【最小费用最大流】N. April Fools' Problem (medium)

http://codeforces.com/contest/802/problem/N

【题解】

方法一：

 #include<bits/stdc++.h>
 using namespace std;

 typedef long long LL;

 #define N 200020

 int to[N], head[N], nxt[N], cost[N], cap[N], cnt;

 void init(){ memset(head, -, sizeof head); }

 void add_Edge(int S, int T, int c, int w){
     nxt[cnt] = head[S], cap[cnt] = c, cost[cnt] = w, to[cnt] = T, head[S] = cnt ++;
     nxt[cnt] = head[T], cap[cnt] = , cost[cnt] = -w, to[cnt] = S, head[T] = cnt ++;
 }

 const LL INF = 1e14;

 LL dist[N];
 int prv[N];
 bool vis[N];

 LL SPFA(int S, int T, int vet){
     queue <int> Q;
     fill(prv, prv + vet, -);
     fill(dist, dist + vet, INF);
     Q.push(S); dist[S] = , vis[S] = true;
     while(!Q.empty()){
         int x = Q.front(); Q.pop(), vis[x] = false;
         for(int id = head[x]; ~id; id = nxt[id]) if( cap[id] ){
             int y = to[id];
             if(dist[y] > dist[x] + cost[id]){
                 dist[y] = dist[x] + cost[id];
                 prv[y] = id;
                 if(!vis[y]) Q.push(y), vis[y] = true;
             }
         }
     }
     if(!~prv[T]) { return INF; }

     for(int cur = T; cur != S; cur = to[cur^]){
         cur = prv[cur];
         cap[cur] --;
         cap[cur ^ ] ++;
     }
     return dist[T];
 }

 int a[N], b[N], n, k;

 int main(){
     //freopen("in.txt", "r", stdin);
     scanf("%d %d", &n, &k);
     for(int i = ; i <= n; i ++) scanf("%d", a + i);
     for(int i = ; i <= n; i ++) scanf("%d", b + i);

     int S = , T =  * n + ;
     init();
     for(int i = ; i <= n; i ++){
         add_Edge(S, i, , a[i]);
         add_Edge(i + n, T, , b[i]);
         add_Edge(i, i + n, n, );
         if(i < n) add_Edge(i + n, i + n + , n - i, );
     }
     LL ans = ;
     for(int i = ; i <= k; i ++) ans += SPFA(S, T, T + );
     cout << ans << endl;
 }

方法二：

 #include <bits/stdc++.h>
 using namespace std;

 #define oo 0x3f3f3f3fLL
 #define pb push_back
 #define mp make_pair
 #define fi first
 #define se second
 #define SUM(x,y) accumulate(x,y,0)
 #define gcd(x,y) (__gcd(x,y))
 #define LOWBIT(x) (x&(-x))
 #define CLR(x,y) memset(x,y,sizeof(x))
 #define PD(x) printf("%d\n",(x))
 #define PF(f) printf("%lf\n",(f))
 #define PS(s) puts(s)

 typedef pair<int, int> pii;
 typedef long long LL;

 template <class T> inline bool chkmin(T& x, T y) { return x > y ? x = y, true : false; }
 template <class T> inline bool chkmax(T& x, T y) { return x < y ? x = y, true : false; }

 template <class T> T reads() {
     T x = ;
     static int f;
     static char c;
     for (f = ; !isdigit(c = getchar()); ) if (c == '-') f = -;
     for (x = ; isdigit(c); c = getchar()) x = x *  + c - ;
     return x * f;
 }
 #define read   reads<int>
 #define readll reads<LL>

 int n, k, v, a[], b[];
 LL l, r, res, m, ans;

 int main(void) {
     scanf("%d%d", &n, &k);
     for (int i = ; i < n; ++i) a[i] = read();
     for (int i = ; i < n; ++i) b[i] = read();
     for(int i=;i<n;i++)
     {
         cout<<a[i]<<" ";
     }
     cout<<endl;
     for(int i=;i<n;i++)
     {
         cout<<b[i]<<" ";
     }
     cout<<endl;
     l = ; r = 1e10; res = 1e18;
     while (l < r) {
         ans = ; m = (l + r) >> ; v = ;
         priority_queue<LL> qa, qb;

         for (int i = ; i < n; ++i) {
             qa.push(-a[i]);

             LL ga = -qa.top() + b[i] - m;
             LL gb = qb.empty() ? LL(1e18) : b[i] - qb.top();
             if (ga <= gb && ga <= ) {
                 ans += ga; v++;
                 qa.pop(); qb.push(b[i]);
             } else if (ga > gb && gb < ) {
                 ans += gb;
                 qb.pop(); qb.push(b[i]);
             }
         }

         if (v >= k) { r = m - ; res = ans + k * m; }
         else l = m + ;
     }
     return cout << res << endl, ;
 }