HDU-3416-MarriageMatch4(最大流,最短路)

链接:

https://vjudge.net/problem/HDU-3416

题意:

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?

思路:

先找出最短路,然后将最短路的每条边加入网络流的图,边权为1,跑最大流即可.
(找最短路SPFAwa了好几次,改成Dij就过了..玄学)(SPFA写错了,真的丢人)

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 1e3+10;
const int INF = 1e9;

struct Edge
{
    int from, to, cap;
};
struct Road
{
    int from, to, len;
};
struct Dj
{
    int to, dis;
    bool operator < (const Dj &that) const
    {
        return this->dis > that.dis;
    }
};
vector<int> G[MAXN];
vector<int> GR[MAXN];
vector<int> GR2[MAXN];
vector<Edge> edges;
vector<Road> roads;
vector<Road> roads2;
int Dis[MAXN], Vis[MAXN], Dis2[MAXN];
int n, m, s, t;

void AddEdge(int from ,int to, int cap)
{
    edges.push_back(Edge{from, to, cap});
    edges.push_back(Edge{to, from, 0});
    G[from].push_back(edges.size()-2);
    G[to].push_back(edges.size()-1);
}

void Dij()
{
    memset(Vis, 0, sizeof(Vis));
    memset(Dis, MINF, sizeof(Dis));
    Dis[s] = 0;
    priority_queue<Dj> que;
    que.push(Dj{s, 0});
    while (!que.empty())
    {
        Dj u = que.top();
        que.pop();
        if (Vis[u.to])
            continue;
        Vis[u.to] = 1;
        for (int i = 0;i < GR[u.to].size();i++)
        {
            Road &r = roads[GR[u.to][i]];
            if (Dis[r.to] > Dis[u.to]+r.len)
            {
                Dis[r.to] = Dis[u.to]+r.len;
                que.push(Dj{r.to, Dis[r.to]});
            }
        }
    }
}

//void SPFA2()
//{
//    memset(Vis, 0, sizeof(Vis));
//    memset(Dis2, MINF, sizeof(Dis2));
//    Dis2[t] = 0;
//    Vis[t] = 1;
//    queue<int> que;
//    que.push(t);
//    while (!que.empty())
//    {
//        int u = que.front();
//        que.pop();
//        for (int i = 0;i < GR2[u].size();i++)
//        {
//            Road &r = roads2[GR2[u][i]];
//            if (Dis2[r.to] > Dis2[u]+r.len)
//            {
//                Dis2[r.to] = Dis2[u]+r.len;
//                if (Vis[r.to] == 0)
//                {
//                    que.push(r.to);
//                    Vis[r.to] = 1;
//                }
//            }
//        }
//    }
//}

bool Bfs()
{
    memset(Dis, -1, sizeof(Dis));
    queue<int> que;
    Dis[s] = 0;
    que.push(s);
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        for (int i = 0;i < G[u].size();i++)
        {
            Edge &e = edges[G[u][i]];
            if (e.cap > 0 && Dis[e.to] == -1)
            {
                Dis[e.to] = Dis[u]+1;
                que.push(e.to);
            }
        }
    }
    return Dis[t] != -1;
}

int Dfs(int u, int flow)
{
    if (u == t)
        return flow;
    int res = 0;
    for (int i = 0;i < G[u].size();i++)
    {
        Edge &e = edges[G[u][i]];
        if (e.cap > 0 && Dis[e.to] == Dis[u]+1)
        {
            int tmp = Dfs(e.to, min(e.cap, flow));
            flow -= tmp;
            e.cap -= tmp;
            res += tmp;
            edges[G[u][i]^1].cap += tmp;
            if (flow == 0)
                break;
        }
    }
    if (res == 0)
        Dis[u] = -1;
    return res;
}

int MaxFlow()
{
    int res = 0;
    while (Bfs())
        res += Dfs(s, INF);
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        cin >> n >> m;
        for (int i = 1;i <= n;i++)
            G[i].clear(), GR[i].clear();
        edges.clear(), roads.clear();
        int u, v, w;
        for (int i = 1;i <= m;i++)
        {
            cin >> u >> v >> w;
            roads.push_back(Road{u, v, w});
            GR[u].push_back(roads.size()-1);
//            roads2.push_back(Road{v, u, w});
//            GR2[v].push_back(roads2.size()-1);
        }
        cin >> s >> t;
        Dij();
//        SPFA2();
        if (Dis[t] == MINF)
        {
            cout << 0 << endl;
            continue;
        }
//        cout << 1 << endl;
        for (int i = 1;i <= n;i++)
        {
            for (int j = 0;j < GR[i].size();j++)
            {
                Road &r = roads[GR[i][j]];
//                cout << Dis[r.from] << ' ' << Dis[r.to] << ' ' << r.len << endl;
                if (Dis[r.from]+r.len == Dis[r.to])
                {
//                    cout << r.from << ' ' << r.to << endl;
                    AddEdge(r.from, r.to, 1);
                }
            }
        }
        int res = MaxFlow();
        cout << res << endl;
    }

    return 0;
}