HDU 3549 Flow Problem

Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
Input
The first line of input contains an integer T, denoting the number of test cases. For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000) Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output
Case 1: 1 Case 2: 2
제목:
사실은 우리 에 게 가장 큰 흐름 의 템 플 릿 문 제 를 구 하 라 는 것 이다.
첫 번 째 방법(Ford-Fulkerson)
생각:
가장 큰 흐름 의 템 플 릿 을 쓰 면 됩 니 다.

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1010;
const int inf = 0x7fffffff;
struct Edge {
    int to;
    int cap;
    int rev;
    Edge () {}
    Edge (int to, int cap, int rev) :  to(to), cap(cap), rev(rev) {}
};
vector g[maxn];
bool use[maxn];
int dfs(int v, int t, int f) {
    if (v == t) return f;
    use[v] = true;
    for (int i = 0; i < g[v].size(); i++) {
        Edge &e = g[v][i];
        if (!use[e.to] && e.cap > 0) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                g[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int main() {
    int n, m, t, from, to, cap;
    scanf("%d", &t);
    for (int Case = 1; Case <= t; Case++) {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++) g[i].clear();
        while (m--) {
            scanf("%d %d %d", &from, &to, &cap);
            g[from].push_back(Edge(to, cap, g[to].size()));
            g[to].push_back(Edge(from, 0, g[from].size() - 1));
        }
        int flow = 0;
        while (1) {
            memset(use, false, sizeof(use));
            int f = dfs(1, n, inf);
            if (f == 0) break;
            flow += f;
        }
        printf("Case %d: %d
", Case, flow);
    }
    return 0;
}

두 번 째 방법(dinic)

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1010;
const int inf = 0x7fffffff;
struct Edge {
    int to;
    int cap;
    int rev;
    Edge () {}
    Edge (int to, int cap, int rev) : to(to), cap(cap), rev(rev) {}
};
vector g[maxn];
int level[maxn];
bool use[maxn];
void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue q;
    level[s] = 0;
    q.push(s);
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        for (int i = 0; i < g[v].size(); i++) {
            Edge &e = g[v][i];
            if (level[e.to] < 0 && e.cap > 0) {
                level[e.to] = level[v] + 1;
                q.push(e.to);
            }
        }
    }
}
int dfs(int v, int t, int f) {
    if (v == t) return f;
    use[v] = true;
    for (int i = 0; i < g[v].size(); i++) {
        Edge &e = g[v][i];
        if (level[v] < level[e.to] && e.cap > 0 && !use[e.to]) {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0) {
                e.cap -= d;
                g[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int main() {
    int t, n, m, cap, to, from;
    scanf("%d", &t);
    for (int Case = 1; Case <= t; Case++) {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++) g[i].clear();
        while (m--) {
            scanf("%d %d %d", &from, &to, &cap);
            g[from].push_back(Edge(to, cap, g[to].size()));
            g[to].push_back(Edge(from, 0, g[from].size() - 1));
        }
        int flow = 0;
        while (1) {
            bfs(1);
            if (level[n] < 0) break;
            memset(use, false, sizeof(use));
            int f = 0;
            if ((f = dfs(1, n, inf)) > 0) {
                flow += f;
            }
        }
        printf("Case %d: %d
", Case, flow);
    }
    return 0;
}