BZOJ 1391 [CEOI] Order - 网络流 最大流

Solution

非常简单的建边！！！

但是刚开始的代码不够体现社会主义的优越性， 于是我 ....

惨痛教训啊。。。 终于到了今天才能够体现社会主义优越性。。。

Code

 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<queue>
 #define rd read()
 #define R register
 #define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
 #define per(i,a,b) for(register int i = (a); i >= (b); --i)
 #define ll long long
 using namespace std;

 const int N = 1e6 + 1e5, M = 4e3;
 const int inf = ~0U >> ;
 int n, m, head[M], tot, cur[M];
 int dep[M], S, T;
 int sum, ans;

 queue<int> q;

 struct edge {
     int nxt, to, val;
 }e[N * ];

 inline int read() {
     int X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 inline void added(int u, int v, int val) {
     e[++tot].to = v;
     e[tot].val = val;
     e[tot].nxt = head[u];
     head[u] = tot;
 }

 inline void add(int u, int v, int val) {
     added(u, v, val); added(v, u, );
 }

 int ch(int x) {
     return ((x + ) ^ ) - ;
 }

 int bfs() {
     rep(i, S, T) dep[i] = ;
     dep[S] = ;
     q.push(S);
     for(R int u; !q.empty();) {
         u = q.front(); q.pop();
         for(R int i = head[u]; i; i = e[i].nxt) {
             R int nt = e[i].to;
             if(!dep[nt] && e[i].val) {
                 dep[nt] = dep[u] + ;
                 q.push(nt);
             }
         }
     }
     return dep[T];
 }

 inline int minn(int A, int B) {
     return A > B ? B : A;
 }

 int dfs(R int u, R int flow) {
     if(u == T || !flow) return flow;
     if(!flow) return ;
     R int re = ;
     for(R int i = cur[u]; i && flow; i = e[i].nxt, cur[u] = i) {
         R int nt = e[i].to;
         if(dep[nt] != dep[u] +  || !e[i].val) continue;
         R int tmp = dfs(nt, minn(flow, e[i].val));
         if(!tmp) continue;
         re += tmp;
         flow -= tmp;
         e[i].val -= tmp;
         e[i ^ ].val += tmp;
         if(!flow) return re;
     }
     return re;
 }

 int main()
 {
     n = rd; m = rd;
     T = n + m + ;
     tot = ;
     rep(i, , n) {
         R int cost = rd;
         add(S, i, cost);
         sum += cost;
         int cnt = rd;
         rep(j, , cnt) {
             int x = rd, v = rd;
             add(i, n + x, v);
         }
     }
     rep(i, , m) {
         R int x = rd;
         add(i + n, T, x);
     }
     for(; bfs();) {
         memcpy(cur, head, sizeof(head));
         ans += dfs(S, inf);
     }
     printf("%d\n", sum - ans);
 }