UVa 11248 - Frequency Hopping - 最大流 - 最小割

题目描述：lrj厚白书第5章第6节第一道例题

题目分析：建图求最大流，如果最大流大于c，那么直接输出possible；如果最大流小于c，那么就要改变属于最小割边的值。

下面是代码：（水过，有时间在优化吧）

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const int maxn_v = 110;
const int maxn_e = 22000;
const int inf = 2100000000 ;
struct edge
{
int from,to,cap,flow;
edge() {}
edge(int f,int t,int cp,int fl)
{
from = f;
to = t;
cap = cp;
flow = fl;
}
};
vector<edge> E;
vector<int> G[maxn_v];
vector<int> cut;
int n,m,c,s,t;
int d[maxn_v];
bool vis[maxn_v];
int cur[maxn_e];
int cmp(edge a,edge b)
{
if(a.from == b.from) return a.to < b.to;
else return a.from < b.from;
}

void add_e(int from,int to,int cap)
{
E.push_back(edge(from,to,cap,0));
E.push_back(edge(to,from,0,0));
int e = E.size();
G[from].push_back(e-2);
G[to].push_back(e-1);
}
void read_graph()
{
s = 1;
t = n;
for(int i = 0; i <= n; i++) G[i].clear();
E.clear();
for(int i = 0; i < m; i++)
{
int from,to,cap;
scanf("%d%d%d",&from,&to,&cap);
add_e(from,to,cap);
}

}
bool bfs()
{
memset(vis,0,sizeof(vis));
queue<int> q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while(!q.empty())
{
int u = q.front(); q.pop();
for(int i = 0; i < G[u].size(); i++)
{
edge& e = E[G[u][i]];
int v = e.to;
if(!vis[v] && e.cap > e.flow)
{
vis[v] = 1;
d[v] = d[u] + 1;
q.push(v);
}
}
}
return vis[t];
}
int dfs(int u,int a)
{
if(u == t || a == 0) return a;
int flow = 0, f;
for(int& i = cur[u]; i < G[u].size(); i++)
{
edge& e = E[G[u][i]];
if(d[u] + 1 == d[e.to] && (f = dfs(e.to,min(a,e.cap-e.flow))) > 0)
{
e.flow += f;
E[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}

int max_flow()
{
int flow = 0;
while(bfs())
{
memset(cur,0,sizeof(cur));
flow += dfs(s,inf);
}
return flow;
}
void min_cut()
{
bfs();
cut.clear();
for(int i = 0; i < E.size(); i++)
{
edge& e = E[i];
if(vis[e.from] && !vis[e.to] && e.cap > 0) cut.push_back(i);
}
}
void reduce()
{
for(int i = 0; i < E.size(); i++)
E[i].cap -= E[i].flow;
}
void clear()
{
for(int i = 0; i < E.size(); i++) E[i].flow = 0;
}
void print()
{
for(int i = 0; i < E.size(); i++)
printf("%d -> %d %d\n",E[i].from,E[i].to,E[i].cap);
}
int main()
{
int cas = 0;
while(scanf("%d%d%d",&n,&m,&c) && n)
{
read_graph();
//print();
printf("Case %d: ",++cas);
int maxflow = max_flow();
if(maxflow >= c || c == 0) printf("possible\n");
else
{
vector<edge> ans;
min_cut();
reduce();
for(int i = 0; i < cut.size(); i++)
{
clear();
edge& e = E[cut[i]];
e.cap += c;
if(maxflow + max_flow() >= c) ans.push_back(E[cut[i]]);
e.cap -= c;
}
if(ans.empty()) printf("not possible\n");
else
{
sort(ans.begin(),ans.end(),cmp);
for(int i = 0; i < ans.size(); i++)
{
if(i == 0) printf("possible option:(%d,%d)",ans[i].from,ans[i].to);
else printf(",(%d,%d)",ans[i].from,ans[i].to);
}
printf("\n");
}
}
}
return 0;
}