vis[v]) { vis[v] = true; q.push(v); } } } } if (pre[sink] == -1) return false; return true; } int MCMF() { int flow = 0; int cost = 0; while (spfa()) { int f = INF; for (int i = pre[sink]; ~i; i = pre[edge[i^1].to]) { dn(f, edge[i].cap - edge[i].flow); } for (int i = pre[sink]; ~i; i = pre[edge[i^1].to]) { edge[i].flow += f; edge[i^1].flow -= f; cost += edge[i].cost * f; } flow += f; } //return flow; return cost; } int Scan() { int res = 0, flag = 0; char ch; if((ch = getchar()) == '-') flag = 1; else if(ch >= '0' && ch <= '9') res = ch - '0'; while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0'); return flag ? -res : res; } int n, m; int G[205][205]; int main() { int T = Scan(); while (T--) { clr(head, -1); cntE = 0; n = Scan(), m = Scan(); int u, v, w; src = 0, sink = n*2+1, tot = n*2+2;; for (int i = 1; i <= n; ++i) { addedge(src, i, 1, 0); addedge(i+n, sink, 1, 0); } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) G[i][j] = INF; while (m--) { u = Scan(), v = Scan(), w = Scan(); G[u][v] = min(G[u][v], w); } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) { if (G[i][j] != INF) { addedge(i, j+n, 1, G[i][j]); } } printf("%d\n", MCMF()); } return 0; }
上面的费用流太慢辣，又找了个快点的。嘿嘿XD
#include <algorithm> #include <iostream> #include <cstring> #include <string> #include <vector> #include <bitset> #include <cstdio> #include <queue> #include <stack> #include <cmath> #include <list> #include <map> #include <set> #define pk(x) printf("%d\n", x) using namespace std; #define PI acos(-1.0) #define EPS 1E-6 #define clr(x,c) memset(x,c,sizeof(x)) //#pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; const int MAXV = 410; const int INF = 1<<30; struct Edge { int to, cap, cost, rev; }; vector<Edge> G[MAXV]; int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV]; queue<int> que; void addedge(int from, int to, int cap, int cost) { G[from].push_back((Edge){to, cap, cost, G[to].size()}); G[to].push_back((Edge){from, 0, -cost, G[from].size()-1}); } int min_cost_max_flow(int s, int t) { //, int f) { int res = 0; int f = 0; while (1) { //f > 0) { for (int i = 0; i <= t; ++i) dist[i] = INF, in[i] = 0; dist[s] = 0; while (!que.empty()) que.pop(); in[s] = 1; que.push(s); while (!que.empty()) { int u = que.front(); que.pop(); in[u] = 0; for (int i = 0; i < G[u].size(); ++i) { Edge &e = G[u][i]; if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) { dist[e.to] = dist[u] + e.cost; prv[e.to] = u; pre[e.to] = i; if (in[e.to] == 0) { in[e.to] = 1; que.push(e.to); } } } } if (dist[t] == INF) break; //return -1; int d = INF; // d = f; for (int v = t; v != s; v = prv[v]) { d = min(d, G[prv[v]][pre[v]].cap); } f += d; res += d * dist[t]; for (int v = t; v != s; v = prv[v]) { Edge &e = G[prv[v]][pre[v]]; e.cap -= d; G[v][e.rev].cap += d; } } return res; } int Scan() { int res = 0, flag = 0; char ch; if((ch = getchar()) == '-') flag = 1; else if(ch >= '0' && ch <= '9') res = ch - '0'; while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0'); return flag ? -res : res; } int n, m; int mp[205][205]; int main() { int T = Scan(); while (T--) { n = Scan(), m = Scan(); int u, v, w; int src = 0, sink = n*2+1; for (int i = 0; i <= sink; ++i) G[i].clear(); for (int i = 1; i <= n; ++i) { addedge(src, i, 1, 0); addedge(i+n, sink, 1, 0); } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) mp[i][j] = INF; while (m--) { u = Scan(), v = Scan(), w = Scan(); mp[u][v] = min(mp[u][v], w); } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) { if (mp[i][j] != INF) { addedge(i, j+n, 1, mp[i][j]); } } printf("%d\n", min_cost_max_flow(src, sink)); } return 0; }
