uva 11248 Frequency Hopping

题目大意：给定一个有向网络，每条边均有一个容量。

问是否存在一个从点1到点N。流量为C的流。假设不存在，能否够恰好改动一条弧的容量，使得存在这种流。

解题思路：先依照题目给出的边建好图，然后跑一发最大流，得到原始最大流C1，假设C1==C<script id="MathJax-Element-126" type="math/tex">C1 == C</script>或者C==0<script id="MathJax-Element-127" type="math/tex">C == 0</script>时。能够直接输出possible。假设不存在这种流。那么開始找割边，将这些割边的容量添加C，再求最大流。假设能够，那么要输出全部的方案。改动全部割边后，仍没有符合条件的流，输出 not possible。

优化：1）第一次跑最大流的每条边的流量情况，能够留着，接下来能够在它的基础上增广。2）每次求最大流，不用求完，当流符合条件即>=C时，就可以返回成功。

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

typedef long long ll;
const int N = 505;
const int M = 20005;
const ll INF = 1e18;
int n, e, s, t ;
ll c;
struct Edge{
    int from, to;
    ll cap, flow; 
};

int cmp(Edge a, Edge b) {
    if (a.from != b.from) return a.from < b.from;
    else return a.to < b.to;
}

struct Dinic{
    vector<Edge> edges, tempOE;
    vector<int> G[M];
    vector<int> MCut;
    Edge rec[M];
    int vis[N], d[N];
    int cur[M];
    ll ans;

    void init() {
        ans = 0;
        for (int i = 0; i < e; i++) G[i].clear();
        edges.clear();
    }

    void addEdge(int from, int to, ll cap) {
        edges.push_back((Edge){from, to, cap, 0});
        edges.push_back((Edge){to, from, 0, 0});
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    } 

    int 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 = 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];
    }

    ll DFS(int u, ll a) {
        if (u == t || a == 0) return a;
        ll flow = 0, f; 
        for (int &i = cur[u]; i < G[u].size(); i++) {
            Edge &e = edges[G[u][i]];
            if (d[u] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;    
                edges[G[u][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    bool MF() {
        while (BFS()) {
            memset(cur, 0, sizeof(cur));
            ans += DFS(s, INF);
            if (ans >= c) return true; //不用完整地求最大流，当流量大于等于c时，该改动就是可行的
        }   
        return false;
    }

    void getMincut() { //求最小割
        MCut.clear();
        for (int i = 0; i < edges.size(); i += 2) {
            if (vis[edges[i].from] && !vis[edges[i].to]) {
                MCut.push_back(i);  
            }   
        }   
    }

    int solve() {
        int cnt = 0;
        tempOE.clear();
        getMincut();
        ll f = ans; //求完初始最大流，记录当前总流量
        int es = edges.size();
        for (int i = 0; i < es; i++) tempOE.push_back(edges[i]); //记录初始最大流每条边的情况
        int cs = MCut.size();
        for (int i = 0; i < cs; i++) {
            edges[MCut[i]].cap = edges[MCut[i]].flow + c;
            if (MF()) {
                rec[cnt++] = edges[MCut[i]]; //若将该边的容量加上c之后。总流量大于等于c。则改动该边是可行方案。记录该边
            }
            ans = f; //每改动一条边。且检验完之后，恢复初始状态
            edges.clear();
            for (int i = 0; i < es; i++) edges.push_back(tempOE[i]);
        }
        return cnt;
    }
}din;

void input() {
    s = 1, t = n;
    int u, v;
    ll cap;
    for (int i = 0; i < e; i++) {
        scanf("%d %d %lld", &u, &v, &cap);      
        din.addEdge(u, v, cap);
    }
}

void solve() {
    if (din.MF() || c == 0) {
        printf("possible\n");   
        return;
    }
    int cnt = din.solve();
    if (!cnt) {
        printf("not possible\n");
        return;
    }
    sort(din.rec, din.rec + cnt, cmp); //记得排序
    printf("possible option:(%d,%d)", din.rec[0].from, din.rec[0].to);  
    for (int i = 1; i < cnt; i++) {
        printf(",(%d,%d)", din.rec[i].from, din.rec[i].to); 
    }
    puts("");
}

int main() {
    int Case = 1;
    while (scanf("%d %d %lld", &n, &e, &c) == 3) {
        if (!n && !e && !c) break;  
        din.init();
        printf("Case %d: ", Case++);
        input();
        solve();
    }
    return 0;
}