hdoj 1853 Cyclic Tour 【最小费用最大流 or KM算法】【构图后可以判断图中是否存在哈密顿环】
2015-08-24 17:11
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Cyclic TourTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1948 Accepted Submission(s): 982 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him? Input There are several test cases in the input. You should process to the end of file (EOF). The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000). Output Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1. Sample Input 6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4 6 5 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 Sample Output 42 -1 Hint In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42. |
思路:把每个点i拆分成左点i和右点i+N
1,超级源点连左点,容量为1,费用为0
2,所有右点连超级汇点,容量为1,费用为0
3,每条单向边—— 起点左点 连 终点右点 容量为1,费用为边权。
最后跑一下最小费用最大流流。 若满流说明存在哈密顿环,否则不存在。
AC代码:
#include <cstdio> #include <cstring> #include <queue> #include <stack> #include <vector> #include <algorithm> #define MAXN 200+10 #define MAXM 30000+10 #define INF 0x3f3f3f3f using namespace std; struct Edge { int from, to, cap, flow, cost, next; }; Edge edge[MAXM]; int head[MAXN], edgenum; int pre[MAXN], dist[MAXN]; bool vis[MAXN]; int N, M; int source, sink;//超级源点 超级汇点 void init() { edgenum = 0; memset(head, -1, sizeof(head)); } void addEdge(int u, int v, int w, int c) { Edge E1 = {u, v, w, 0, c, head[u]}; edge[edgenum] = E1; head[u] = edgenum++; Edge E2 = {v, u, 0, 0, -c, head[v]}; edge[edgenum] = E2; head[v] = edgenum++; } void getMap() { int a, b, c; source = 0, sink = 2*N+1; for(int i = 1; i <= N; i++) addEdge(source, i, 1, 0),//超级源点 连左点 addEdge(i + N, sink, 1, 0);//右点 连超级汇点 while(M--) { scanf("%d%d%d", &a, &b, &c); addEdge(a, b+N, 1, c);//左点 连 右点 } } bool SPFA(int s, int t) { queue<int> Q; memset(dist, INF, sizeof(dist)); memset(vis, false, sizeof(vis)); memset(pre, -1, sizeof(pre)); dist[s] = 0; vis[s] = true; Q.push(s); while(!Q.empty()) { int u = Q.front(); Q.pop(); vis[u] = false; for(int i = head[u]; i != -1; i = edge[i].next) { Edge E = edge[i]; if(dist[E.to] > dist[u] + E.cost && E.cap > E.flow) { dist[E.to] = dist[u] + E.cost; pre[E.to] = i; if(!vis[E.to]) { vis[E.to] = true; Q.push(E.to); } } } } return pre[t] != -1; } void MCMF(int s, int t, int &cost, int &flow) { cost = flow = 0; while(SPFA(s, t)) { int Min = INF; for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) { Edge E = edge[i]; Min = min(Min, E.cap-E.flow); } for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) { edge[i].flow += Min; edge[i^1].flow -= Min; cost += edge[i].cost * Min; } flow += Min; } } int main() { while(scanf("%d%d", &N, &M) != EOF) { init(); getMap(); int cost, flow; MCMF(source, sink, cost, flow); if(flow == N)//满流 printf("%d\n", cost); else printf("-1\n"); } return 0; }
KM算法:重刷
注意KM算法结束后,要判断是否完美匹配。
#include <cstdio> #include <cstring> #include <algorithm> #define INF 0x3f3f3f3f using namespace std; int lx[110], ly[110]; int Map[110][110]; bool visx[110], visy[110]; int slack[110]; int match[110]; int N, M; void getMap() { for(int i = 1; i <= N; i++) { for(int j = 1; j <= N; j++) Map[i][j] = -INF; } int a, b, c; while(M--) { scanf("%d%d%d", &a, &b, &c); if(-c > Map[a][b]) Map[a][b] = -c; } } int DFS(int x) { visx[x] = true; for(int y = 1; y <= N; y++) { if(visy[y]) continue; int t = lx[x] + ly[y] - Map[x][y]; if(t == 0) { visy[y] = true; if(match[y] == -1 || DFS(match[y])) { match[y] = x; return 1; } } else if(slack[y] > t) slack[y] = t; } return 0; } void KM() { memset(match, -1, sizeof(match)); memset(ly, 0, sizeof(ly)); for(int x = 1; x <= N; x++) { lx[x] = -INF; for(int y = 1; y <= N; y++) lx[x] = max(lx[x], Map[x][y]); } for(int x = 1; x <= N; x++) { for(int i = 1; i <= N; i++) slack[i] = INF; while(1) { memset(visx, false, sizeof(visx)); memset(visy, false, sizeof(visy)); if(DFS(x)) break; int d = INF; for(int i = 1; i <= N; i++) { if(!visy[i] && slack[i] < d) d = slack[i]; } for(int i = 1; i <= N; i++) { if(visx[i]) lx[i] -= d; } for(int i = 1; i <= N; i++) { if(visy[i]) ly[i] += d; else slack[i] -= d; } } } //判断是否存在完美匹配 int ans = 0; bool flag = true; for(int i = 1; i <= N; i++) { if(match[i] == -1 || Map[match[i]][i] == -INF) { flag = false; break; } ans += Map[match[i]][i]; } if(flag) printf("%d\n", -ans); else printf("-1\n"); } int main() { while(scanf("%d%d", &N, &M) != EOF) { getMap(); KM(); } return 0; }
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