Marriage Match IV HDU

题意

给你n个点，m条边，要求每条边只能走一次的S到T的最短路径的个数

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题解

在我又WA又TLE还RE时，yyb大佬告诉我说要跑两遍SPFA，还说我写的一遍SPFA是错的，然而

啪啪打脸。。。
而且他的

比我跑得慢，2333
接下来讲一下方法
首先一遍SPFA（或dijkstra）从S跑一遍到所有点的最短路，重新建图时对于每对u， v 若 dis[u] + w[u][v] == dis[v] 则加入这条边，容量为1（还要加反边），最后跑最大流即可，最大流我用的是Dinic，然后注意手打队列，系统的会TLE

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常熟巨大的丑陋代码

# include <bits/stdc++.h># define RG register# define IL inline# define ll long long# define mem(a, b) memset(a, b, sizeof(a))# define Min(a, b) (((a) > (b)) ? (b) : (a))# define Max(a, b) (((a) < (b)) ? (b) : (a))using namespace std;IL int Get(){    RG char c = '!'; RG int x = 0, z = 1;    for(; c > '9' || c < '0'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c <= '9' && c >= '0'; c = getchar()) x = x * 10 + c - '0';    return x * z;}const int MAXN = 1001, MAXM = 200001, INF = 2147483647;int n, m, ft[MAXN], cnt, ans, dis[MAXN], vis[MAXN], _ft[MAXN], level[MAXN], Q[MAXM];struct Edge{    int to, f, nt;} edge[MAXM], _edge[MAXM];IL void Add(RG int u, RG int v, RG int f){    edge[cnt] = (Edge){v, f, ft[u]}; ft[u] = cnt++;}IL void Add2(RG int u, RG int v, RG int f){    _edge[cnt] = (Edge){v, f, _ft[u]}; _ft[u] = cnt++;}IL void SPFA(RG int S, RG int T){    RG int head = 0, tail = 0;    Q[0] = S; vis[S] = 1; dis[S] = 0;    while(head <= tail){        RG int u = Q[head++]; vis[u] = 0;        for(RG int e = ft[u]; e != -1; e = edge[e].nt){            RG int v = edge[e].to, f = edge[e].f + dis[u];            if(dis[v] > f){                dis[v] = f;                if(!vis[v]) vis[v] = 1, Q[++tail] = v;            }        }    }}IL bool Bfs(RG int S, RG int T){    mem(level, 0);    RG int head = 0, tail = 0;    Q[0] = S; level[S] = 1;    while(head <= tail){        RG int u = Q[head++];        if(u == T) return 1;        for(RG int e = _ft[u]; e != -1; e = _edge[e].nt){            RG int v = _edge[e].to, f = _edge[e].f;            if(f && !level[v]){                level[v] = level[u] + 1;                Q[++tail] = v;            }        }    }    return 0;}IL int Dfs(RG int u, RG int T, RG int maxf){    if(u == T) return maxf;    RG int res = 0;    for(RG int e = _ft[u]; e != -1; e = _edge[e].nt){        RG int v = _edge[e].to, f = _edge[e].f;        if(level[u] + 1 == level[v] && f){            f = Dfs(v, T, Min(f, maxf - res));            _edge[e].f -= f; _edge[e ^ 1].f += f;            res += f;            if(res == maxf) break;        }    }    return res;}int main(){    RG int T = Get();    while(T--){        n = Get(); m = Get();        mem(ft, -1); mem(dis, 63); mem(_ft, -1); ans = cnt = 0;        for(RG int i = 1; i <= m; i++){            RG int u = Get(), v = Get(), f = Get();            if(u == v) continue;            Add(u, v, f);        }        RG int S = Get(), T = Get(); cnt = 0;        SPFA(S, T);        for(RG int i = 1; i <= n; i++)            for(RG int e = ft[i]; e != -1; e = edge[e].nt)                if(dis[i] + edge[e].f == dis[edge[e].to])                    Add2(i, edge[e].to, 1), Add2(edge[e].to, i, 0);        while(Bfs(S, T)) ans += Dfs(S, T, INF);        printf("%d\n", ans);    }    return 0;}