学步园

开始最大流判断容量是否大于C，如果不大于，然后去找最小割中的弧，枚举容量为C，TLE。因为最小割中的弧增加的容量一定会引起最大流的增加，于是我们可以再原有最大流的基础上枚举最小割容量中的弧，然后判断是否大于等于C即可。

楼主渣代码。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 110;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, flow;
    bool operator < (const Edge &e) const
    {
		if(e.from != from) return from < e.from;
		else return to < e.to;
	}
    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}
};

struct Dinic
{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    
    void init(int n)
    {
        this->n = n;
        for(int i = 0; i <= n; i++) G[i].clear();
        edges.clear();
    }
    
    void ClearFlow ()
    {
        for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
    }
    
    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()
    {
        memset(vis, 0, sizeof(vis));
        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)
    {
        this->s = s; this->t = t;
        int flow = 0;
        while(BFS())
        {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }
    
    void Reduce()
    {
		for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
	}
	
    void Mincut(vector<int> &cut)
    {
		cut.clear();
		for(int i = 0; i < edges.size(); i++)
		{
			Edge& e = edges[i];
			if(vis[e.from] && !vis[e.to] && e.cap > 0) cut.push_back(i);
		}
	}
};

void readint(int &x)
{
    char c;
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c))
    {
        x = x*10 + c-'0';
        c = getchar();
    }
}

void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

////////////////////

Dinic g;
int n, m, s, t;
int c;

int times;
vector<int> cut;

vector<Edge> ans;

void prosess()
{
	g.init(n+3);
	s = 1, t = n;
	while(m--)
	{
		int x, y, z;
		readint(x), readint(y), readint(z);
		g.AddEdge(x, y, z);
	}
	printf("Case %d: ", ++times);
	int flow = g.Maxflow(s, t);
	if(flow >= c) { printf("possible\n"); return ;}
	else
	{
		g.Mincut(cut);
		ans.clear();
		g.Reduce();
		for(int i = 0; i < cut.size(); i++)
		{
			Edge&e = g.edges[cut[i]];
			e.cap = c;
			g.ClearFlow();
			if(flow + g.Maxflow(s, t) >= c) ans.push_back(e);
			e.cap = 0;
		}
	}
	if(ans.empty()) { printf("not possible\n"); return ;}
	sort(ans.begin(), ans.end());
	int first = 1;
	for(int i = 0; i < ans.size(); i++)
	{
		Edge& e = ans[i];
		if(first) { printf("possible option:(%d,%d)", e.from, e.to); first = 0; }
		else printf(",(%d,%d)", e.from, e.to);
	}
	printf("\n");
}

int main()
{
	times = 0;
	while(~scanf("%d%d%d", &n, &m, &c) && n)
	{
		prosess();
	}
	return 0;
}