HDU3549 Flow Problem 【最大流量】

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 8387    Accepted Submission(s): 3908

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

 

Author

HyperHexagon

 

Source

field=problem&key=HyperHexagon%27s+Summer+Gift+%28Original+tasks%29&source=1&searchmode=source">HyperHexagon's Summer Gift (Original tasks)

水题。

#include <stdio.h>
#include <string.h>

#define maxn 20
#define maxm 2010
#define inf 0x3f3f3f3f

int head[maxn], n, m, source, sink, id; // n个点m条边
struct Node {
    int u, v, c, next;
} E[maxm];
int que[maxn], pre[maxn], Layer[maxn];
bool vis[maxn];

void addEdge(int u, int v, int c) {
    E[id].u = u; E[id].v = v;
    E[id].c = c; E[id].next = head[u];
    head[u] = id++;

    E[id].u = v; E[id].v = u;
    E[id].c = 0; E[id].next = head[v];
    head[v] = id++;
}

void getMap() {
    int u, v, c; id = 0;
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof(int) * (n + 1));
    source = 1; sink = n;
    while(m--) {
        scanf("%d%d%d", &u, &v, &c);
        addEdge(u, v, c);
    }
}

bool countLayer() {
    memset(Layer, 0, sizeof(int) * (n + 1));
    int id = 0, front = 0, u, v, i;
    Layer[source] = 1; que[id++] = source;
    while(front != id) {
        u = que[front++];
        for(i = head[u]; i != -1; i = E[i].next) {
            v = E[i].v;
            if(E[i].c && !Layer[v]) {
                Layer[v] = Layer[u] + 1;
                if(v == sink) return true;
                else que[id++] = v;
            }
        }
    }
    return false;
}

int Dinic() {
    int i, u, v, minCut, maxFlow = 0, pos, id = 0;
    while(countLayer()) {
        memset(vis, 0, sizeof(bool) * (n + 1));
        memset(pre, -1, sizeof(int) * (n + 1));
        que[id++] = source; vis[source] = 1;
        while(id) {
            u = que[id - 1];
            if(u == sink) {
                minCut = inf;
                for(i = pre[sink]; i != -1; i = pre[E[i].u])
                    if(minCut > E[i].c) {
                        minCut = E[i].c; pos = E[i].u;
                    }
                maxFlow += minCut;
                for(i = pre[sink]; i != -1; i = pre[E[i].u]) {
                    E[i].c -= minCut;
                    E[i^1].c += minCut;
                }
                while(que[id-1] != pos)
                    vis[que[--id]] = 0;
            } else {
                for(i = head[u]; i != -1; i = E[i].next)
                    if(E[i].c && Layer[u] + 1 == Layer[v = E[i].v] && !vis[v]) {
                        vis[v] = 1; que[id++] = v; pre[v] = i; break;
                    }
                if(i == -1) --id;
            }
        }
    }
    return maxFlow;
}

void solve(int i) {
    printf("Case %d: %d\n", i, Dinic());
}

int main() {
    int t, cas;
    scanf("%d", &t);
    for(cas = 1; cas <= t; ++cas) {
        getMap();
        solve(cas);
    }
}

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