题意：与正常的网络流一样，不过给定的第一行的最后一个数C的意思是能能否在给定的图里求出修改某一条边或者不修改某一条边是的这个图的流变成C，如果没修改就能有C，那么输出possible，通过修改能得到C输出possible+能修改的边集合，否则输出no possible
思路：（自己的是死暴力方法，直接爆了，，想了很多法子都来不起，最后参照白书的思路来起了）可以先求出最大流，然后求出最小割里的弧，依次修改最小割里的弧，看能求出的最大流是否大于C
PS：不优化的话很容易超时，第一个优化是把第一次求得得最大流保存下来，在第一次次最大流得基础上增广；第二个优化是求出当前的流到达C就可以了
AC代码：
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>using namespace std;#define maxn 110#define maxm 22010typedef long long ll;#define INF 0x3f3f3f3f#define cl(x,v); memset(x,v,sizeof(x));struct Edge{int from,to,cap,flow;};bool cmp(const Edge& a,const Edge& b){return a.from < b.from||(a.from==b.from&&a.to<b.to);}struct Dinic{int n,m,s,t;vector<Edge>edges;vector<int>g[maxn];bool vis[maxn];int d[maxn];int cur[maxm];void init(int n){this->n=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});m=edges.size();g[from].push_back(m-2);g[to].push_back(m-1);}bool BFS(){cl(vis,0);queue<int>q;q.push(s);d[s]=0;vis[s]=1;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(!vis[e.to]&&e.cap>e.flow){vis[e.to]=1;d[e.to]=d[x]+1;q.push(e.to);}}}return vis[t];}int DFS(int x,int a){if(x==t||a==0)return a;int flow=0,f;for(int& i=cur[x];i<g[x].size();i++){Edge& e=edges[g[x][i]];if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){e.flow+=f;edges[g[x][i]^1].flow-=f;flow+=f;a-=f;if(a==0)break;}}return flow;}int Maxflow(int s,int t,int need){this->s=s;this->t=t;int flow=0;while(BFS()){cl(cur,0);flow+=DFS(s,INF);if(flow>need)return flow;}return flow;}vector<int> Mincut(){BFS();vector<int>ans;for(int i=0;i<edges.size();i++){Edge& e=edges[i];if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i);}return ans;}void Reduce(){for(int i=0;i<edges.size();i++)edges[i].cap-=edges[i].flow;}void ClearFlow(){for(int i=0;i<edges.size();i++)edges[i].flow=0;}};Dinic solver;int main(){int N,E,C,cas=0;while(scanf("%d%d%d",&N,&E,&C)!=EOF){if(N==0)break;solver.init(N);int a,b,c;while(E–){scanf("%d%d%d",&a,&b,&c);solver.AddEdge(a,b,c);}int flow=solver.Maxflow(1,N,INF);printf("Case %d: ",++cas);if(flow>=C)printf("possible\n");else{vector<int>cut=solver.Mincut();solver.Reduce();vector<Edge>ans;for(int i=0;i<cut.size();i++){Edge& e=solver.edges[cut[i]];int temp=e.cap;e.cap=C;solver.ClearFlow();if(flow+solver.Maxflow(1,N,C-flow)>=C)ans.push_back(e);e.cap=temp;}if(ans.empty())printf("not possible\n");else{sort(ans.begin(),ans.end(),cmp);printf("possible option:(%d,%d)",ans[0].from,ans[0].to);for(int i=1;i<ans.size();i++)printf(",(%d,%d)",ans[i].from,ans[i].to);printf("\n");}}}return 0;}
爱的力量大到可以使人忘记一切，
