HDU 3416 Marriage Match IV

Posted 2021-01-13 mingsd

Marriage Match IV

题意：现在有n城市，m条单向路，现在男生在A城市，女生在B城市，现在男生要从A -> B 去看女生，每条路只能走一次，并且男生很懒，他第一次走的是A -> B的最短路，以后每次走的路都不能超过这个，求

男生从A->B次数最多有多少次。

题解：就是求有多少个不相干的最短路径数，我们正向跑一遍spfa 跑出最短路，再反向跑一边跑出最短路， 对于一条边来说，如果 dis1[u] + dis2[v] + w = dis1[t] 那么这条边就是构成最短路的一条边，我们把这个边加入网络流中，最后跑出最大流就是结果了。

代码：

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
  4 #define LL long long
  5 #define ULL unsigned LL
  6 #define fi first
  7 #define se second
  8 #define pb push_back
  9 #define lson l,m,rt<<1
 10 #define rson m+1,r,rt<<1|1
 11 #define lch(x) tr[x].son[0]
 12 #define rch(x) tr[x].son[1]
 13 #define max3(a,b,c) max(a,max(b,c))
 14 #define min3(a,b,c) min(a,min(b,c))
 15 typedef pair<int,int> pll;
 16 const int inf = 0x3f3f3f3f;
 17 const LL INF = 0x3f3f3f3f3f3f3f3f;
 18 const LL mod =  (int)1e9 + 7;
 19 const int N = 2000;
 20 const int M = 1000000;
 21 int head[N], deep[N], cur[N];
 22 int w[M], to[M], nx[M];
 23 int tot;
 24 void add(int u, int v, int val){
 25     w[tot]  = val; to[tot] = v;
 26     nx[tot] = head[u]; head[u] = tot++;
 27 
 28     w[tot] = 0; to[tot] = u;
 29     nx[tot] = head[v]; head[v] = tot++;
 30 }
 31 int bfs(int s, int t){
 32     queue<int> q;
 33     memset(deep, 0, sizeof(deep));
 34     q.push(s);
 35     deep[s] = 1;
 36     while(!q.empty()){
 37         int u = q.front();
 38         q.pop();
 39         for(int i = head[u]; ~i; i = nx[i]){
 40             if(w[i] > 0 && deep[to[i]] == 0){
 41                 deep[to[i]] = deep[u] + 1;
 42                 q.push(to[i]);
 43             }
 44         }
 45     }
 46     return deep[t] > 0;
 47 }
 48 int Dfs(int u, int t, int flow){
 49     if(u == t) return flow;
 50     for(int &i = cur[u]; ~i; i = nx[i]){
 51         if(deep[u]+1 == deep[to[i]] && w[i] > 0){
 52             int di = Dfs(to[i], t, min(w[i], flow));
 53             if(di > 0){
 54                 w[i] -= di, w[i^1] += di;
 55                 return di;
 56             }
 57         }
 58     }
 59     return 0;
 60 }
 61 
 62 int Dinic(int s, int t){
 63     int ans = 0, tmp;
 64     while(bfs(s, t)){
 65         for(int i = 0; i <= t; i++) cur[i] = head[i];
 66         while(tmp = Dfs(s, t, inf)) ans += tmp;
 67     }
 68     return ans;
 69 }
 70 
 71 void init(){
 72     memset(head, -1, sizeof(head));
 73     tot = 0;
 74 }
 75 struct Node{
 76     int to;
 77     int w;
 78     int op;
 79 };
 80 vector<Node> e[N];
 81 int dis1[N], dis2[N], vis[N];
 82 void spfa(int s, int t, int dd[], int op){
 83     dd[s] = 0;
 84     queue<int> q;
 85     q.push(s);
 86     vis[s] = 1;
 87     while(!q.empty()){
 88         int u = q.front();
 89         vis[u] = 0;
 90         q.pop();
 91         for(int i = 0; i < e[u].size(); i++){
 92             if(e[u][i].op != op) continue;
 93             if(dd[u] + e[u][i].w < dd[e[u][i].to]){
 94                 dd[e[u][i].to] = dd[u] + e[u][i].w;
 95                 if(!vis[e[u][i].to]) {
 96                     vis[e[u][i].to] = 1;
 97                     q.push(e[u][i].to);
 98                 }
 99             }
100         }
101     }
102 
103 }
104 int main(){
105     int T, n, m, u, v, dis;
106     scanf("%d", &T);
107     while(T--){
108         scanf("%d%d", &n, &m);
109         for(int i = 1; i <= n; i++) e[i].clear();
110         for(int i = 1; i <= m; i++){
111             scanf("%d%d%d", &u, &v, &dis);
112             e[u].pb({v, dis, 1});
113             e[v].pb({u, dis, 2});
114         }
115         int a, b;
116         scanf("%d%d", &a, &b);
117         memset(dis1, inf, sizeof(dis1));
118         spfa(a, b, dis1, 1);
119         memset(dis2, inf, sizeof(dis2));
120         spfa(b, a, dis2, 2);
121         init();
122         int s = 0, t = n + 1;
123         for(int i = 1; i <= n; i++)
124             for(int j = 0; j < e[i].size(); j++){
125                 int v = e[i][j].to, w = e[i][j].w;
126                 if(dis1[i] + w + dis2[v] == dis1[b] && e[i][j].op == 1){
127                         add(i,v,1);
128                 }
129         }
130         add(s, a, inf);
131         add(b, t, inf);
132         printf("%d
",Dinic(s,t));
133     }
134     return 0;
135 }

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