uva 11248

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=565&page=show_problem&problem=2205

题意：见大白书368页

要注意的地方在大白书上面也都说了，不过做的时候被求割点卡了。。之前没做过- -。长了见识。

求割的那部分是这样的：因为最后已经走不过去了，所以如果某条边起点能走到，终点走不到，那说明它就是在最小割的集合里面,否则就还有增广路。

/* * ThinkingLion.cpp * *  Created on: 2014年11月25日 *      Author: dell */#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfs(a) printf("%s\n",a)#define pfI(a) printf("%I64d\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define clr1(a) memset(a,0,sizeof(a))#define clr2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) {return b ? gcd(b, a % b) : a;}template<class T> T lcm(T a, T b) {return a / gcd(a, b) * b;}template<class T> inline T Min(T a, T b) {return a < b ? a : b;}template<class T> inline T Max(T a, T b) {return a > b ? a : b;}using namespace std;int n, e, c;#define N 110struct Edge {int from, to, cap, flow;};struct Dinic {int n, m, s, t;vector<Edge> edges;vector<Edge> edtmp;vector<int> G[N];vector<int> cut;Edge able[N * N * 2];int p[N];bool vis[N];int d[N];int cur[N];int k;int flow;void Init(int n, int m, int s, int t) {this->n = n;this->m = m;this->s = s;this->t = t;for (int i = 0; i <= N; i++)G[i].clear();edges.clear();flow = 0;}void addEdge(int u, int v, int cap) {edges.push_back((Edge ) { u, v, cap, 0 });edges.push_back((Edge ) { v, u, 0, 0 });m = edges.size();G[u].push_back(m - 2);G[v].push_back(m - 1);}bool BFS() {memset(vis, 0, sizeof vis);queue<int> q;q.push(s);vis[s] = 1;d[s] = 0;while (!q.empty()) {int u = q.front();q.pop();for (int i = 0; i < (int) G[u].size(); i++) {Edge &e = edges[G[u][i]];if (e.cap > e.flow && !vis[e.to]) {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 < (int) 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() {while (BFS()) {memset(cur, 0, sizeof cur);flow += DFS(s, INF);if (flow >= c)return true;}return false;}void getcut() {BFS();cut.clear();for (int i = 0; i < (int) edges.size(); i += 2) {if (vis[edges[i].from] && !vis[edges[i].to] && edges[i].cap > 0) {cut.push_back(i);}}}void getTable() {getcut();edtmp.clear();for (int i = 0; i < (int) edges.size(); i++)edtmp.push_back(edges[i]);int lastflow = flow;k = 0;for (int i = 0; i < (int) cut.size(); i++) {edges[cut[i]].cap = edges[cut[i]].flow + c;if (maxFlow())able[k++] = edges[cut[i]];flow = lastflow;edges.clear();for (int j = 0; j < (int) edtmp.size(); j++)edges.push_back(edtmp[j]);}}};Dinic solver;bool cmp(Edge a, Edge b) {if (a.from < b.from)return true;else if (a.from == b.from && a.to < b.to)return true;return false;}int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifint kas = 1;while (scanf("%d%d%d", &n, &e, &c) != EOF && n + e + c) {solver.Init(n, e, 1, n);int u, v, cap;for (int i = 0; i < e; i++) {scanf("%d%d%d", &u, &v, &cap);solver.addEdge(u, v, cap);}printf("Case %d: ", kas++);if (solver.maxFlow() || c == 0)printf("possible\n");else {solver.getTable();if (solver.k <= 0)printf("not possible\n");else {sort(solver.able, solver.able + solver.k, cmp);printf("possible option:(%d,%d)", solver.able[0].from,solver.able[0].to);for (int i = 1; i < solver.k; i++) {printf(",(%d,%d)", solver.able[i].from, solver.able[i].to);}printf("\n");}}}return 0;}