HDU Problem - 6214 Smallest Minimum Cut（最小割边，两种方法）

题目链接

Problem Description

Consider a network G=(V,E)G=(V,E)G=(V,E) with source sss and sink t" role="presentation">ttt. An s-t cut is a partition of nodes set VVV into two parts such that s" role="presentation">sss and ttt belong to different parts. The cut set is the subset of E" role="presentation">EEE with all edges connecting nodes in different parts. A minimum cut is the one whose cut set has the minimum summation of capacities. The size of a cut is the number of edges in the cut set. Please calculate the smallest size of all minimum cuts.

Input

The input contains several test cases and the first line is the total number of cases T (1≤T≤300)T(1≤T≤300)T~(1\le T\le 300).Each case describes a network GGG, and the first line contains two integers n (2≤n≤200)" role="presentation">n (2≤n≤200)n (2≤n≤200)n~(2\le n\le 200) and m (0≤m≤1000)m(0≤m≤1000)m~(0\le m\le 1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from 111 to n" role="presentation">nnn.The second line contains two different integers sss and t (1≤s,t≤n)" role="presentation">t (1≤s,t≤n)t (1≤s,t≤n)t~(1\le s,t\le n) corresponding to the source and sink.Each of the next mmm lines contains three integers u,v" role="presentation">u,vu,vu,v and w (1≤w≤255)w(1≤w≤255)w~(1\le w\le 255) describing a directed edge from node uuu to v" role="presentation">vvv with capacity ww<script type="math/tex" id="MathJax-Element-21">w</script>.

Output

For each test case, output the smallest size of all minimum cuts in a line.

Sample Input

Sample Output

2
3

AC

方法一：先跑一次最大流，将满流边的流量加1反向流量不变，不满流的边流量变为inf，反向流量不变，然后再跑一次最大流
方法二：建图的时候把流量C，改为C * (M + 1) + 1;
然后跑一个最大流，最大流 = maxflow / (M + 1) 最小割边 = maxflow % (M + 1)

// 方法一
#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define N 205
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;
struct ac{int v, c, pre;
}edge[8000001];
int head[N], dis[N], curedge[N], cnt;
int inf = 0x3f3f3f3f;
void addedge(int u, int v, int c) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v);edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].pre = head[u];head[u] = cnt++;
}
bool bfs(int s, int e) {queue<int> que;que.push(s);memset(dis, 0, sizeof(dis));dis[s] = 1;while (!que.empty()) {int t = que.front();que.pop();for (int i = head[t]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] || edge[i].c == 0)   continue;dis[edge[i].v] = dis[t] + 1;que.push(edge[i].v);} }return dis[e] != 0;
}
int dfs(int s, int e, int flow) {if (s == e) return flow;for (int &i = curedge[s]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] == dis[s] + 1 && edge[i].c) {int d = dfs(edge[i].v, e, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i ^ 1].c += d;return d; }}}return 0;
}
int n, d, m;
int solve(int s, int e) {int sum = 0;while (bfs(s, e)) {for (int i = 1; i < N ; ++i) {curedge[i] = head[i];}int d;while (d = dfs(s, e, inf)) {sum += d;}}return sum;
}
int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint t;scanf("%d", &t);while (t--) {memset(head, -1, sizeof(head));cnt = 0;int n, m, s, e;scanf("%d%d%d%d", &n, &m, &s, &e);for (int i = 0; i < m; ++i) {int x, y, c;scanf("%d%d%d", &x, &y, &c);addedge(x, y, c);}solve(s, e);for (int i = 0; i < cnt; i += 2) {if (edge[i].c == 0) edge[i].c = 1;else    edge[i].c = inf;}int ans = solve(s, e);printf("%d\n", ans);} return 0;
}

// 方法二
#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define N 205
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;
struct ac{int v, c, pre;
}edge[8000001];
int head[N], dis[N], curedge[N], cnt;
int inf = 0x3f3f3f3f;
void addedge(int u, int v, int c) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v);edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].pre = head[u];head[u] = cnt++;
}
bool bfs(int s, int e) {queue<int> que;que.push(s);memset(dis, 0, sizeof(dis));dis[s] = 1;while (!que.empty()) {int t = que.front();que.pop();for (int i = head[t]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] || edge[i].c == 0)   continue;dis[edge[i].v] = dis[t] + 1;que.push(edge[i].v);} }return dis[e] != 0;
}
int dfs(int s, int e, int flow) {if (s == e) return flow;for (int &i = curedge[s]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] == dis[s] + 1 && edge[i].c) {int d = dfs(edge[i].v, e, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i ^ 1].c += d;return d; }}}return 0;
}
int n, d, m;
int solve(int s, int e) {int sum = 0;while (bfs(s, e)) {for (int i = 1; i < N ; ++i) {curedge[i] = head[i];}int d;while (d = dfs(s, e, inf)) {sum += d;}}return sum;
}
int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint t;scanf("%d", &t);while (t--) {memset(head, -1, sizeof(head));cnt = 0;int n, m, s, e;scanf("%d%d%d%d", &n, &m, &s, &e);for (int i = 0; i < m; ++i) {int x, y, c;scanf("%d%d%d", &x, &y, &c);addedge(x, y, c * (m + 1) + 1);}int ans = solve(s, e) % (m + 1);printf("%d\n", ans);} return 0;
}