Marriage Match IV HDU

Marriage Match IV

思路：属于最短路径上的边应该满足：dis_A[u] + dis_B[v] + w == dis_A[B]，dis_A是出发点到其他点的距离，dis_B是终点到其他点的距离，u,v是边的两个端点，w是权值。题目说每条边只能用一次，我们可以用最大流算法来求最短路径有几条。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 1010;
const int M = 1e5;
const int INF = 1e9;
struct node{
    int x, d;
    bool friend operator<(const node& a, const node& b){
        return a.d > b.d;
    }
};
struct edge{
    int to, nxt, flow;
}e[M << 1];
vector<pair<int, int> > E[2][N];
int dis[2][N];
priority_queue<node > que; //最短路
queue<int > Q;//网络流分层
int d[N];//层级
int n, m, A, B;
int head[N], cur[N], vis[N], id[N]; //链式，当前弧优化，...，编号记录
int tot, cnt;

void dijkstra(int now, int pos)
{
    for(int i = 0; i <= n; ++i) dis[pos][i] = INF;
    while(!que.empty()) que.pop();
    dis[pos][now] = 0;
    que.push({now, 0});

    while(!que.empty()){
        int u = que.top().x;
        que.pop();

        for(auto e : E[pos][u]){
            int v = e.fi;
            int w = e.se;

            if(dis[pos][v] > dis[pos][u] + w){
                dis[pos][v] = dis[pos][u] + w;
                que.push({v, dis[pos][v]});
            }
        }
    }
    
}

inline void add(int u, int v)
{
    e[tot].to = v; e[tot].flow = 1;
    e[tot].nxt = head[u]; head[u] = tot++;
    e[tot].to = u; e[tot].flow = 0;
    e[tot].nxt = head[v]; head[v] = tot++;
}

void build()
{
    for(int i = 0; i <= n; ++i) head[i] = -1; tot = 0;
    for(int x = 1; x <= n; ++x){
        for(auto e : E[0][x]){
            int y = e.fi;
            int w = e.se;
            if(dis[0][x] + dis[1][y] + w == dis[0][B] || dis[0][y] + dis[1][x] + w == dis[0][B]){
                if(!vis[x]) id[cnt++] = x, vis[x] = 1;
                if(!vis[y]) id[cnt++] = y, vis[y] = 1;
                add(x, y);
            }
        }   
    }
}

bool bfs()
{
    while(!Q.empty()) que.pop();
    for(int i = 0; i < cnt; ++i) {
        d[id[i]] = 0;
    }
    d[A] = 1;
    Q.push(A);

    while(!Q.empty()){
        int now = Q.front();

        Q.pop();

        for(int o = head[now]; ~o; o = e[o].nxt){
            int to = e[o].to;
            int flow = e[o].flow;

            if(flow && !d[to]){
                d[to] = d[now] + 1;
                Q.push(to);
            }
        }
    }

    if(!d[B]) return false;
    else return true;
}

int dfs(int now, int pre_flow){
    if(now == B) return pre_flow;
    int sum = 0;
    for(int& o = cur[now]; ~o; o = e[o].nxt){
        int to = e[o].to;
        int flow = e[o].flow;

        if(flow && d[to] == d[now] + 1){
            int tmp = dfs(to, min(pre_flow - sum, flow));
            if(tmp == 0) continue;
            e[o].flow -= tmp;
            e[o ^ 1].flow += tmp;
            sum += tmp;
            if(sum == pre_flow) return sum;
        }
    }
    return sum;
}

void dinic()
{
    int mf = 0;
    while(bfs()){
        for(int i = 0; i < cnt; ++i) cur[id[i]] = head[id[i]];
        mf += dfs(A, INF);
    }

    printf("%d
", mf);
}

void init(){
    for(int i = 0; i <= n; ++i){
        E[0][i].clear();
        E[1][i].clear();
    }
    for(int i = 0; i < cnt; ++i) vis[id[i]] = 0;
    cnt = 0;
}

void solve()
{
    int T;
    scanf("%d", &T);

    while(T--){
        scanf("%d%d", &n, &m);
        
        init();

        int u, v, w;
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d", &u, &v, &w);
            if(u == v) continue;
            E[0][u].pb(make_pair(v, w));
            E[1][v].pb(make_pair(u, w));
        }
        scanf("%d%d", &A, &B);
        dijkstra(A, 0);
        dijkstra(B, 1);

        build();
        dinic(); 
    }
}

int main()
{

    solve();

    return 0;
}