[SWUST1738] 最小路径覆盖问题(最大流,最小路径覆盖)
题目链接:https://www.oj.swust.edu.cn/problem/show/1738
把每一个数拆成两个点,建图跑最大流,结论是满足最小路径覆盖的路径数=总点数-最小割,即总点数-最大流。
打印路径dfs一下,非递归的时候PE了,但是这个OJ会报WA,哎。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef struct Edge { 5 int u, v, w, next; 6 }Edge; 7 8 const int inf = 0x7f7f7f7f; 9 const int maxn = 9090; 10 11 int cnt, dhead[maxn]; 12 int cur[maxn], dd[maxn]; 13 Edge dedge[maxn<<3]; 14 int S, T, N; 15 16 void init() { 17 memset(dhead, -1, sizeof(dhead)); 18 for(int i = 0; i < maxn; i++) dedge[i].next = -1; 19 S = 0; cnt = 0; 20 } 21 22 void adde(int u, int v, int w, int c1=0) { 23 dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 24 dedge[cnt].next = dhead[u]; dhead[u] = cnt++; 25 dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 26 dedge[cnt].next = dhead[v]; dhead[v] = cnt++; 27 } 28 29 bool bfs(int s, int t, int n) { 30 queue<int> q; 31 for(int i = 0; i < n; i++) dd[i] = inf; 32 dd[s] = 0; 33 q.push(s); 34 while(!q.empty()) { 35 int u = q.front(); q.pop(); 36 for(int i = dhead[u]; ~i; i = dedge[i].next) { 37 if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) { 38 dd[dedge[i].v] = dd[u] + 1; 39 if(dedge[i].v == t) return 1; 40 q.push(dedge[i].v); 41 } 42 } 43 } 44 return 0; 45 } 46 47 int dinic(int s, int t, int n) { 48 int st[maxn], top; 49 int u; 50 int flow = 0; 51 while(bfs(s, t, n)) { 52 for(int i = 0; i < n; i++) cur[i] = dhead[i]; 53 u = s; top = 0; 54 while(cur[s] != -1) { 55 if(u == t) { 56 int tp = inf; 57 for(int i = top - 1; i >= 0; i--) { 58 tp = min(tp, dedge[st[i]].w); 59 } 60 flow += tp; 61 for(int i = top - 1; i >= 0; i--) { 62 dedge[st[i]].w -= tp; 63 dedge[st[i] ^ 1].w += tp; 64 if(dedge[st[i]].w == 0) top = i; 65 } 66 u = dedge[st[top]].u; 67 } 68 else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) { 69 st[top++] = cur[u]; 70 u = dedge[cur[u]].v; 71 } 72 else { 73 while(u != s && cur[u] == -1) { 74 u = dedge[st[--top]].u; 75 } 76 cur[u] = dedge[cur[u]].next; 77 } 78 } 79 } 80 return flow; 81 } 82 83 int n, m; 84 vector<int> path; 85 bool vis[maxn]; 86 87 void dfs(int u) { 88 vis[u] = 1; 89 path.push_back(u); 90 for(int j = dhead[u]; ~j; j=dedge[j].next) { 91 if(vis[dedge[j].v]) continue; 92 if(dedge[j].w || !dedge[j].v) continue; 93 dfs(dedge[j].v-n); 94 } 95 } 96 97 int main() { 98 // freopen("in", "r", stdin); 99 int u, v, w; 100 while(~scanf("%d%d",&n,&m)) { 101 init(); 102 S = 0; T = 2 * n + 1; N = T + 1; 103 for(int i = 1; i <= n; i++) adde(S, i, 1); 104 for(int i = 1; i <= n; i++) adde(n+i, T, 1); 105 for(int i = 0; i < m; i++) { 106 scanf("%d%d",&u,&v); 107 adde(u, n+v, 1); 108 } 109 int ret = dinic(S, T, N); 110 memset(vis, 0, sizeof(vis)); 111 for(int i = 1; i <= n; i++) { 112 if(vis[i]) continue; 113 path.clear(); 114 dfs(i); 115 for(int j = 0; j < path.size(); j++) { 116 printf("%d%c", path[j], j==path.size()-1?'\n':' '); 117 } 118 } 119 printf("%d\n", n - ret); 120 } 121 return 0; 122 }