网络流24题1 飞行员配对方案问题
Time Limit:10000MS Memory Limit:512000K
Total Submit:41 Accepted:10
Case Time Limit:1000MS
Description
Input
Output
Sample Input
Sample Output
Hint
不用输出方案
Source
题解:最大流模版
#include <vector>
#include<iostream>
#include <queue>#include <cstdio>
#include <cstring>
using namespace std;
const int INF = 200;
struct edge{
int from,to,cap,flow;
};
vector<edge> edges;
vector<int> g[maxn];
int d[maxn],cur[maxn];
bool vst[maxn];
int n,m,s,t;
void Init()
{
for(int i = 0;i < maxn;++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});
int sz = edges.size();
G[from].push_back(sz-2);
G[to].push_back(sz-1);
}
bool bfs()
{
memset(vst,false,sizeof(vst));
queue<int> q;
q.push(s);
d[s] = 0;
vst[s] = true;
while(!q.empty()){
int u = q.front();
q.pop();
for(int i = 0;i < (int)g[u].size();++i){
Edge& e = edges[g[u][i]];
if(!vst[e.to]&&e.cap > e.flow){
d[e.to] = d[u]+1;
vst[e.to] = true;
q.push(e.to);
}
}
}
return vst[t];
}
int dfs(int u,int a)
{
if(u==t||a==0)
return a;
int f,flow = 0;
for(int& i = cur[u];i < (int)g[u].size();++i){
edge& e = edges[G[u][i]];
if(d[e.to]==d[u]+1&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0){
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(a==0)break;
}
}
return flow;
}
int maxflow()
{
int flow=0;
while(bfs()){
memset(cur,0,sizeof(cur));
flow+=dfs(s,INF);
}
return flow;
}
int main()
{
while(~scanf("%d%d",&n,&m)){
Init();
int x,y;
while(scanf("%d%d",&x,&y),(x!=-1&&y!=-1)){
addedge(x,y,1);
}
s = 0,t = m+1;
for(int i = 1;i <= n;++i)addedge(s,i,1);
for(int i = n+1;i <= m;++i)addedge(i,t,1);
int res = maxflow();
if(res<=0)
puts("No Solution!");
else{
printf("%d\n",res);
int sz = edges.size();
for(int i = 0;i < sz;i += 2){
edge& e = edges[i];
if(e.flow == 1&&e.from!=s&&e.to!=t)
printf("%d %d\n",e.from,e.to);
}
}
}
return 0;
}