题意：
给你一个有向图，问你从1到n的最大流是多少？如果流量大于等于C那么直接输出一个串，否则输出只扩充一条边的流量就可以达到1->n大于等于C的所有边，如果扩充不了就
输出另一个串。
Sample Input
4 4 5
1 2 5
1 3 5
2 4 5
3 4 5
4 4 5
1 2 1
1 3 5
2 4 5
3 4 1
4 4 5
1 2 1
1 3 1
2 4 1
3 4 1
0 0 0
Output for Sample Input
Case 1: possible
Case 2: possible option:(1,2),(3,4)
Case 3: not possible
思路：
很容易想到的一点就是扩展后有作用的点肯定是割边，那么我们可以先跑一遍最大流把割边找出来，然后枚举割边，扩充割边流量，看最大流是否大于等于C，但是这样会TLE，有两个比较有用的优化，就是每次都在残余网络上改流量，然后加上残余网络之前跑出来的流量，还有一个优化就是跑最大流的时候，如果当前流量大于等于C了就已经满足了，没必要再跑了。
#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N_node 100 + 5
#define N_edge 20000 + 10
#define INF 2005000000
using namespace std;
typedef struct
{
int from ,to ,next;
long long cost;
}STAR;
typedef struct
{
int x ,t;
}DEP;
typedef struct
{
int a ,b;
}EDGE;
STAR E[N_edge] ,mkE[N_edge];
EDGE edge[N_edge] ,Ans_Edge[N_edge];
DEP xin ,tou;
int list[N_node] ,list2[N_node] ,mklist[N_node] ,tot;
int deep[N_node];
void add(int a ,int b ,long long c)
{
E[++tot].from = a;
E[tot].to = b;
E[tot].cost = c;
E[tot].next = list[a];
list[a] = tot;
E[++tot].from = b;
E[tot].to = a;
E[tot].cost = 0;
E[tot].next = list[b];
list[b] = tot;
}
bool camp(EDGE a ,EDGE b)
{
return a.a < b.a || a.a == b.a && a.b < b.b;
}
long long minn(long long a ,long long b)
{
return a < b ? a : b;
}
bool BFS_Deep(int s ,int t ,int n)
{
memset(deep ,255 ,sizeof(deep));
xin.x = s ,xin.t = 0;
deep[s] = 0;
queue<DEP>q;
q.push(xin);
while(!q.empty())
{
tou = q.front();
q.pop();
for(int k = list[tou.x] ;k ;k = E[k].next)
{
xin.x = E[k].to;
xin.t = tou.t + 1;
if(deep[xin.x] != -1 || !E[k].cost)
continue;
deep[xin.x] = xin.t;
q.push(xin);
}
}
for(int i = 0 ;i <= n ;i ++)
list2[i] = list[i];
return deep[t] != -1;
}
long long DFS_Flow(int s ,int t ,long long flow ,long long C)
{
if(s == t) return flow;
long long nowflow = 0;
for(int k = list2[s] ;k ;k = E[k].next)
{
list2[s] = k;
int to = E[k].to;
long long c = E[k].cost;
if(deep[to] != deep[s] + 1 || !c) continue;
long long tmp = DFS_Flow(to ,t ,minn(c ,flow - nowflow) ,C);
nowflow += tmp;
E[k].cost -= tmp;
E[k^1].cost += tmp;
if(nowflow == flow) break;
}
if(!nowflow) deep[s] = 0;
return nowflow;
}
long long DINIC(int s ,int t ,int n ,long long C)
{
long long Ans = 0;
while(BFS_Deep(s ,t ,n) && Ans < C)
{
Ans += DFS_Flow(s ,t ,INF ,C);
}
return Ans;
}
int main ()
{
int n ,m ,C ,cas = 1;
int a ,b ,c ,i ,j;
while(~scanf("%d %d %d" ,&n ,&m ,&C) && n + m + C)
{
memset(list ,0 ,sizeof(list)) ,tot = 1;
for(i = 1 ;i <= m ;i ++)
{
scanf("%d %d %d" ,&a ,&b ,&c);
add(a ,b ,(long long)c);
}
long long Ans = DINIC(1 ,n ,n ,C);
printf("Case %d: " ,cas ++);
if(Ans >= C)
{
printf("possible
");
continue;
}
int nowid = 0;
for(i = 2 ;i <= tot ;i += 2)
{
mkE[i] = E[i] ,mkE[i+1] = E[i+1];
if(!E[i].cost)
{
edge[++nowid].a = E[i].from;
edge[nowid].b = E[i].to;
}
}
for(i = 1 ;i <= n ;i ++)
mklist[i] = list[i];
int Ans_Id = 0;
int mktot = tot;
for(i = 1 ;i <= nowid ;i ++)
{
add(edge[i].a ,edge[i].b ,C);
long long tmp = DINIC(1 ,n ,n ,C);
if(tmp + Ans >= C) Ans_Edge[++Ans_Id] = edge[i];
tot = mktot;
for(j = 2 ;j <= tot ;j ++) E[j] = mkE[j];
for(j = 1 ;j <= n ;j ++) list[j] = mklist[j];
}
if(!Ans_Id)
{
puts("not possible");
continue;
}
sort(Ans_Edge + 1 ,Ans_Edge + Ans_Id + 1 ,camp);
printf("possible option:");
for(i = 1 ;i <= Ans_Id ;i ++)
if(i != 1) printf(",(%d,%d)" ,Ans_Edge[i].a ,Ans_Edge[i].b);
else printf("(%d,%d)" ,Ans_Edge[i].a ,Ans_Edge[i].b);
puts("");
}
return 0;
}