hdu 3416 Marriage Match IV （最短路+最大流）

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，将该边加入最大流的图中，容量为1，建完图后，以A为源点，B为汇点跑最大流即可。

#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];     //每个结点出发的边编号（从0开始编号）      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) {           //如果是无向图，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到所有点的距离          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;}