題目鏈接

Problem Description

Consider a network G=(V,E)G=(V,E) with source ss and sink tt. An s-t cut is a partition of nodes set VV into two parts such that ss and tt belong to different parts. The cut set is the subset of EE 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).Each case describes a network GG, and the first line contains two integers n (2≤n≤200)n (2≤n≤200) and m?(0≤m≤1000)m(0≤m≤1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from 11 to nn.The second line contains two different integers ss and t (1≤s,t≤n)t (1≤s,t≤n) corresponding to the source and sink.Each of the next mm lines contains three integers u,vu,v and w?(1≤w≤255)w(1≤w≤255) describing a directed edge from node uu to vv 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

2 4 5 1 4 1 2 3 1 3 1 2 3 1 2 4 1 3 4 2 4 5 1 4 1 2 3 1 3 1 2 3 1 2 4 1 3 4 3

Sample Output

2 3

AC

• 方法一：先跑一次最大流，將滿流邊的流量加1反向流量不變，不滿流的邊流量變為inf，反向流量不變，然后再跑一次最大流
• 方法二：建圖的時候把流量C， 改為C * (M + 1) + 1;
  然后跑一個最大流，最大流 = maxflow / (M + 1) 最小割邊 = 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; }

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