洛谷 P1344 [USACO4.4]追查坏牛奶Pollutant Control
https://www.luogu.org/problemnew/show/P1344
题目要求最小割和边数最小的最小割的边数
前者容易，重点是后者
方法：https://blog.csdn.net/chenzhenyu123456/article/details/48103229
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<vector> 5 #include<queue> 6 #include<iostream> 7 using namespace std; 8 #define fi first 9 #define se second 10 #define mp make_pair 11 #define pb push_back 12 typedef long long ll; 13 typedef unsigned long long ull; 14 typedef pair<int,int> pii; 15 namespace F 16 { 17 18 struct E 19 { 20 int to,nxt,from,cap,flow; 21 }e[400100]; 22 int f1[60100],ne=1; 23 int S,T,n; 24 int d[60100]; 25 bool bfs() 26 { 27 int k,u; 28 memset(d,0,sizeof(int)*(n+1)); 29 queue<int> q; 30 q.push(S);d[S]=1; 31 while(!q.empty()) 32 { 33 u=q.front();q.pop(); 34 for(k=f1[u];k;k=e[k].nxt) 35 if(!d[e[k].to]&&e[k].cap>e[k].flow) 36 { 37 d[e[k].to]=d[u]+1; 38 //if(e[k].to==T) return 1; 39 q.push(e[k].to); 40 } 41 } 42 //return 0; 43 return d[T]; 44 } 45 int cur[60100]; 46 int dfs(int u,int x) 47 { 48 if(u==T||x==0) return x; 49 int flow=0,f; 50 for(int &k=cur[u];k;k=e[k].nxt) 51 if(e[k].cap>e[k].flow&&d[e[k].to]==d[u]+1) 52 { 53 f=dfs(e[k].to,min(x-flow,e[k].cap-e[k].flow)); 54 e[k].flow+=f;e[k^1].flow-=f;flow+=f; 55 if(flow==x) return flow; 56 } 57 return flow; 58 } 59 int solve() 60 { 61 int flow=0; 62 while(bfs()) 63 { 64 memcpy(cur,f1,sizeof(int)*(n+1)); 65 flow+=dfs(S,0x3f3f3f3f); 66 } 67 return flow; 68 } 69 void me(int a,int b,int c) 70 { 71 e[++ne].to=b;e[ne].nxt=f1[a];f1[a]=ne; 72 e[ne].from=a;e[ne].cap=c;e[ne].flow=0; 73 e[++ne].to=a;e[ne].nxt=f1[b];f1[b]=ne; 74 e[ne].from=b;e[ne].cap=0;e[ne].flow=0; 75 } 76 77 78 } 79 80 int n,m; 81 int main() 82 { 83 int i,a,b,c; 84 scanf("%d%d",&n,&m);F::n=n;F::S=1;F::T=n; 85 for(i=1;i<=m;i++) 86 { 87 scanf("%d%d%d",&a,&b,&c); 88 F::me(a,b,c); 89 } 90 int an=F::solve(); 91 for(i=2;i<=F::ne;i+=2) 92 { 93 using F::e; 94 if(e[i].cap==e[i].flow) 95 { 96 e[i].cap=1;e[i^1].cap=0; 97 } 98 else 99 { 100 e[i].cap=0x3f3f3f3f;e[i^1].cap=0; 101 } 102 e[i].flow=0;e[i^1].flow=0; 103 } 104 printf("%d %d",an,F::solve()); 105 return 0; 106 }
