hdu 3549 Flow Problem-优快云博客

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=3549

Flow 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

裸的最大流dinic，测模板。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using std::min;
using std::find;
using std::sort;
using std::pair;
using std::queue;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1100;
const int INF = 0x3f3f3f3f;
struct Dinic {
    struct edge { int to, cap, next, rev; }G[N << 2];
    int s, t, tot, level[N], ite[N], head[N];
    inline void init() {
        tot = 0, cls(head, -1);
    }
    inline void add_edge(int u, int v, int cap) {
        G[tot] = (edge){ v, cap, head[u], tot + 1 }; head[u] = tot++;
        G[tot] = (edge){ u,   0, head[v], tot - 1 }; head[v] = tot++;
    }
    inline void built(int n, int m) {
        int u, v, f;
        s = 1, t = n;
        while(m--) {
            scanf("%d %d %d", &u, &v, &f);
            add_edge(u, v, f);
        }
    }
    inline void bfs(int s) {
        cls(level, -1);
        queue<int> q;
        q.push(s);
        level[s] = 0;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            for(int i = head[u]; ~i; i = G[i].next) {
                edge &e = G[i];
                if(e.cap > 0 && level[e.to] < 0) {
                    level[e.to] = level[u] + 1;
                    q.push(e.to);
                }
            }
        }
    }
    inline int dfs(int u, int t, int f) {
        if(u == t) return f;
        for(int &i = ite[u]; ~i; i = G[i].next) {
            edge &e = G[i];
            if(e.cap > 0 && level[u] < level[e.to]) {
                int d = dfs(e.to, t, min(e.cap, f));
                if(d > 0) {
                    e.cap -= d;
                    G[e.rev].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }
    inline int max_flow() {
        int flow = 0;
        while(true) {
            bfs(s);
            if(level[t] < 0) break;
            int f;
            rep(i, t) ite[i] = head[i];
            while((f = dfs(s, t, INF)) > 0) {
                flow += f;
            }
        }
        return flow;
    }
    inline void solve(int n, int m) {
        static int k = 1;
        init(), built(n, m);
        printf("Case %d: %d\n", k++, max_flow());
    }
}go;
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int t, n, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        go.solve(n, m);
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4792931.html