[2019杭电多校第一场][hdu6582]Path(最短路&&最小割)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6582

题意：删掉边使得1到n的最短路改变，删掉边的代价为该边的边权。求最小代价。

比赛时一片浆糊，赛后听到dinic瞬间思维通透XD

大致做法就是先跑一遍最短路，然后再保留所有满足dis[i]+w==dis[j]的边，在这些边上跑最小割(dinic)。

代码写的异常丑陋，见谅QAQ

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cstring>
 #include<string>
 #include<queue>
 using namespace std;
 typedef long long ll;
 const ll maxn = 3e5 + ;
 const ll inf = 1e18;
 struct node {
     ll s, e, next;
     ll w;
 }edge[maxn * ], edge2[maxn * ];
 ll head[maxn], head2[maxn], len, len2;
 void init() {
     memset(head, -, sizeof(head));
     memset(head2, -, sizeof(head2));
     len = len2 = ;
 }
 void add(ll s, ll e, ll w) {
     edge[len].s = s;
     edge[len].e = e;
     edge[len].w = w;
     edge[len].next = head[s];
     head[s] = len++;
 }
 void add2(ll s, ll e, ll w) {
     edge2[len2].s = s;
     edge2[len2].e = e;
     edge2[len2].w = w;
     edge2[len2].next = head2[s];
     head2[s] = len2++;
 }
 struct p {
     ll dis, num;
     bool operator<(const p&a)const {
         return a.dis < dis;
     }
 };
 ll dis[maxn];
 ll vis[maxn];
 void dij(ll n) {
     priority_queue<p>q;
     for (ll i = ; i <= n; i++)
         dis[i] = inf, vis[i] = ;
     q.push({ 0ll,1ll });
     dis[] = 0ll;
     while (!q.empty()) {
         ll x = q.top().num;
         q.pop();
         if (vis[x])
             continue;
         vis[x] = ;
         for (ll i = head[x]; i != -; i = edge[i].next) {
             ll y = edge[i].e;
             if (dis[y] > dis[x] + edge[i].w) {
                 dis[y] = dis[x] + edge[i].w;
                 q.push({ dis[y],y });
             }
         }
     }
 }
 ll d[maxn];
 bool bfs(ll s, ll t) {
     queue<ll>q;
     memset(d, , sizeof(d));
     d[s] = 1ll;
     q.push(s);
     while (!q.empty()) {
         ll x = q.front(); q.pop();
         for (ll i = head2[x]; i != -; i = edge2[i].next) {
             ll y = edge2[i].e;
             if (edge2[i].w && !d[y]) {
                 d[y] = d[x] + 1ll;
                 q.push(y);
             }
         }
     }
     return d[t];
 }
 ll dfs(ll x, ll t, ll limit) {
     if (x == t)
         return limit;
     ll add, ans = ;
     for (ll i = head2[x]; i != -; i = edge2[i].next) {
         ll y = edge2[i].e;
         if (d[y] == d[x] +  && edge2[i].w) {
             add = dfs(y, t, min(limit, edge2[i].w));
             edge2[i].w -= add;
             edge2[i ^ ].w += add;
             ans += add;
             limit -= add;
             if (!limit)
                 break;
         }
     }
     if (!ans)
         d[x] = -;
     return ans;
 }
 ll dinic(ll s, ll t) {
     ll ans = ;
     while (bfs(s, t))
         ans += dfs(s, t, inf);
     return ans;
 }
 int main() {
     ll t;
     scanf("%lld", &t);
     while (t--) {
         ll n, m;
         init();
         scanf("%lld%lld", &n, &m);
         ll x, y, z;
         for (ll i = ; i <= m; i++) {
             scanf("%lld%lld%lld", &x, &y, &z);
             add(x, y, z);
         }
         dij(n);
         for (ll i = ; i <= n; i++)
             for (ll j = head[i]; j != -; j = edge[j].next)
                 if (dis[edge[j].s] + edge[j].w == dis[edge[j].e])
                     add2(edge[j].s, edge[j].e, edge[j].w), add2(edge[j].e, edge[j].s, 0ll);
         printf("%lld\n", dinic(, n));
     }
     return ;
 }