补一下昨天的博客 J - Sabotage 网络流
题目链接:https://vjudge.net/contest/299467#problem/J
这个题目是一个裸的最小割问题,就不多赘述了。
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#include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <queue> #include <vector> #include <algorithm>
#define inf 0x3f3f3f3f
using namespace std; typedef long long ll; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; struct edge { int u, v, c, f; edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {} }; vector<edge>e; vector<int>G[maxn]; int level[maxn];//BFS分层,表示每个点的层数
int iter[maxn];//当前弧优化
int m; void init(int n) { for (int i = 0; i <= n; i++)G[i].clear(); e.clear(); } void add(int u, int v, int c) { e.push_back(edge(u, v, c, 0)); e.push_back(edge(v, u, c, 0)); m = e.size(); G[u].push_back(m - 2); G[v].push_back(m - 1); } void BFS(int s)//预处理出level数组 //直接BFS到每个点
{ memset(level, -1, sizeof(level)); queue<int>q; level[s] = 0; q.push(s); while (!q.empty()) { int u = q.front(); q.pop(); for (int v = 0; v < G[u].size(); v++) { edge& now = e[G[u][v]]; if (now.c > now.f && level[now.v] < 0) { level[now.v] = level[u] + 1; q.push(now.v); } } } } ll dfs(int u, int t, int f)//DFS寻找增广路
{ if (u == t)return f;//已经到达源点,返回流量f
for (int &v = iter[u]; v < G[u].size(); v++) //这里用iter数组表示每个点目前的弧,这是为了防止在一次寻找增广路的时候,对一些边多次遍历 //在每次找增广路的时候,数组要清空
{ edge &now = e[G[u][v]]; if (now.c - now.f > 0 && level[u] < level[now.v]) //now.c - now.f > 0表示这条路还未满 //level[u] < level[now.v]表示这条路是最短路,一定到达下一层,这就是Dinic算法的思想
{ ll d = dfs(now.v, t, min(f, now.c - now.f)); if (d > 0) { now.f += d;//正向边流量加d
e[G[u][v] ^ 1].f -= d; //反向边减d,此处在存储边的时候两条反向边可以通过^操作直接找到
return d; } } } return 0; } ll Maxflow(int s, int t) { ll flow = 0; for (;;) { BFS(s); if (level[t] < 0)return flow;//残余网络中到达不了t,增广路不存在
memset(iter, 0, sizeof(iter));//清空当前弧数组
int f;//记录增广路的可增加的流量
while ((f = dfs(s, t, INF)) > 0) { flow += f; } } return flow; } vector<pair<int, int>>vec; void solve(int u) { printf("u=%d\n", u); for(int i=0;i<G[u].size();i++) { edge now = e[G[u][i]]; edge non = e[G[u][i] ^ 1]; if(now.c==now.f) { printf("now.u=%d now.v=%d\n", now.u, now.v); vec.push_back(make_pair(now.u, now.v)); solve(now.v); } } } int cx[maxn], cy[maxn]; int main() { int n, m; while (scanf("%d%d",&n,&m)!=EOF&&(m+n)) { init(n); int s = 1, t = 2; for(int i=1;i<=m;i++) { int x, y, z; scanf("%d%d%d", &x, &y, &z); add(x, y, z); cx[i] = x; cy[i] = y; } ll ans = Maxflow(s, t); for(int i=1;i<=m;i++) { if((level[cx[i]]!=-1&&level[cy[i]]==-1)||(level[cx[i]]==-1&&level[cy[i]]!=-1)) { printf("%d %d\n", cx[i], cy[i]); } } printf("\n"); } return 0; }

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