题意:裸的最大流,什么是最大流,参考别的博客
运用复杂度最高的EK算法 O(M*N),模板来自紫书
#include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #include <algorithm> #include <iostream> #include <sstream> #include <string> #include <queue> #define INF 0x3f3f3f3f #define oo 0x13131313 const int maxn=200+5; using namespace std; struct Edge { int from,to,cap,flow; Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){} }; struct EdmondsKarp{ int n,m; vector<Edge> edges; //边集数组,边数*2 vector<int> G[maxn]; //邻接表 int a[maxn]; //起点到i的可改进量 int p[maxn]; //最短路树上p的弧编号 void init(int n){ for(int i = 0;i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap){ edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); //反向弧 m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } int Maxflow(int s,int t) { int flow=0; for(;;){ memset(a,0,sizeof(a)); queue<int> Q; // while(!Q.empty()) Q.pop(); Q.push(s); a[s]=INF; while(!Q.empty()) { int x=Q.front();Q.pop(); for(int i=0;i<G[x].size();i++) { Edge &e=edges[G[x][i]]; if(!a[e.to]&&(e.cap>e.flow)) { p[e.to]=G[x][i]; a[e.to]=min(a[x],e.cap-e.flow); Q.push(e.to); } } if(a[t]) break; } if(!a[t]) break; for(int u=t;u!=s;u=edges[p[u]].from){ edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; } flow+=a[t]; } return flow; } };完整代码:
#include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #include <algorithm> #include <iostream> #include <sstream> #include <string> #include <queue> #define INF 0x3f3f3f3f #define oo 0x13131313 const int maxn=200+5; using namespace std; struct Edge { int from,to,cap,flow; Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){} }; struct EdmondsKarp{ int n,m; vector<Edge> edges; //边集数组,边数*2 vector<int> G[maxn]; //邻接表 int a[maxn]; //起点到i的可改进量 int p[maxn]; //最短路树上p的弧编号 void init(int n){ for(int i = 0;i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap){ edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); //反向弧 m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } int Maxflow(int s,int t) { int flow=0; for(;;){ memset(a,0,sizeof(a)); queue<int> Q; // while(!Q.empty()) Q.pop(); Q.push(s); a[s]=INF; while(!Q.empty()) { int x=Q.front();Q.pop(); for(int i=0;i<G[x].size();i++) { Edge &e=edges[G[x][i]]; if(!a[e.to]&&(e.cap>e.flow)) { p[e.to]=G[x][i]; a[e.to]=min(a[x],e.cap-e.flow); Q.push(e.to); } } if(a[t]) break; } if(!a[t]) break; for(int u=t;u!=s;u=edges[p[u]].from){ edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; } flow+=a[t]; } return flow; } }; EdmondsKarp A; int N; void File() { freopen("a.in","r",stdin); freopen("a.out","w",stdout); } void input() { int a,b,c; for(int i=1;i<=N;i++) { scanf("%d%d%d",&a,&b,&c); A.AddEdge(a,b,c); } } void solve() { int ans=0; ans=A.Maxflow(1,A.n); printf("%d\n",ans); } int main() { // File(); A.init(maxn); while(scanf("%d%d",&N,&A.n)!=EOF) { input(); solve(); A.init(maxn); //注意用maxn 去初始化! } return 0; }