hdu 3416 Marriage Match IV

Description

Do not sincere non-interference。

Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it’s said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.

So, under a good RP, starvae may have many chances to get to city B. But he don’t know how many chances at most he can make a data with the girl he likes . Could you help starvae?

Input

The first line is an integer T indicating the case number.(1<=T<=65)

For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0

题目大意：有n个城市m条路，从城市A到城市B的最短路径有几条。

解题思路：先正向反向求最短路，获得起点到每点的最短距离d1[]。 终点到每点的最短距离d2[]，最短路Min。然后遍历每一条边。当d1[edges.from]+edges.dis+d2[edges.to]==Min时。将该边增加最大流的图中，容量为1，建完图后，以A为源点，B为汇点跑最大流就可以。

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <cstdlib>

#include <queue>

using namespace std;

const int INF = 0x3f3f3f3f;

const int N = 2005;

const int M = 200005;

typedef long long ll;

int n, m, s, t, Min;

struct Edge{

    int from,to;

    ll dist;

};

struct HeapNode{

    int d,u;

    bool operator < (const HeapNode& rhs) const{

        return d > rhs.d;

    }

};  

struct Dinic{

    int ec, head[N], first[N], que[N], lev[N];

    int Next[M], to[M], v[M];

    void init() {

        ec = 0;

        memset(first, -1, sizeof(first));

    }

    void addEdge(int a,int b,int c) {

        to[ec] = b;

        v[ec] = c;

        Next[ec] = first[a];

        first[a] = ec++;

        to[ec] = a;

        v[ec] = 0;

        Next[ec] = first[b];

        first[b] = ec++;

    }

    int BFS() {

        int kid, now, f = 0, r = 1, i;

        memset(lev, 0, sizeof(lev));

        que[0] = s, lev[s] = 1;

        while (f < r) {

            now = que[f++];

            for (i = first[now]; i != -1; i = Next[i]) {

                kid = to[i];

                if (!lev[kid] && v[i]) {

                    lev[kid] = lev[now] + 1;

                    if (kid == t) return 1;

                    que[r++] = kid;

                }

            }

        }

        return 0;

    }

    int DFS(int now, int sum) {

        int kid, flow, rt = 0;

        if (now == t) return sum;

        for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {

            head[now] = i;

            kid = to[i];

            if (lev[kid] == lev[now] + 1 && v[i]) {

                flow = DFS(kid, min(sum - rt, v[i]));

                if (flow) {

                    v[i] -= flow;

                    v[i^1] += flow;

                    rt += flow;

                } else lev[kid] = -1;

            }

        }

        return rt;

    }

    int dinic() {

        int ans = 0;

        while (BFS()) {

            for (int i = 0; i <= n; i++) {

                head[i] = first[i];

            }

            ans += DFS(s, INF);

        }

        return ans;

    }

}din;

struct Dijkstra{

    int n,m;               //点数和边数

    vector<Edge> edges;    //边列表

    vector<int> G[M];     //每一个结点出发的边编号（从0開始编号）

    bool done[N];         //是否已永久标号

    int d[N];             //s到各个点的距离

    ll L; 

    void init(int n) {

        this->n = n;

        for(int i = 0; i <= m * 2; i++) G[i].clear();//清空邻接表

        edges.clear();//清空边表

    }  

    void addEdge(int from, int to, ll dist) {

        //假设是无向图。每条无向边需调用两次AddEdge

        edges.push_back((Edge){from, to, dist});

        m = edges.size();

        G[from].push_back(m - 1);

    }  

    void dijkstra(int s) {//求s到全部点的距离

        priority_queue<HeapNode> Q;

        for(int i = 0; i <= n; i++) d[i] = INF;

        d[s] = 0;

        memset(done, 0, sizeof(done));

        Q.push((HeapNode){0, s});

        while(!Q.empty()){

            HeapNode x = Q.top(); Q.pop();

            int u = x.u;

            if(done[u]) continue;

            done[u] = true;

            for(int i = 0; i < G[u].size(); i++){

                Edge& e = edges[G[u][i]];

                if(d[e.to] > d[u] + e.dist){

                    d[e.to] = d[u] + e.dist;

                    Q.push((HeapNode){d[e.to], e.to});

                }

            }

        }

    }  

}dij, dij2;  

void input() {

    int u, v, d;

    scanf("%d %d", &n, &m);

    dij.init(n);

    dij2.init(n);

    for (int i = 0; i < m; i++) {

        scanf("%d %d %d", &u, &v, &d);

        dij.addEdge(u, v, d);

        dij2.addEdge(v, u, d);

    }

    scanf("%d %d", &s, &t);

    dij.dijkstra(s);

    dij2.dijkstra(t);

    Min = dij.d[t];

}

void solve() {

    din.init();

    for (int i = 0; i < m; i++) {

        if (dij.d[dij.edges[i].from] + dij2.d[dij.edges[i].to] + dij.edges[i].dist == Min) {

            din.addEdge(dij.edges[i].from, dij.edges[i].to, 1);

        }

    }

    printf("%d\n", din.dinic());

}

int main() {

    int T;

    scanf("%d", &T);

    while (T--) {

        input();

        solve();

    }

    return 0;

}