POJ 3204 Ikki's Story I-Road Reconstruction (网络流关键边)

【题意】给定一个N个节点M条边的网络流，求有多少条边，使得当增其中加任何一个边的容量后，整个网络的流将增加. 挺好的一道题，考察对网络流和增广路的理解。 【思路】 首先关键边一定是满流边。那么对于一个满流边<x,y>来说，如果残余网络中从起点到x和从y到终点都有路径可达的话，那么这条边的容量增加时，在残量网络上将会产生一条增广路，最大流的值一定会发生改变。 则算法如下： 求最大流，得到残余网络 枚举每条满流边，DFS判断是否分别从源点和到汇点可达，如果可达则加1。  

#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 505;
const int MAXE = 20005;
const int oo = 0x3fffffff;
struct node{
    int u, v, flow;
    int opp;
    int next;
};
struct Dinic{
    node arc[MAXE];
    int vn, en, head[MAXV];     //vn点个数(包括源点汇点),en边个数
    int cur[MAXV];              //当前弧
    int q[MAXV];                //bfs建层次图时的队列
    int path[MAXE], top;        //存dfs当前最短路径的栈
    int dep[MAXV];              //各节点层次
    void init(int n){
        vn = n;
        en = 0;
        mem(head, -1);
    }
    void insert_flow(int u, int v, int flow){
        arc[en].u = u;
        arc[en].v = v;
        arc[en].flow = flow;
        arc[en].opp = en + 1;
        arc[en].next = head[u];
        head[u] = en ++;

        arc[en].u = v;
        arc[en].v = u;
        arc[en].flow = 0;       //反向弧
        arc[en].opp = en - 1;
        arc[en].next = head[v];
        head[v] = en ++;
    }
    bool bfs(int s, int t){
        mem(dep, -1);
        int lq = 0, rq = 1;
        dep[s] = 0;
        q[lq] = s;
        while(lq < rq){
            int u = q[lq ++];
            if (u == t){
                return true;
            }
            for (int i = head[u]; i != -1; i = arc[i].next){
                int v = arc[i].v;
                if (dep[v] == -1 && arc[i].flow > 0){
                    dep[v] = dep[u] + 1;
                    q[rq ++] = v;
                }
            }
        }
        return false;
    }
    int solve(int s, int t){
        int maxflow = 0;
        while(bfs(s, t)){
            int i, j;
            for (i = 1; i <= vn; i ++)  cur[i] = head[i];
            for (i = s, top = 0;;){
                if (i == t){
                    int mink;
                    int minflow = 0x3fffffff;
                    for (int k = 0; k < top; k ++)
                        if (minflow > arc[path[k]].flow){
                            minflow = arc[path[k]].flow;
                            mink = k;
                        }
                    for (int k = 0; k < top; k ++)
                        arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
                    maxflow += minflow;
                    top = mink;		//arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
                    i = arc[path[top]].u;
                }
                for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
                    int v = arc[j].v;
                    if (arc[j].flow && dep[v] == dep[i] + 1)
                        break;
                }
                if (j != -1){
                    path[top ++] = j;
                    i = arc[j].v;
                }
                else{
                    if (top == 0)   break;
                    dep[i] = -1;
                    i = arc[path[-- top]].u;
                }
            }
        }
        return maxflow;
    }
}dinic;
bool vis[MAXV];
bool reach(int u, int p){
    vis[u] = 1;
    if (u == p)
        return true;
    for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){
        int v = dinic.arc[i].v;
        if (vis[v] || dinic.arc[i].flow <= 0) continue;
        if (reach(v, p))
            return true;
    }
    return false;
}
int work(int n){
    int res = 0;
    for (int i = 0; i < dinic.en; i += 2){
        if (dinic.arc[i].flow == 0){
            mem(vis, 0);
            int u = dinic.arc[i].u;
            int v = dinic.arc[i].v;
            if (reach(1, u) && reach(v, n)){
                res ++;
            }
        }
    }
    return res;
}
int main(){
	//freopen("test.in", "r", stdin);
	//freopen("test.out", "w", stdout);
    int n, m;
    scanf("%d %d", &n, &m);
    dinic.init(n);
    for (int i = 0; i < m; i ++){
        int u,v,w;
        scanf("%d %d %d", &u, &v, &w);
        dinic.insert_flow(u+1, v+1, w);
    }
    dinic.solve(1, n);
    printf("%d
", work(n));
	return 0;
}