文章地址:http://henuly.top/?p=784
Description:
The regime of a small but wealthy dictatorship has been abruptly overthrown by an unexpected rebellion. Because of the enormous disturbances this is causing in world economy, an imperialist military
super power has decided to invade the country and reinstall the old regime.
For this operation to be successful, communication between the capital and the largest city must
be completely cut. This is a difficult task, since all cities in the country are connected by a computer
network using the Internet Protocol, which allows messages to take any path through the network.
Because of this, the network must be completely split in two parts, with the capital in one part and
the largest city in the other, and with no connections between the parts.
There are large differences in the costs of sabotaging different connections, since some are much
more easy to get to than others.
Write a program that, given a network specification and the costs of sabotaging each connection,
determines which connections to cut in order to separate the capital and the largest city to the lowest
possible cost.
Input:
Input file contains several sets of input. The description of each set is given below.
The first line of each set has two integers, separated by a space: First one the number of cities, n in
the network, which is at most 50. The second one is the total number of connections, m, at most 500.
The following m lines specify the connections. Each line has three parts separated by spaces: The
first two are the cities tied together by that connection (numbers in the range 1 − n). Then follows the
cost of cutting the connection (an integer in the range 1 to 40000000). Each pair of cites can appear
at most once in this list.
Input is terminated by a case where values of n and m are zero. This case should not be processed.
For every input set the capital is city number 1, and the largest city is number 2.
Output:
For each set of input you should produce several lines of output. The description of output for each set
of input is given below:
The output for each set should be the pairs of cities (i.e. numbers) between which the connection
should be cut (in any order), each pair on one line with the numbers separated by a space. If there is
more than one solution, any one of them will do.
Print a blank line after the output for each set of input.
Sample Input:
5 8
1 4 30
1 3 70
5 3 20
4 3 5
4 5 15
5 2 10
3 2 25
2 4 50
5 8
1 4 30
1 3 70
5 3 20
4 3 5
4 5 15
5 2 10
3 2 25
2 4 50
0 0
Sample Output:
4 1
3 4
3 5
3 2
4 1
3 4
3 5
3 2
题目链接
题目要求最小割使源点无法到达汇点，建图Dinic跑最大流，之后再跑一边DFS，深搜能搜到的点即为可以和源点相连的点，无法搜索到的点即为和汇点相连的点，枚举建边时建立的边，若边的两个顶点中有一个顶点可以由源点到达而另一个不能则此边为最小割中的一条割边。
AC代码:
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout<<#x<<"="<<x<<endl;
#define ArrayDebug(x,i) cout<<#x<<"["<<i<<"]="<<x[i]<<endl;
#define print(x) out(x);putchar('\n')
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
char c;
int sgn;
if (c = getchar(), c == EOF) {
return false;
}
while (c != '-' && (c < '0' || c > '9')) {
c = getchar();
}
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0');
}
ret *= sgn;
return true;
}
template <class T>
inline void out(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) {
out(x / 10);
}
putchar(x % 10 + '0');
}
int N, M;
int Book[maxn];
int Depth[maxn];
int Adj[maxn][maxn];
bool Bfs(int Start, int End) {
std::queue<int> Que;
memset(Depth, -1, sizeof(Depth));
Depth[Start] = 0;
Que.push(Start);
while (!Que.empty()) {
int Vertex = Que.front();
Que.pop();
for (int i = 1; i <= N; ++i) {
if (Depth[i] == -1 && Adj[Vertex][i]) {
Depth[i] = Depth[Vertex] + 1;
Que.push(i);
}
}
}
return Depth[End] > 0;
}
int Dfs(int Vertex, int End, int NowFlow) {
Book[Vertex] = 1;
if (Vertex == End) {
return NowFlow;
}
int FindFlow = 0;
for (int i = 1; i <= N; ++i) {
if (Adj[Vertex][i] && Depth[i] == Depth[Vertex] + 1) {
FindFlow = Dfs(i, End, std::min(NowFlow, Adj[Vertex][i]));
if (FindFlow) {
Adj[Vertex][i] -= FindFlow;
Adj[i][Vertex] += FindFlow;
return FindFlow;
}
}
}
if (!FindFlow) {
Depth[Vertex] = -2;
}
return false;
}
int Dinic(int Start, int End) {
int MaxFlow = 0;
while (Bfs(Start, End)) {
MaxFlow += Dfs(Start, End, INF);
}
return MaxFlow;
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
while (read(N) && read(M) && N + M) {
mem(Adj, 0);
vector<int> U(M), V(M);
for (int i = 0, C; i < M; ++i) {
read(U[i]); read(V[i]); read(C);
Adj[U[i]][V[i]] = Adj[V[i]][U[i]] = C;
}
Dinic(1, 2);
mem(Book, 0);
Dfs(1, 2, INF);
for (int i = 0; i < M; ++i) {
if ((!Book[U[i]] && Book[V[i]]) || (Book[U[i]] && !Book[V[i]])) {
printf("%d %d\n", U[i], V[i]);
}
}
putchar('\n');
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}