题意：给定上一个网络，每个边有一个容量，问你能不能从 1 到 n，使得流量为 c，如果不能，那么是不是可以修改一条边，使得达到。
析：背景就是一个网络流，如果原图能跑出来，那么就不用了，就肯定能达到，如果不能，那么修改的边肯定是最小割里的边，那么就枚举这最小割里的边，这样可能会超时，所以就优化，其中一个优化就是每次不是从0开始跑，而是在第一次的基础再走，把两次的流量加起来如果超过c了，那么就能，再就是可以每次不用跑出最大流，如果流量超过c了，就可以结束了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int c;
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];
void init(int n){
this-> n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
}
void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m - 2);
G[to].pb(m - 1);
}
bool bfs(){
ms(vis, 0); vis[s] = 1;
d[s] = 0;
queue<int> q;
q.push(s);
while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[u] + 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].sz; ++i){
Edge &e = edges[G[x][i]];
if(d[e.to] == d[x] + 1 && (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;
if(flow >= c) return flow;
}
}
return flow;
}
int maxflow(int s, int t){
this->s = s; this->t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
vector<int> mincut;
void getmincut(){
mincut.cl;
for(int i = 0; i < edges.sz; i += 2){
Edge &e = edges[i];
if(vis[e.from] && !vis[e.to] && e.cap > 0) mincut.push_back(i);
}
}
void clearflow(){
for(int i = 0; i < edges.sz; ++i) edges[i].flow = 0;
}
void solve(int s, int t){
int flow = maxflow(s, t);
if(flow >= c){ puts("possible"); return ; }
c -= flow;
getmincut();
for(int i = 0; i < edges.sz; ++i)
edges[i].cap -= edges[i].flow;
vector<P> ans;
for(int i = 0; i < mincut.sz; ++i){
Edge &e = edges[mincut[i]];
e.cap = c;
clearflow();
if(maxflow(s, t) >= c) ans.push_back(P(e.from, e.to));
e.cap = 0;
}
if(ans.empty()) printf("not possible
");
else{
sort(ans.begin(), ans.end());
printf("possible option:(%d,%d)", ans[0].fi, ans[0].se);
for(int i = 1; i < ans.sz; ++i)
printf(",(%d,%d)", ans[i].fi, ans[i].se);
printf("
");
}
}
};
Dinic dinic;
int main(){
int kase = 0;
while(scanf("%d %d %d", &n, &m, &c) == 3 && n){
dinic.init(n + 5);
for(int i = 0; i < m; ++i){
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
dinic.addEdge(u, v, c);
}
printf("Case %d: ", ++kase);
dinic.solve(1, n);
}
return 0;
}